Kompleksinis skaičius yra dviejų realiųjų skaičių pora z:
z
=
(
a
,
b
)
=
a
+
b
⋅
i
=
R
e
(
z
)
+
i
I
m
(
z
)
{\displaystyle z=(a,b)=a+b\cdot i=Re(z)+iIm(z)}
,
kur a ir b – realieji skaičiai ,
o
i
=
(
0
,
1
)
{\displaystyle i=(0,1)}
– menamasis vienetas tenkinantis sąlygą:
i
2
=
−
1
{\displaystyle i^{2}=-1}
Nors priimta, kad
i
=
−
1
{\displaystyle i={\sqrt {-1}}}
, tačiau ši išraiška turi būti taikoma su tam tikromis išlygomis.
Skaičius a vadinamas realiąja z dalimi, žymima a = Re(z), skaičius b vadinamas menamąja z dalimi, žymima b = Im(z).
Kompleksinių skaičių aibė žymima C :
C
=
{
a
+
b
⋅
i
;
a
,
b
∈
R
}
{\displaystyle \mathbb {C} =\{a+b\cdot i;a,b\in \mathbb {R} \}}
Aritmetinės operacijos su kompleksiniais skaičiais
keisti
Sudėtis
(
a
,
b
)
+
(
c
,
d
)
=
(
a
+
b
i
)
+
(
c
+
d
i
)
=
(
a
+
c
)
+
(
b
+
d
)
⋅
i
=
(
a
+
c
,
b
+
d
)
{\displaystyle (a,b)+(c,d)=(a+bi)+(c+di)=(a+c)+(b+d)\cdot i=(a+c,b+d)\,}
Atimtis
(
a
,
b
)
−
(
c
,
d
)
=
(
a
+
b
i
)
−
(
c
+
d
i
)
=
(
a
−
c
)
+
(
b
−
d
)
⋅
i
=
(
a
−
c
,
b
−
d
)
{\displaystyle (a,b)-(c,d)=(a+bi)-(c+di)=(a-c)+(b-d)\cdot i=(a-c,b-d)}
,
Daugyba
(
a
,
b
)
⋅
(
c
,
d
)
=
(
a
+
b
i
)
(
c
+
d
i
)
=
(
a
c
−
b
d
)
+
(
a
d
+
b
c
)
⋅
i
=
(
a
c
−
b
d
,
a
d
+
b
c
)
{\displaystyle (a,b)\cdot (c,d)=(a+bi)(c+di)=(ac-bd)+(ad+bc)\cdot i=(ac-bd,ad+bc)\,}
a
⋅
(
1
,
0
)
=
(
a
,
0
)
⋅
(
1
,
0
)
=
(
a
,
0
)
=
a
{\displaystyle a\cdot (1,0)=(a,0)\cdot (1,0)=(a,0)=a}
b
⋅
(
0
,
1
)
=
(
b
,
0
)
⋅
(
0
,
1
)
=
(
b
+
0
)
⋅
(
0
+
i
)
=
0
+
b
i
=
(
0
,
b
)
=
b
i
{\displaystyle b\cdot (0,1)=(b,0)\cdot (0,1)=(b+0)\cdot (0+i)=0+bi=(0,b)=bi}
Dalyba
(
a
,
b
)
(
c
,
d
)
=
a
+
b
i
c
+
d
i
=
a
c
+
b
d
c
2
+
d
2
+
(
b
c
−
a
d
)
c
2
+
d
2
⋅
i
=
(
a
c
+
b
d
c
2
+
d
2
,
b
c
−
a
d
c
2
+
d
2
)
,
{\displaystyle {\frac {(a,b)}{(c,d)}}={\frac {a+bi}{c+di}}={\frac {ac+bd}{c^{2}+d^{2}}}+{\frac {(bc-ad)}{c^{2}+d^{2}}}\cdot i=({\frac {ac+bd}{c^{2}+d^{2}}},{\frac {bc-ad}{c^{2}+d^{2}}}),}
(
a
,
b
)
(
a
,
b
)
=
a
+
b
i
a
+
b
i
=
1
+
0
i
=
(
1
,
0
)
=
1
{\displaystyle {\frac {(a,b)}{(a,b)}}={\frac {a+bi}{a+bi}}=1+0i=(1,0)=1}
.
1
(
c
,
d
)
=
(
1
,
0
)
(
c
,
d
)
=
1
+
0
i
c
+
d
i
=
c
c
2
+
d
2
+
(
−
d
c
2
+
d
2
)
i
=
(
c
c
2
+
d
2
,
−
d
c
2
+
d
2
)
{\displaystyle {\frac {1}{(c,d)}}={\frac {(1,0)}{(c,d)}}={\frac {1+0i}{c+di}}={\frac {c}{c^{2}+d^{2}}}+({\frac {-d}{c^{2}+d^{2}}})i=({\frac {c}{c^{2}+d^{2}}},{\frac {-d}{c^{2}+d^{2}}})}
.
Sudėtis
(
a
,
b
)
+
(
c
,
d
)
=
(
a
+
b
i
)
+
(
c
+
d
i
)
=
a
+
b
i
+
c
+
d
i
=
a
+
c
+
(
b
+
d
)
⋅
i
=
(
a
+
c
,
b
+
d
)
{\displaystyle (a,b)+(c,d)=(a+bi)+(c+di)=a+bi+c+di=a+c+(b+d)\cdot i=(a+c,b+d)\,}
Atimtis
(
a
,
b
)
−
(
c
,
d
)
=
(
a
+
b
i
)
−
(
c
+
d
i
)
=
a
+
b
i
−
c
−
d
i
=
a
−
c
+
(
b
−
d
)
⋅
i
=
(
a
−
c
,
b
−
d
)
{\displaystyle (a,b)-(c,d)=(a+bi)-(c+di)=a+bi-c-di=a-c+(b-d)\cdot i=(a-c,b-d)}
,
Daugyba
(
a
,
b
)
⋅
(
c
,
d
)
=
(
a
+
b
i
)
(
c
+
d
i
)
=
a
c
+
a
d
i
+
c
b
i
+
b
i
d
i
=
a
c
−
b
d
+
(
a
d
+
b
c
)
i
=
(
a
c
−
b
d
,
a
d
+
b
c
)
{\displaystyle (a,b)\cdot (c,d)=(a+bi)(c+di)=ac+adi+cbi+bidi=ac-bd+(ad+bc)i=(ac-bd,ad+bc)\,}
a
⋅
(
1
,
0
)
=
(
a
,
0
)
⋅
(
1
,
0
)
=
(
a
+
0
)
⋅
(
1
+
0
)
=
(
a
,
0
)
=
a
{\displaystyle a\cdot (1,0)=(a,0)\cdot (1,0)=(a+0)\cdot (1+0)=(a,0)=a}
b
⋅
(
0
,
1
)
=
(
b
,
0
)
⋅
(
0
,
1
)
=
(
b
+
0
)
⋅
(
0
+
i
)
=
(
b
⋅
0
+
b
⋅
i
+
0
⋅
0
+
0
⋅
i
)
=
0
+
b
i
=
(
0
,
b
)
=
b
i
{\displaystyle b\cdot (0,1)=(b,0)\cdot (0,1)=(b+0)\cdot (0+i)=(b\cdot 0+b\cdot i+0\cdot 0+0\cdot i)=0+bi=(0,b)=bi}
Dalyba
(
a
,
b
)
(
c
,
d
)
=
a
+
b
i
c
+
d
i
=
a
c
+
b
d
c
2
+
d
2
+
(
b
c
−
a
d
)
i
c
2
+
d
2
=
a
c
+
b
d
+
b
c
i
−
a
d
i
c
2
+
d
2
=
(
a
c
+
b
d
c
2
+
d
2
,
b
c
−
a
d
c
2
+
d
2
)
,
{\displaystyle {\frac {(a,b)}{(c,d)}}={\frac {a+bi}{c+di}}={\frac {ac+bd}{c^{2}+d^{2}}}+{\frac {(bc-ad)i}{c^{2}+d^{2}}}={\frac {ac+bd+bci-adi}{c^{2}+d^{2}}}=({\frac {ac+bd}{c^{2}+d^{2}}},{\frac {bc-ad}{c^{2}+d^{2}}}),}
(
a
,
b
)
(
a
,
b
)
=
a
+
b
i
a
+
b
i
=
1
+
0
i
=
(
1
,
0
)
=
1
{\displaystyle {\frac {(a,b)}{(a,b)}}={\frac {a+bi}{a+bi}}=1+0i=(1,0)=1}
.
1
(
c
,
d
)
=
(
1
,
0
)
(
c
,
d
)
=
1
+
0
i
c
+
d
i
=
1
⋅
c
+
0
⋅
d
c
2
+
d
2
+
(
0
⋅
c
−
1
⋅
d
)
i
c
2
+
d
2
=
c
c
2
+
d
2
+
(
−
d
c
2
+
d
2
)
i
=
(
c
c
2
+
d
2
,
−
d
c
2
+
d
2
)
{\displaystyle {\frac {1}{(c,d)}}={\frac {(1,0)}{(c,d)}}={\frac {1+0i}{c+di}}={\frac {1\cdot c+0\cdot d}{c^{2}+d^{2}}}+{\frac {(0\cdot c-1\cdot d)i}{c^{2}+d^{2}}}={\frac {c}{c^{2}+d^{2}}}+({\frac {-d}{c^{2}+d^{2}}})i=({\frac {c}{c^{2}+d^{2}}},{\frac {-d}{c^{2}+d^{2}}})}
.
Dalybos:
(
c
+
d
i
)
a
c
+
b
d
+
b
c
i
−
a
d
i
c
2
+
d
2
=
a
c
2
+
b
c
d
+
b
c
2
i
−
a
c
d
i
+
a
c
d
i
+
b
d
2
i
−
b
c
d
+
a
d
2
c
2
+
d
2
=
{\displaystyle (c+di){\frac {ac+bd+bci-adi}{c^{2}+d^{2}}}={\frac {ac^{2}+bcd+bc^{2}i-acdi+acdi+bd^{2}i-bcd+ad^{2}}{c^{2}+d^{2}}}=}
=
a
c
2
+
b
c
2
i
+
b
d
2
i
+
a
d
2
c
2
+
d
2
=
c
2
(
a
+
b
i
)
+
d
2
(
b
i
+
a
)
c
2
+
d
2
=
(
c
2
+
d
2
)
(
a
+
b
i
)
c
2
+
d
2
=
a
+
b
i
{\displaystyle ={\frac {ac^{2}+bc^{2}i+bd^{2}i+ad^{2}}{c^{2}+d^{2}}}={\frac {c^{2}(a+bi)+d^{2}(bi+a)}{c^{2}+d^{2}}}={\frac {(c^{2}+d^{2})(a+bi)}{c^{2}+d^{2}}}=a+bi}
.
a
+
b
i
c
+
d
i
=
(
a
+
b
i
)
(
c
−
d
i
)
(
c
+
d
i
)
(
c
−
d
i
)
=
(
a
c
+
b
d
)
+
(
b
c
−
a
d
)
i
c
2
+
d
2
=
a
c
+
b
d
c
2
+
d
2
+
b
c
−
a
d
c
2
+
d
2
i
{\displaystyle {\frac {a+bi}{c+di}}={\frac {(a+bi)(c-di)}{(c+di)(c-di)}}={\frac {(ac+bd)+(bc-ad)i}{c^{2}+d^{2}}}={\frac {ac+bd}{c^{2}+d^{2}}}+{\frac {bc-ad}{c^{2}+d^{2}}}i}
.
Pavyzdys.
z
1
=
(
2
+
5
i
)
,
{\displaystyle z_{1}=(2+5i),}
z
2
=
(
4
+
3
i
)
{\displaystyle z_{2}=(4+3i)}
.
z
=
z
1
z
2
=
(
2
+
5
i
)
(
4
+
3
i
)
=
8
+
6
i
+
20
i
−
15
=
−
7
+
26
i
{\displaystyle z=z_{1}z_{2}=(2+5i)(4+3i)=8+6i+20i-15=-7+26i}
.
Čia gauto vektoriaus z koordinatės (x; y)=(-7, 26).
Gauto vektoriaus z ilgis :
r
=
(
−
7
)
2
+
26
2
=
725
≈
26.92582404
{\displaystyle r={\sqrt {(-7)^{2}+26^{2}}}={\sqrt {725}}\approx 26.92582404}
.
r
1
=
2
2
+
5
2
=
29
≈
5.385
{\displaystyle r_{1}={\sqrt {2^{2}+5^{2}}}={\sqrt {29}}\approx 5.385}
,
r
2
=
4
2
+
3
2
=
25
=
5
{\displaystyle r_{2}={\sqrt {4^{2}+3^{2}}}={\sqrt {25}}=5}
.
r
=
r
1
r
2
=
5
29
≈
26.92582404
{\displaystyle r=r_{1}r_{2}=5{\sqrt {29}}\approx 26.92582404}
.
Kaip matome naujo vektoriaus z ilgį galima surasti ir be menamojo vieneto i .
ϕ
1
=
arccos
a
r
1
=
arcsin
b
r
1
=
arccos
2
29
≈
1.19028995
{\displaystyle \phi _{1}=\arccos {\frac {a}{r_{1}}}=\arcsin {\frac {b}{r_{1}}}=\arccos {\frac {2}{\sqrt {29}}}\approx 1.19028995}
ϕ
2
=
arccos
c
r
2
=
arcsin
d
r
2
=
arccos
4
5
≈
0.643501108
{\displaystyle \phi _{2}=\arccos {\frac {c}{r_{2}}}=\arcsin {\frac {d}{r_{2}}}=\arccos {\frac {4}{5}}\approx 0.643501108}
z
=
z
1
z
2
=
r
1
(
cos
ϕ
1
+
i
sin
ϕ
1
)
r
2
(
cos
ϕ
2
+
i
sin
ϕ
2
)
=
r
(
cos
(
ϕ
1
+
ϕ
2
)
+
i
sin
(
ϕ
1
+
ϕ
2
)
)
=
26.92582404
(
cos
(
1.19028995
+
0.643501108
)
+
i
sin
(
1.833791058
)
)
=
{\displaystyle z=z_{1}z_{2}=r_{1}(\cos \phi _{1}+i\sin \phi _{1})r_{2}(\cos \phi _{2}+i\sin \phi _{2})=r(\cos(\phi _{1}+\phi _{2})+i\sin(\phi _{1}+\phi _{2}))=26.92582404(\cos(1.19028995+0.643501108)+i\sin(1.833791058))=}
=
26.92582404
(
−
0.259973473
+
0.965615758
i
)
=
−
6.999999989
+
25.99999999
i
{\displaystyle =26.92582404(-0.259973473+0.965615758i)=-6.999999989+25.99999999i}
.
Kaip matome praktiškai visiškai tiksliai suradome naujo vektoriaus z koordinates (-7, 26), be panaudojimo menamojo vieneto savybių (i2 =-1) ir menamasis vienetas atliko tik žymeklio vaidmenį.
Čia mes sudėjome du kampus per kuriuos buvo pasukti nuo x ašies abu vektoriai ((a, b) ir (c, d)). Pavyzdžiui vektorius (a, b) su x ašimi sudaro
ϕ
1
{\displaystyle \phi _{1}}
=~68.2 laipsnių kampą, o vektorius (c, d) su x ašimi sudaro
ϕ
2
{\displaystyle \phi _{2}}
=~36.87 laipsnių kampą. Kai mes sudauginome
z
1
z
2
{\displaystyle z_{1}z_{2}}
, tai kampai susidėjo ir atsirado naujas vektorius kurio ilgis
r
=
r
1
r
2
=
26.92582404
{\displaystyle r=r_{1}r_{2}=26.92582404}
ir kuris su x ašimi sudaro
ϕ
=
ϕ
1
+
ϕ
2
{\displaystyle \phi =\phi _{1}+\phi _{2}}
=36.87+68.2=~105.07 laipsnių kampą. Kaip matome kampus galima sudėti ir be kompleksinių skaičių, o sudaugintų vektorių ilgius (
r
1
{\displaystyle r_{1}}
ir
r
2
)
{\displaystyle r_{2})}
bei naujo atsiradusio vektoriaus
r
=
r
1
r
2
{\displaystyle r=r_{1}r_{2}}
ilgį taip pat rasti be kompleksinių skaičių (o tiksliau be menamojo vieneto i ).
Kompleksiniai skaičiai trigonometrijoje
keisti
Kompleksinių skaičių daugyba .
Aukščiau nustatėme, kad dviejų kompleksinių skaičių
z
1
=
(
x
1
,
y
1
)
{\displaystyle z_{1}=(x_{1},\;y_{1})}
ir
z
2
=
(
x
2
,
y
2
)
{\displaystyle z_{2}=(x_{2},\;y_{2})}
sandauga yra lygi
z
=
z
1
⋅
z
2
=
(
x
1
x
2
−
y
1
y
2
,
x
1
y
2
+
x
2
y
1
)
.
(
7.2
)
{\displaystyle z=z_{1}\cdot z_{2}=(x_{1}x_{2}-y_{1}y_{2},\;x_{1}y_{2}+x_{2}y_{1}).\quad (7.2)}
Sakykime, duoti du bet kokie kompleksiniai skaičiai
z
1
=
(
x
1
,
y
1
)
=
(
ρ
1
cos
θ
1
,
ρ
1
sin
θ
1
)
{\displaystyle z_{1}=(x_{1},\;y_{1})=(\rho _{1}\cos \theta _{1},\;\rho _{1}\sin \theta _{1})}
ir
z
2
=
(
x
2
,
y
2
)
=
(
ρ
2
cos
θ
2
,
ρ
2
sin
θ
2
)
.
{\displaystyle z_{2}=(x_{2},\;y_{2})=(\rho _{2}\cos \theta _{2},\;\rho _{2}\sin \theta _{2}).}
Pagal daugybos apibrėžimą ((7.2) formulė)
z
1
⋅
z
2
=
(
x
1
x
2
−
y
1
y
2
,
x
1
y
2
+
x
2
y
1
)
=
(
ρ
1
ρ
2
cos
θ
1
cos
θ
2
−
ρ
1
ρ
2
sin
θ
1
sin
θ
2
,
ρ
1
ρ
2
cos
θ
1
sin
θ
2
+
ρ
1
ρ
2
sin
θ
1
cos
θ
2
)
=
{\displaystyle z_{1}\cdot z_{2}=(x_{1}x_{2}-y_{1}y_{2},\;x_{1}y_{2}+x_{2}y_{1})=(\rho _{1}\rho _{2}\cos \theta _{1}\cos \theta _{2}-\rho _{1}\rho _{2}\sin \theta _{1}\sin \theta _{2},\;\rho _{1}\rho _{2}\cos \theta _{1}\sin \theta _{2}+\rho _{1}\rho _{2}\sin \theta _{1}\cos \theta _{2})=}
=
[
(
ρ
1
ρ
2
)
cos
(
θ
1
+
θ
2
)
,
(
ρ
1
ρ
2
)
sin
(
θ
1
+
θ
2
)
]
.
(
7.7
)
{\displaystyle =[(\rho _{1}\rho _{2})\cos(\theta _{1}+\theta _{2}),\;(\rho _{1}\rho _{2})\sin(\theta _{1}+\theta _{2})].\quad (7.7)}
Pasinaudojome trigonometrinėmis formulėmis:
cos
(
A
+
B
)
=
cos
A
cos
B
−
sin
A
sin
B
,
{\displaystyle \cos(A+B)=\cos A\cos B-\sin A\sin B,}
sin
(
A
+
B
)
=
sin
A
cos
B
+
cos
A
sin
B
.
{\displaystyle \sin(A+B)=\sin A\cos B+\cos A\sin B.}
Vadinasi, sudauginus du kompleksinius skaičius, jų argumentai (pasisukimo nuo ašies Ox kampai) sudedami.
Kai kampai
θ
1
=
θ
2
=
θ
,
ρ
1
=
ρ
2
=
ρ
,
{\displaystyle \theta _{1}=\theta _{2}=\theta ,\;\rho _{1}=\rho _{2}=\rho ,}
gauname:
(
ρ
cos
θ
,
ρ
sin
θ
)
2
=
(
ρ
2
cos
(
2
θ
)
,
ρ
2
sin
(
2
θ
)
)
.
{\displaystyle (\rho \cos \theta ,\;\rho \sin \theta )^{2}=(\rho ^{2}\cos(2\theta ),\;\rho ^{2}\sin(2\theta )).}
Kai kampai
θ
1
=
θ
2
=
θ
{\displaystyle \theta _{1}=\theta _{2}=\theta }
ir
ρ
=
1
{\displaystyle \rho =1}
gauname Muavro formulę, kai n=2:
(
cos
θ
,
sin
θ
)
2
=
(
cos
(
2
θ
)
,
sin
(
2
θ
)
)
.
{\displaystyle (\cos \theta ,\;\sin \theta )^{2}=(\cos(2\theta ),\;\sin(2\theta )).}
Indukcijos metodu galima įrodyti Muavro formulę bet kokiam n :
(
cos
θ
,
sin
θ
)
n
=
(
cos
n
θ
,
sin
n
θ
)
.
{\displaystyle (\cos \theta ,\;\sin \theta )^{n}=(\cos n\theta ,\;\sin n\theta ).}
Pastarąją forumulę galima užrašyti ir šitaip:
(
cos
θ
+
i
sin
θ
)
n
=
cos
n
θ
+
i
sin
n
θ
.
{\displaystyle (\cos \theta +i\sin \theta )^{n}=\cos n\theta +i\sin n\theta .}
Muavro formulės įrodymas indukcijos metodu .
Remdamiesi kompleksinių skaičių sandaugos (7.7) formule, turime
(
cos
θ
+
i
sin
θ
)
(
cos
θ
+
i
sin
θ
)
=
cos
(
θ
+
θ
)
+
i
sin
(
θ
+
θ
)
{\displaystyle (\cos \theta +i\sin \theta )(\cos \theta +i\sin \theta )=\cos(\theta +\theta )+i\sin(\theta +\theta )}
arba
(
cos
θ
+
i
sin
θ
)
2
=
cos
2
θ
+
i
sin
2
θ
,
{\displaystyle (\cos \theta +i\sin \theta )^{2}=\cos 2\theta +i\sin 2\theta ,}
(
cos
θ
+
i
sin
θ
)
2
(
cos
θ
+
i
sin
θ
)
=
(
cos
2
θ
+
i
sin
2
θ
)
(
cos
θ
+
i
sin
θ
)
=
cos
(
2
θ
+
θ
)
+
i
sin
(
2
θ
+
θ
)
{\displaystyle (\cos \theta +i\sin \theta )^{2}(\cos \theta +i\sin \theta )=(\cos 2\theta +i\sin 2\theta )(\cos \theta +i\sin \theta )=\cos(2\theta +\theta )+i\sin(2\theta +\theta )}
arba
(
cos
θ
+
i
sin
θ
)
3
=
cos
3
θ
+
i
sin
3
θ
,
{\displaystyle (\cos \theta +i\sin \theta )^{3}=\cos 3\theta +i\sin 3\theta ,}
(
cos
θ
+
i
sin
θ
)
3
(
cos
θ
+
i
sin
θ
)
=
(
cos
3
θ
+
i
sin
3
θ
)
(
cos
θ
+
i
sin
θ
)
=
cos
(
3
θ
+
θ
)
+
i
sin
(
3
θ
+
θ
)
{\displaystyle (\cos \theta +i\sin \theta )^{3}(\cos \theta +i\sin \theta )=(\cos 3\theta +i\sin 3\theta )(\cos \theta +i\sin \theta )=\cos(3\theta +\theta )+i\sin(3\theta +\theta )}
arba
(
cos
θ
+
i
sin
θ
)
4
=
cos
4
θ
+
i
sin
4
θ
,
{\displaystyle (\cos \theta +i\sin \theta )^{4}=\cos 4\theta +i\sin 4\theta ,}
..........
(
cos
θ
+
i
sin
θ
)
n
−
1
(
cos
θ
+
i
sin
θ
)
=
(
cos
(
(
n
−
1
)
θ
)
+
i
sin
(
(
n
−
1
)
θ
)
)
(
cos
θ
+
i
sin
θ
)
=
cos
(
(
n
−
1
)
θ
+
θ
)
+
i
sin
(
(
n
−
1
)
θ
+
θ
)
{\displaystyle (\cos \theta +i\sin \theta )^{n-1}(\cos \theta +i\sin \theta )=(\cos((n-1)\theta )+i\sin((n-1)\theta ))(\cos \theta +i\sin \theta )=\cos((n-1)\theta +\theta )+i\sin((n-1)\theta +\theta )}
arba
(
cos
θ
+
i
sin
θ
)
n
=
cos
n
θ
+
i
sin
n
θ
.
{\displaystyle (\cos \theta +i\sin \theta )^{n}=\cos n\theta +i\sin n\theta .}
Kompleksinių skaičių dalyba .
Dviejų kompleksinių skaičių
z
1
=
(
x
1
,
y
1
)
{\displaystyle z_{1}=(x_{1},\;y_{1})}
ir
z
2
=
(
x
2
,
y
2
)
{\displaystyle z_{2}=(x_{2},\;y_{2})}
dalmuo yra lygus
z
=
z
1
z
2
=
(
x
1
x
2
+
y
1
y
2
x
2
2
+
y
2
2
,
x
2
y
1
−
x
1
y
2
x
2
2
+
y
2
2
)
.
(
7.4
)
{\displaystyle z={\frac {z_{1}}{z_{2}}}=({\frac {x_{1}x_{2}+y_{1}y_{2}}{x_{2}^{2}+y_{2}^{2}}},\;{\frac {x_{2}y_{1}-x_{1}y_{2}}{x_{2}^{2}+y_{2}^{2}}}).\quad (7.4)}
Analogiškai remdamiesi (7.4) formule, įsitikiname, kad kompleksinių skaičių
z
1
=
(
x
1
,
y
1
)
=
(
ρ
1
cos
θ
1
,
ρ
1
sin
θ
1
)
{\displaystyle z_{1}=(x_{1},\;y_{1})=(\rho _{1}\cos \theta _{1},\;\rho _{1}\sin \theta _{1})}
ir
z
2
=
(
x
2
,
y
2
)
=
(
ρ
2
cos
θ
2
,
ρ
2
sin
θ
2
)
{\displaystyle z_{2}=(x_{2},\;y_{2})=(\rho _{2}\cos \theta _{2},\;\rho _{2}\sin \theta _{2})}
dalmuo
z
1
z
2
{\displaystyle {z_{1} \over z_{2}}}
išreiškiamas šitaip (
z
2
{\displaystyle z_{2}}
nelygus nuliui, t. y.
ρ
2
≠
0
{\displaystyle \rho _{2}\neq 0}
):
z
1
z
2
=
(
x
1
x
2
+
y
1
y
2
x
2
2
+
y
2
2
,
x
2
y
1
−
x
1
y
2
x
2
2
+
y
2
2
)
=
{\displaystyle {\frac {z_{1}}{z_{2}}}={\Big (}{\frac {x_{1}x_{2}+y_{1}y_{2}}{x_{2}^{2}+y_{2}^{2}}},\;{\frac {x_{2}y_{1}-x_{1}y_{2}}{x_{2}^{2}+y_{2}^{2}}}{\Big )}=}
=
(
ρ
1
cos
(
θ
1
)
ρ
2
cos
(
θ
2
)
+
ρ
1
sin
(
θ
1
)
ρ
2
sin
(
θ
2
)
ρ
2
2
cos
2
θ
2
+
ρ
2
2
sin
2
θ
2
,
ρ
2
cos
(
θ
2
)
ρ
1
sin
(
θ
1
)
−
ρ
1
cos
(
θ
1
)
ρ
2
sin
(
θ
2
)
ρ
2
2
cos
2
θ
2
+
ρ
2
2
sin
2
θ
2
)
=
{\displaystyle ={\Big (}{\frac {\rho _{1}\cos(\theta _{1})\rho _{2}\cos(\theta _{2})+\rho _{1}\sin(\theta _{1})\rho _{2}\sin(\theta _{2})}{\rho _{2}^{2}\cos ^{2}\theta _{2}+\rho _{2}^{2}\sin ^{2}\theta _{2}}},\;{\frac {\rho _{2}\cos(\theta _{2})\rho _{1}\sin(\theta _{1})-\rho _{1}\cos(\theta _{1})\rho _{2}\sin(\theta _{2})}{\rho _{2}^{2}\cos ^{2}\theta _{2}+\rho _{2}^{2}\sin ^{2}\theta _{2}}}{\Big )}=}
=
(
ρ
1
ρ
2
cos
(
θ
1
)
cos
(
θ
2
)
+
ρ
1
ρ
2
sin
(
θ
1
)
sin
(
θ
2
)
ρ
2
2
,
ρ
1
ρ
2
cos
(
θ
2
)
sin
(
θ
1
)
−
ρ
1
ρ
2
cos
(
θ
1
)
sin
(
θ
2
)
ρ
2
2
)
=
{\displaystyle ={\Big (}{\frac {\rho _{1}\rho _{2}\cos(\theta _{1})\cos(\theta _{2})+\rho _{1}\rho _{2}\sin(\theta _{1})\sin(\theta _{2})}{\rho _{2}^{2}}},\;{\frac {\rho _{1}\rho _{2}\cos(\theta _{2})\sin(\theta _{1})-\rho _{1}\rho _{2}\cos(\theta _{1})\sin(\theta _{2})}{\rho _{2}^{2}}}{\Big )}=}
=
[
ρ
1
ρ
2
(
cos
θ
1
cos
θ
2
+
sin
θ
1
sin
θ
2
)
,
ρ
1
ρ
2
(
cos
θ
2
sin
θ
1
−
cos
θ
1
sin
θ
2
)
]
=
{\displaystyle ={\Big [}{\frac {\rho _{1}}{\rho _{2}}}(\cos \theta _{1}\cos \theta _{2}+\sin \theta _{1}\sin \theta _{2}),\;{\frac {\rho _{1}}{\rho _{2}}}(\cos \theta _{2}\sin \theta _{1}-\cos \theta _{1}\sin \theta _{2}){\Big ]}=}
=
[
ρ
1
ρ
2
cos
(
θ
1
−
θ
2
)
,
ρ
1
ρ
2
sin
(
θ
1
−
θ
2
)
]
.
(
7.8
)
{\displaystyle ={\Big [}{\frac {\rho _{1}}{\rho _{2}}}\cos(\theta _{1}-\theta _{2}),\;{\frac {\rho _{1}}{\rho _{2}}}\sin(\theta _{1}-\theta _{2}){\Big ]}.\quad (7.8)}
Čia pasinaudojome trigonometrinėmis formulėmis:
cos
(
A
−
B
)
=
cos
A
cos
B
+
sin
A
sin
B
,
{\displaystyle \cos(A-B)=\cos A\cos B+\sin A\sin B,}
sin
(
A
−
B
)
=
sin
A
cos
B
−
cos
A
sin
B
.
{\displaystyle \sin(A-B)=\sin A\cos B-\cos A\sin B.}
Kėlimui laipsniu yra naudojama Muavro formulė:
z
n
=
(
r
e
i
φ
)
n
=
r
n
e
i
n
φ
=
r
n
(
cos
n
φ
+
i
sin
n
φ
)
.
{\displaystyle z^{n}={\big (}r\,e^{i\varphi }{\big )}^{n}=r^{n}\,e^{in\varphi }=r^{n}(\cos {n\varphi }+i\sin {n\varphi }).}
Didelę reikšmę turi vienetinio ilgio kompleksiniai skaičiai, kai r=1.
Kai
φ
=
π
4
{\displaystyle \varphi ={\frac {\pi }{4}}}
, tai
e
i
φ
=
cos
φ
+
i
sin
φ
=
1
2
+
1
2
i
{\displaystyle e^{i\varphi }=\cos \varphi +i\sin \varphi ={\frac {1}{\sqrt {2}}}+{\frac {1}{\sqrt {2}}}i}
.
(
cos
φ
+
i
sin
φ
)
2
=
cos
2
φ
−
sin
2
φ
+
2
i
sin
φ
cos
φ
=
cos
2
φ
+
i
sin
2
φ
=
0
+
i
{\displaystyle (\cos \varphi +i\sin \varphi )^{2}=\cos ^{2}\varphi -\sin ^{2}\varphi +2i\sin \varphi \cos \varphi =\cos 2\varphi +i\sin 2\varphi =0+i}
(
cos
φ
+
i
sin
φ
)
3
=
cos
3
φ
+
i
sin
3
φ
=
−
1
2
+
1
2
i
{\displaystyle (\cos \varphi +i\sin \varphi )^{3}=\cos 3\varphi +i\sin 3\varphi =-{\frac {1}{\sqrt {2}}}+{\frac {1}{\sqrt {2}}}i}
ir t.t. Vienetinio ilgio vektorius kaskart nuo x ašies pasisuka po 45 laipsnius prieš laikrodžio rodyklę (kai kampas
φ
=
45
{\displaystyle \varphi =45}
laipsniai).
Pavyzdžiui turime vieno taško
x
1
{\displaystyle x_{1}}
koordinatę 0.6 ir turime
x
2
{\displaystyle x_{2}}
koordinatę 0.8. Tada
y
1
=
1
−
0.6
2
=
0.64
=
0.8
{\displaystyle y_{1}={\sqrt {1-0.6^{2}}}={\sqrt {0.64}}=0.8}
, o
y
2
=
1
−
0.8
=
0.36
=
0.6
{\displaystyle y_{2}={\sqrt {1-0.8}}={\sqrt {0.36}}=0.6}
. Taigi turime
(
x
1
;
y
1
)
=
0.6
+
0.8
i
{\displaystyle (x_{1};y_{1})=0.6+0.8i}
ir
(
x
2
;
y
2
)
=
0.8
+
0.6
i
{\displaystyle (x_{2};y_{2})=0.8+0.6i}
. Sudauginus turime: (0.6+0.8i)(0.8+0.6i)=0.48+0.36i+0.64i-0.48=i. Taigi gavome, kad x ašis yra 0, o y ašis yra 1. Taigi gavome tą patį tarsi sudėję du kampus, kur pirmas kampas yra
ϕ
1
=
arccos
0.6
=
arcsin
0.8
=
0.927295218
=
53.13010235
{\displaystyle \phi _{1}=\arccos 0.6=\arcsin 0.8=0.927295218=53.13010235}
laipsnio, o antras kampas yra
ϕ
2
=
arccos
x
2
=
arccos
0.8
=
0.927295218
=
36.86989765
{\displaystyle \phi _{2}=\arccos x_{2}=\arccos 0.8=0.927295218=36.86989765}
laipnsio.
Pavyzdys. Duotas kampas
ϕ
1
=
60
{\displaystyle \phi _{1}=60}
laipsnių, kas yra lygu
ϕ
1
=
π
3
=
1.047197551
{\displaystyle \phi _{1}={\frac {\pi }{3}}=1.047197551}
. Ir duotas kampas
ϕ
2
=
15
{\displaystyle \phi _{2}=15}
laipsnių arba
ϕ
2
=
π
12
=
0.261799387
{\displaystyle \phi _{2}={\frac {\pi }{12}}=0.261799387}
. Žinome, kad pirmo taško ant vienetinio apskritimo (kurio spindulys r=1, o centras O =(0; 0)) koordinatės yra
(
x
1
;
y
1
)
=
(
0.5
;
0.866025403
)
{\displaystyle (x_{1};\;y_{1})=(0.5;\;0.866025403)}
, o antro taško koordinatės yra
(
x
2
;
y
2
)
=
(
0
,
965925826
;
0
,
258819045
)
{\displaystyle (x_{2};\;y_{2})=(0,965925826;\;0,258819045)}
. Žinome, kad
cos
ϕ
1
=
cos
π
3
=
0.5
{\displaystyle \cos \phi _{1}=\cos {\frac {\pi }{3}}=0.5}
ir
sin
ϕ
1
=
sin
π
3
=
0.866025403
{\displaystyle \sin \phi _{1}=\sin {\frac {\pi }{3}}=0.866025403}
. Taip pat
cos
ϕ
2
=
cos
π
12
=
0.965925826
{\displaystyle \cos \phi _{2}=\cos {\frac {\pi }{12}}=0.965925826}
ir
sin
ϕ
2
=
sin
π
12
=
0.258819045
{\displaystyle \sin \phi _{2}=\sin {\frac {\pi }{12}}=0.258819045}
.
Įrodysime, kad kosinuso ir sinuso formulės gautos Teiloro eilutės pagalba yra teisingos (nes kalkuliatorius kosinuso ir sinuso reikšmes skaičiuoja naudodamsis Teiloro eilute, bet galima patikrinti ir betarpiškai įstačius
ϕ
{\displaystyle \phi }
reikšmę į teiloro eilutę kosinuso, tik ilgas tikrinimas gausis).
Sudėsime kampus 60 laispnių ir 15 laipnsių ir tokiu budu gausime kampą 60+15=75 laipsnių. Arba kampus galime sudėti taip:
ϕ
=
ϕ
1
+
ϕ
2
=
π
3
+
π
12
=
5
π
12
=
1.308996939
{\displaystyle \phi =\phi _{1}+\phi _{2}={\frac {\pi }{3}}+{\frac {\pi }{12}}={\frac {5\pi }{12}}=1.308996939}
.
Dabar sudauginsime kompleksinius skaičius (sudauginsime du taškus):
(
x
;
y
)
=
(
x
1
+
y
1
i
)
(
x
2
+
y
2
i
)
=
(
0.5
+
0.866025403
i
)
(
0.965925826
+
0.258819045
i
)
=
{\displaystyle (x;y)=(x_{1}+y_{1}i)(x_{2}+y_{2}i)=(0.5+0.866025403i)(0.965925826+0.258819045i)=}
=
0.5
⋅
0.965925826
+
0.5
⋅
0.258819045
i
+
0.866025403
i
⋅
0.965925826
+
0.866025403
i
⋅
0.258819045
i
=
{\displaystyle =0.5\cdot 0.965925826+0.5\cdot 0.258819045i+0.866025403i\cdot 0.965925826+0.866025403i\cdot 0.258819045i=}
=
0.482962913
+
0.129409522
i
+
0.836516303
i
−
0.224143868
=
0.258819045
+
i
0.965925825
=
(
0.258819045
;
0.965925825
)
.
{\displaystyle =0.482962913+0.129409522i+0.836516303i-0.224143868=0.258819045+i0.965925825=(0.258819045;0.965925825).}
x
2
+
y
2
=
0.258819045
2
+
0.965925825
2
=
0.066987298
+
0.933012699
=
0.999999997
{\displaystyle x^{2}+y^{2}=0.258819045^{2}+0.965925825^{2}=0.066987298+0.933012699=0.999999997}
.
x
=
cos
ϕ
=
cos
5
π
12
=
cos
(
1.308996939
)
=
0.258819045.
{\displaystyle x=\cos \phi =\cos {\frac {5\pi }{12}}=\cos(1.308996939)=0.258819045.}
y
=
sin
ϕ
=
sin
5
π
12
=
sin
(
1.308996939
)
=
0.965925826.
{\displaystyle y=\sin \phi =\sin {\frac {5\pi }{12}}=\sin(1.308996939)=0.965925826.}
Kampas
ϕ
{\displaystyle \phi }
taipogi gali būti surastas taip:
ϕ
=
75
π
180
=
0.416666667
⋅
3.141592654
=
1.308996939.
{\displaystyle \phi ={\frac {75\pi }{180}}=0.416666667\cdot 3.141592654=1.308996939.}
Pavyzdys . Pirmo taško koordinatės yra
(
x
1
;
y
1
)
=
(
0.3
;
1
−
0.3
2
)
=
(
0.3
;
1
−
0.09
)
=
(
0.3
;
0.91
)
=
(
0.3
;
0.953939201
)
.
{\displaystyle (x_{1};\;y_{1})=(0.3;\;{\sqrt {1-0.3^{2}}})=(0.3;\;{\sqrt {1-0.09}})=(0.3;\;{\sqrt {0.91}})=(0.3;\;0.953939201).}
Antro taško koordinatės yra
(
x
2
;
y
2
)
=
(
0
,
8
;
0
,
6
)
{\displaystyle (x_{2};y_{2})=(0,8;0,6)}
.
Rasime trečio taško koordinates, kai susidės šie du kampai.
(
x
3
;
y
3
)
=
(
0.3
+
i
0.953939201
)
(
0.8
+
i
0.6
)
=
0.24
+
0.18
i
+
0.763151361
i
−
0.57236352
=
−
0.33236352
+
0.943151361
i
=
(
−
0.33236352
;
0.943151361
)
.
{\displaystyle (x_{3};y_{3})=(0.3+i0.953939201)(0.8+i0.6)=0.24+0.18i+0.763151361i-0.57236352=-0.33236352+0.943151361i=(-0.33236352;0.943151361).}
x
3
2
+
y
3
2
=
(
−
0.33236352
)
2
+
0.943151361
2
=
0.110465509
+
0.889534489
=
0.999999999
{\displaystyle x_{3}^{2}+y_{3}^{2}=(-0.33236352)^{2}+0.943151361^{2}=0.110465509+0.889534489=0.999999999}
.
ϕ
3
=
arccos
x
3
=
arccos
(
−
0.33236352
)
=
1.909604781
{\displaystyle \phi _{3}=\arccos x_{3}=\arccos(-0.33236352)=1.909604781}
, tai yra lygu 109.4122945 laipsnio.
ϕ
3
=
arcsin
y
3
=
arcsin
(
0.943151361
)
=
1.231987872
{\displaystyle \phi _{3}=\arcsin y_{3}=\arcsin(0.943151361)=1.231987872}
tai yra lygu 70.58770545 laipsnio. Ir
ϕ
3
=
90
+
(
90
−
70.58770545
)
=
90
+
19.41229455
=
109.4122946
{\displaystyle \phi _{3}=90+(90-70.58770545)=90+19.41229455=109.4122946}
.
ϕ
1
=
arccos
x
1
=
arccos
0.3
=
1.266103673
{\displaystyle \phi _{1}=\arccos x_{1}=\arccos 0.3=1.266103673}
, tai yra lygu 72.54239688 laipsnio.
ϕ
1
=
arcsin
y
1
=
arcsin
0.953939201
=
1.266103673
{\displaystyle \phi _{1}=\arcsin y_{1}=\arcsin 0.953939201=1.266103673}
, tai yra lygu 72.54239688 laipsnio.
ϕ
2
=
arccos
x
2
=
arccos
0.8
=
0.643501108
{\displaystyle \phi _{2}=\arccos x_{2}=\arccos 0.8=0.643501108}
, tai yra lygu 36.86989765 laipsnio.
ϕ
2
=
arcsin
y
2
=
arcsin
0.6
=
0.643501108
{\displaystyle \phi _{2}=\arcsin y_{2}=\arcsin 0.6=0.643501108}
, tai yra lygu 36.86989765 laipsnio.
ϕ
3
=
ϕ
1
+
ϕ
2
=
1.266103673
+
0.643501108
=
1.909604782
{\displaystyle \phi _{3}=\phi _{1}+\phi _{2}=1.266103673+0.643501108=1.909604782}
arba
ϕ
3
=
ϕ
1
+
ϕ
2
=
72.54239688
+
36.86989765
=
109.4122945
{\displaystyle \phi _{3}=\phi _{1}+\phi _{2}=72.54239688+36.86989765=109.4122945}
laipsnio.
Pavyzdys . Pirmo taško koordinatės yra
(
x
1
;
y
1
)
=
(
0.9
;
1
−
0.9
2
)
=
(
0.9
;
1
−
0.81
)
=
(
0.9
;
0.19
)
=
(
0.9
;
0.435889894
)
.
{\displaystyle (x_{1};\;y_{1})=(0.9;\;{\sqrt {1-0.9^{2}}})=(0.9;\;{\sqrt {1-0.81}})=(0.9;\;{\sqrt {0.19}})=(0.9;\;0.435889894).}
Antro taško koordinatės yra
(
x
2
;
y
2
)
=
(
0
,
8
;
0
,
6
)
{\displaystyle (x_{2};y_{2})=(0,8;0,6)}
.
Rasime trečio taško koordinates, kai susidės šie du kampai.
(
x
3
;
y
3
)
=
(
0.9
+
i
0.435889894
)
(
0.8
+
i
0.6
)
=
0.72
+
0.54
i
+
0.348711915
i
−
0.261533936
=
0.458466063
+
0.888711915
i
=
(
0.458466063
;
0.888711915
)
.
{\displaystyle (x_{3};y_{3})=(0.9+i0.435889894)(0.8+i0.6)=0.72+0.54i+0.348711915i-0.261533936=0.458466063+0.888711915i=(0.458466063;0.888711915).}
x
3
2
+
y
3
2
=
0.458466063
2
+
0.888711915
2
=
0.21019113
+
0.789808867
=
0.999999998
{\displaystyle x_{3}^{2}+y_{3}^{2}=0.458466063^{2}+0.888711915^{2}=0.21019113+0.789808867=0.999999998}
.
ϕ
3
=
arccos
(
x
3
)
=
arccos
(
0.458466063
)
=
1.094527921
{\displaystyle \phi _{3}=\arccos(x_{3})=\arccos(0.458466063)=1.094527921}
, tai yra lygu 62.71183043 laipsnio.
ϕ
3
=
arcsin
(
y
3
)
=
arcsin
(
0.888711915
)
=
1.09452792
{\displaystyle \phi _{3}=\arcsin(y_{3})=\arcsin(0.888711915)=1.09452792}
tai yra lygu 62.71183035 laipsnio.
ϕ
1
=
arccos
x
1
=
arccos
0.9
=
0.451026811
{\displaystyle \phi _{1}=\arccos x_{1}=\arccos 0.9=0.451026811}
, tai yra lygu 25.84193276 laipsnio.
ϕ
1
=
arcsin
y
1
=
arcsin
0.435889894
=
0.451026811
{\displaystyle \phi _{1}=\arcsin y_{1}=\arcsin 0.435889894=0.451026811}
, tai yra lygu 25.84193276 laipsnio.
ϕ
2
=
arccos
x
2
=
arccos
0.8
=
0.643501108
{\displaystyle \phi _{2}=\arccos x_{2}=\arccos 0.8=0.643501108}
, tai yra lygu 36.86989765 laipsnio.
ϕ
2
=
arcsin
y
2
=
arcsin
0.6
=
0.643501108
{\displaystyle \phi _{2}=\arcsin y_{2}=\arcsin 0.6=0.643501108}
, tai yra lygu 36.86989765 laipsnio.
ϕ
3
=
ϕ
1
+
ϕ
2
=
0.451026811
+
0.643501108
=
1.094527921
{\displaystyle \phi _{3}=\phi _{1}+\phi _{2}=0.451026811+0.643501108=1.094527921}
arba
ϕ
3
=
ϕ
1
+
ϕ
2
=
25.84193276
+
36.86989765
=
62.71183041
{\displaystyle \phi _{3}=\phi _{1}+\phi _{2}=25.84193276+36.86989765=62.71183041}
laipsnio.
Pavyzdys . Pirmo taško koordinatės yra
(
x
1
;
y
1
)
=
(
0.9
;
1
−
0.9
2
)
=
(
0.9
;
1
−
0.81
)
=
(
0.9
;
0.19
)
=
(
0.9
;
0.435889894
)
.
{\displaystyle (x_{1};\;y_{1})=(0.9;\;{\sqrt {1-0.9^{2}}})=(0.9;\;{\sqrt {1-0.81}})=(0.9;\;{\sqrt {0.19}})=(0.9;\;0.435889894).}
Antro taško koordinatės yra
(
x
2
;
y
2
)
=
(
0
,
6
;
0
,
8
)
{\displaystyle (x_{2};y_{2})=(0,6;0,8)}
.
Rasime trečio taško koordinates, kai susidės šie du kampai.
(
x
3
;
y
3
)
=
(
0.9
+
i
0.435889894
)
(
0.6
+
i
0.8
)
=
0.54
+
0.72
i
+
0.261533936
i
−
0.348711915
=
0.191288084
+
0.981533936
i
=
(
0.191288084
;
0.981533936
)
.
{\displaystyle (x_{3};y_{3})=(0.9+i0.435889894)(0.6+i0.8)=0.54+0.72i+0.261533936i-0.348711915=0.191288084+0.981533936i=(0.191288084;0.981533936).}
x
3
2
+
y
3
2
=
0.191288084
2
+
0.981533936
2
=
0
,
036591131
+
0
,
963408867
=
0.999999998
{\displaystyle x_{3}^{2}+y_{3}^{2}=0.191288084^{2}+0.981533936^{2}=0,036591131+0,963408867=0.999999998}
.
ϕ
3
=
arccos
(
x
3
)
=
arccos
(
0.191288084
)
=
1.37832203
{\displaystyle \phi _{3}=\arccos(x_{3})=\arccos(0.191288084)=1.37832203}
, tai yra lygu 78,97203515 laipsnio.
ϕ
3
=
arcsin
(
y
3
)
=
arcsin
(
0.981533936
)
=
1.378322027
{\displaystyle \phi _{3}=\arcsin(y_{3})=\arcsin(0.981533936)=1.378322027}
tai yra lygu 78,97203493 laipsnio.
ϕ
1
=
arccos
x
1
=
arccos
0.9
=
0.451026811
{\displaystyle \phi _{1}=\arccos x_{1}=\arccos 0.9=0.451026811}
, tai yra lygu 25.84193276 laipsnio.
ϕ
1
=
arcsin
y
1
=
arcsin
0.435889894
=
0.451026811
{\displaystyle \phi _{1}=\arcsin y_{1}=\arcsin 0.435889894=0.451026811}
, tai yra lygu 25.84193276 laipsnio.
ϕ
2
=
arccos
x
2
=
arccos
0.6
=
0.927295218
{\displaystyle \phi _{2}=\arccos x_{2}=\arccos 0.6=0.927295218}
, tai yra lygu 53,13010235 laipsnio.
ϕ
2
=
arcsin
y
2
=
arcsin
0.8
=
0.927295218
{\displaystyle \phi _{2}=\arcsin y_{2}=\arcsin 0.8=0.927295218}
, tai yra lygu 53,13010235 laipsnio.
ϕ
3
=
ϕ
1
+
ϕ
2
=
0.451026811
+
0.927295218
=
1.37832203
{\displaystyle \phi _{3}=\phi _{1}+\phi _{2}=0.451026811+0.927295218=1.37832203}
arba
ϕ
3
=
ϕ
1
+
ϕ
2
=
25.84193276
+
53.13010235
=
78.97203512
{\displaystyle \phi _{3}=\phi _{1}+\phi _{2}=25.84193276+53.13010235=78.97203512}
laipsnio.
Pavyzdys. Sudėsime 3 kampus
ϕ
1
=
30
{\displaystyle \phi _{1}=30}
laipsnių. Kampas
ϕ
1
{\displaystyle \phi _{1}}
taip pat lygus
ϕ
1
=
π
6
=
0.523598775
{\displaystyle \phi _{1}={\frac {\pi }{6}}=0.523598775}
radiano. Kampo
ϕ
1
{\displaystyle \phi _{1}}
galai yra taškai
(
x
0
;
y
0
)
=
(
0
;
0
)
{\displaystyle (x_{0};y_{0})=(0;0)}
ir
(
x
1
;
y
1
)
=
(
3
2
;
0.5
)
=
(
0.866025403
;
0.5
)
.
{\displaystyle (x_{1};\;y_{1})=({\frac {\sqrt {3}}{2}};\;0.5)=(0.866025403;\;0.5).}
Ir
0.866025403
2
+
0.5
2
=
0.75
+
0.25
=
1
{\displaystyle 0.866025403^{2}+0.5^{2}=0.75+0.25=1}
.
(
x
2
;
y
2
)
=
(
x
1
+
i
y
1
)
2
=
(
3
2
+
i
0.5
)
2
=
0.75
+
2
⋅
3
2
⋅
i
2
−
0.25
=
0.5
+
i
3
2
=
(
0.5
;
0.866025403
)
.
{\displaystyle (x_{2};\;y_{2})=(x_{1}+iy_{1})^{2}=({\frac {\sqrt {3}}{2}}+i0.5)^{2}=0.75+2\cdot {\frac {\sqrt {3}}{2}}\cdot {\frac {i}{2}}-0.25=0.5+{\frac {i{\sqrt {3}}}{2}}=(0.5;\;0.866025403).}
(
x
3
;
y
3
)
=
(
x
2
+
i
y
2
)
(
x
1
+
i
y
1
)
=
(
0.5
+
i
0.866025403
)
(
0.866025403
+
i
0.5
)
=
0.433012701
+
i
0.25
+
i
0.75
−
0.433012701
=
i
=
(
0
;
1
)
.
{\displaystyle (x_{3};y_{3})=(x_{2}+iy_{2})(x_{1}+iy_{1})=(0.5+i0.866025403)(0.866025403+i0.5)=0.433012701+i0.25+i0.75-0.433012701=i=(0;1).}
ϕ
1
=
arccos
x
1
=
arccos
3
2
=
0.523598775
{\displaystyle \phi _{1}=\arccos x_{1}=\arccos {\frac {\sqrt {3}}{2}}=0.523598775}
, tai yra 30 laipsnių.
ϕ
2
=
arccos
x
2
=
arccos
0
,
5
=
1
,
047197551
{\displaystyle \phi _{2}=\arccos x_{2}=\arccos 0,5=1,047197551}
, tai yra 60 laipsnių.
ϕ
3
=
arccos
x
3
=
arccos
(
0
)
=
1
,
570796327
{\displaystyle \phi _{3}=\arccos x_{3}=\arccos(0)=1,570796327}
, tai yra 90 laipsnių.
Pavyzdys . Pasinaudodami kompleksinių skaičių dalybos taisykle
a
+
b
i
c
+
d
i
=
(
a
+
b
i
)
(
c
−
d
i
)
(
c
+
d
i
)
(
c
−
d
i
)
=
(
a
c
+
b
d
)
+
(
a
c
−
a
d
)
i
c
2
+
d
2
=
a
c
+
b
d
c
2
+
d
2
+
b
c
−
a
d
c
2
+
d
2
i
{\displaystyle {\frac {a+bi}{c+di}}={\frac {(a+bi)(c-di)}{(c+di)(c-di)}}={\frac {(ac+bd)+(ac-ad)i}{c^{2}+d^{2}}}={\frac {ac+bd}{c^{2}+d^{2}}}+{\frac {bc-ad}{c^{2}+d^{2}}}i}
, atimsime vieną kampą iš kito. Pirmo taško koordinatės yra
(
x
1
;
y
1
)
=
(
0.6
;
0.8
)
{\displaystyle (x_{1};y_{1})=(0.6;0.8)}
, o antro taško koordinatės yra
(
x
2
;
y
2
)
=
(
0.8
;
0.6
)
{\displaystyle (x_{2};y_{2})=(0.8;0.6)}
. Atimsime antrą kampą iš pirmo.
(
x
3
;
y
3
)
=
(
x
1
+
i
y
1
)
/
(
x
2
+
i
y
2
)
=
(
0.6
+
i
0.8
)
/
(
0.8
+
i
0.6
)
=
{\displaystyle (x_{3};y_{3})=(x_{1}+iy_{1})/(x_{2}+iy_{2})=(0.6+i0.8)/(0.8+i0.6)=}
=
x
1
x
2
+
y
1
y
2
x
2
2
+
y
2
2
+
y
1
x
2
−
x
1
y
2
x
2
2
+
y
2
2
⋅
i
=
0.6
⋅
0.8
+
0.8
⋅
0.6
0.6
2
+
0.8
2
+
0.8
⋅
0.8
−
0.6
⋅
0.6
0.8
2
+
0.6
2
⋅
i
=
{\displaystyle ={\frac {x_{1}x_{2}+y_{1}y_{2}}{x_{2}^{2}+y_{2}^{2}}}+{\frac {y_{1}x_{2}-x_{1}y_{2}}{x_{2}^{2}+y_{2}^{2}}}\cdot i={\frac {0.6\cdot 0.8+0.8\cdot 0.6}{0.6^{2}+0.8^{2}}}+{\frac {0.8\cdot 0.8-0.6\cdot 0.6}{0.8^{2}+0.6^{2}}}\cdot i=}
=
0.48
+
0.48
0.36
+
0.64
+
0.64
−
0.36
0.64
+
0.36
⋅
i
=
0.96
1
+
0.28
1
⋅
i
=
(
0.96
;
0.28
)
.
{\displaystyle ={\frac {0.48+0.48}{0.36+0.64}}+{\frac {0.64-0.36}{0.64+0.36}}\cdot i={\frac {0.96}{1}}+{\frac {0.28}{1}}\cdot i=(0.96;\;0.28).}
(
x
3
)
2
+
(
y
3
)
2
=
0.96
2
+
0.28
2
=
0.9216
+
0.0784
=
1
{\displaystyle (x_{3})^{2}+(y_{3})^{2}=0.96^{2}+0.28^{2}=0.9216+0.0784=1}
.
ϕ
3
=
arccos
(
x
3
)
=
arccos
(
0.96
)
=
0.283794109
{\displaystyle \phi _{3}=\arccos(x_{3})=\arccos(0.96)=0.283794109}
, tai yra lygu 16.26020471 laipsnio.
ϕ
3
=
arcsin
(
y
3
)
=
arcsin
(
0.28
)
=
0.283794109
{\displaystyle \phi _{3}=\arcsin(y_{3})=\arcsin(0.28)=0.283794109}
tai yra lygu 16.26020471 laipsnio.
ϕ
1
=
arccos
x
1
=
arccos
0.6
=
0.927295218
{\displaystyle \phi _{1}=\arccos x_{1}=\arccos 0.6=0.927295218}
, tai yra lygu 53.13010235 laipsnio.
ϕ
1
=
arcsin
y
1
=
arcsin
0.8
=
0.927295218
{\displaystyle \phi _{1}=\arcsin y_{1}=\arcsin 0.8=0.927295218}
, tai yra lygu 53.13010235 laipsnio.
ϕ
2
=
arccos
x
2
=
arccos
0.8
=
0.643501108
{\displaystyle \phi _{2}=\arccos x_{2}=\arccos 0.8=0.643501108}
, tai yra lygu 36.86989765 laipsnio.
ϕ
2
=
arcsin
y
2
=
arcsin
0.6
=
0.643501108
{\displaystyle \phi _{2}=\arcsin y_{2}=\arcsin 0.6=0.643501108}
, tai yra lygu 36.86989765 laipsnio.
ϕ
3
=
ϕ
1
−
ϕ
2
=
0.927295218
−
0.643501108
=
0.283794109
{\displaystyle \phi _{3}=\phi _{1}-\phi _{2}=0.927295218-0.643501108=0.283794109}
arba
ϕ
3
=
ϕ
1
−
ϕ
2
=
53.13010235
−
36.86989765
=
16.26020471
{\displaystyle \phi _{3}=\phi _{1}-\phi _{2}=53.13010235-36.86989765=16.26020471}
laipsnio.
Pritaikę vektorių formulę dvimatėms koordinatėms, galime patikrinti, kad:
x
3
=
x
1
⋅
x
2
+
y
1
⋅
y
2
x
1
2
+
y
1
2
⋅
x
2
2
+
y
2
2
=
0.6
⋅
0.8
+
0.8
⋅
0.6
0.6
2
+
0.8
2
⋅
0.8
2
+
0.6
2
=
0.48
+
0.48
0.36
+
0.64
⋅
0.64
+
0.36
=
0.96
1
⋅
1
=
0.96
;
{\displaystyle x_{3}={\frac {x_{1}\cdot x_{2}+y_{1}\cdot y_{2}}{{\sqrt {x_{1}^{2}+y_{1}^{2}}}\cdot {\sqrt {x_{2}^{2}+y_{2}^{2}}}}}={\frac {0.6\cdot 0.8+0.8\cdot 0.6}{{\sqrt {0.6^{2}+0.8^{2}}}\cdot {\sqrt {0.8^{2}+0.6^{2}}}}}={\frac {0.48+0.48}{{\sqrt {0.36+0.64}}\cdot {\sqrt {0.64+0.36}}}}={\frac {0.96}{{\sqrt {1}}\cdot {\sqrt {1}}}}=0.96;}
y
3
=
|
x
1
⋅
y
2
−
x
2
⋅
y
1
x
1
2
+
y
1
2
⋅
x
2
2
+
y
2
2
|
=
|
0.6
⋅
0.6
−
0.8
⋅
0.8
0.6
2
+
0.8
2
⋅
0.8
2
+
0.6
2
|
=
|
0.36
−
0.64
0.36
+
0.64
⋅
0.64
+
0.36
|
=
|
−
0.28
1
⋅
1
|
=
|
−
0.28
|
=
0.28.
{\displaystyle y_{3}=|{\frac {x_{1}\cdot y_{2}-x_{2}\cdot y_{1}}{{\sqrt {x_{1}^{2}+y_{1}^{2}}}\cdot {\sqrt {x_{2}^{2}+y_{2}^{2}}}}}|=|{\frac {0.6\cdot 0.6-0.8\cdot 0.8}{{\sqrt {0.6^{2}+0.8^{2}}}\cdot {\sqrt {0.8^{2}+0.6^{2}}}}}|=|{\frac {0.36-0.64}{{\sqrt {0.36+0.64}}\cdot {\sqrt {0.64+0.36}}}}|=|{\frac {-0.28}{{\sqrt {1}}\cdot {\sqrt {1}}}}|=|-0.28|=0.28.}
Šios vektrorių formulės dvimatėms koordinatėms (skirtos kampui atimti) išplaukia iš vektorių formulių, surasti kampui tarp dviejų vektorių:
ϕ
=
arccos
x
1
⋅
x
2
+
y
1
⋅
y
2
+
z
1
⋅
z
2
x
1
2
+
y
1
2
+
z
1
2
⋅
x
2
2
+
y
2
2
+
z
2
2
;
{\displaystyle \phi =\arccos {\frac {x_{1}\cdot x_{2}+y_{1}\cdot y_{2}+z_{1}\cdot z_{2}}{{\sqrt {x_{1}^{2}+y_{1}^{2}+z_{1}^{2}}}\cdot {\sqrt {x_{2}^{2}+y_{2}^{2}+z_{2}^{2}}}}};}
ϕ
=
arcsin
(
x
1
⋅
y
2
−
x
2
⋅
y
1
)
2
+
(
x
1
⋅
z
2
−
x
2
⋅
z
1
)
2
+
(
y
1
⋅
z
2
−
y
2
⋅
z
1
)
2
x
1
2
+
y
1
2
+
z
1
2
⋅
x
2
2
+
y
2
2
+
z
2
2
.
{\displaystyle \phi =\arcsin {\frac {\sqrt {(x_{1}\cdot y_{2}-x_{2}\cdot y_{1})^{2}+(x_{1}\cdot z_{2}-x_{2}\cdot z_{1})^{2}+(y_{1}\cdot z_{2}-y_{2}\cdot z_{1})^{2}}}{{\sqrt {x_{1}^{2}+y_{1}^{2}+z_{1}^{2}}}\cdot {\sqrt {x_{2}^{2}+y_{2}^{2}+z_{2}^{2}}}}}.}
sin
n
α
{\displaystyle \sin n\alpha }
ir
cos
n
α
{\displaystyle \cos n\alpha }
su dideliais n patogu nustatynėti, naudojantis Muavro formule kompleksiniams skaičiams:
cos
n
α
+
i
sin
n
α
=
(
cos
n
α
+
i
sin
n
α
)
n
=
{\displaystyle \cos n\alpha +i\sin n\alpha =(\cos n\alpha +i\sin n\alpha )^{n}=}
=
cos
n
α
+
i
n
cos
n
−
1
α
sin
α
−
C
n
2
cos
n
−
2
α
sin
2
α
−
i
C
n
3
cos
n
−
3
α
sin
3
α
+
C
n
4
cos
n
−
4
α
sin
4
α
+
.
.
.
,
{\displaystyle =\cos ^{n}\alpha +in\cos ^{n-1}\alpha \sin \alpha -C_{n}^{2}\cos ^{n-2}\alpha \sin ^{2}\alpha -iC_{n}^{3}\cos ^{n-3}\alpha \sin ^{3}\alpha +C_{n}^{4}\cos ^{n-4}\alpha \sin ^{4}\alpha +...,}
iš kur
cos
n
α
=
cos
n
α
−
C
n
2
cos
n
−
2
α
sin
2
α
+
C
n
4
cos
n
−
4
α
sin
4
α
−
C
n
6
cos
n
−
6
α
sin
6
α
+
.
.
.
,
{\displaystyle \cos n\alpha =\cos ^{n}\alpha -C_{n}^{2}\cos ^{n-2}\alpha \sin ^{2}\alpha +C_{n}^{4}\cos ^{n-4}\alpha \sin ^{4}\alpha -C_{n}^{6}\cos ^{n-6}\alpha \sin ^{6}\alpha +...,}
sin
n
α
=
n
cos
n
−
1
α
sin
α
−
C
n
3
cos
n
−
3
α
sin
3
α
+
C
n
5
cos
n
−
5
α
sin
5
α
−
.
.
.
.
{\displaystyle \sin n\alpha =n\cos ^{n-1}\alpha \sin \alpha -C_{n}^{3}\cos ^{n-3}\alpha \sin ^{3}\alpha +C_{n}^{5}\cos ^{n-5}\alpha \sin ^{5}\alpha -...\;.}
Reiškinį
(
cos
n
α
+
i
sin
n
α
)
n
{\displaystyle (\cos n\alpha +i\sin n\alpha )^{n}}
išdėstėme pagal Binomo formulę .
Pavyzdys. Išreikšime
cos
3
φ
{\displaystyle \cos 3\varphi }
ir
sin
3
φ
{\displaystyle \sin 3\varphi }
sinuso ir kosinuso laipsniais.
Pakėlę
cos
φ
+
i
sin
φ
{\displaystyle \cos \varphi +i\sin \varphi }
kubu, gauname
(
cos
φ
+
i
sin
φ
)
3
=
cos
3
φ
+
i
sin
3
φ
,
{\displaystyle (\cos \varphi +i\sin \varphi )^{3}=\cos 3\varphi +i\sin 3\varphi ,}
arba
cos
3
φ
+
3
i
cos
2
φ
sin
φ
−
3
cos
φ
sin
2
φ
−
i
sin
3
φ
=
cos
3
φ
+
i
sin
3
φ
.
{\displaystyle \cos ^{3}\varphi +3i\cos ^{2}\varphi \sin \varphi -3\cos \varphi \sin ^{2}\varphi -i\sin ^{3}\varphi =\cos 3\varphi +i\sin 3\varphi .}
Atskyrę realiąją ir menamąją dalį, turėsime
cos
3
φ
=
cos
3
φ
−
3
cos
φ
sin
2
φ
=
cos
3
φ
−
3
cos
φ
(
1
−
cos
2
φ
)
=
4
cos
3
φ
−
3
cos
φ
;
{\displaystyle \cos 3\varphi =\cos ^{3}\varphi -3\cos \varphi \sin ^{2}\varphi =\cos ^{3}\varphi -3\cos \varphi (1-\cos ^{2}\varphi )=4\cos ^{3}\varphi -3\cos \varphi ;}
sin
3
φ
=
3
cos
2
φ
sin
φ
−
sin
3
φ
=
3
(
1
−
sin
2
φ
)
sin
φ
−
sin
3
φ
=
3
sin
φ
−
4
sin
3
φ
.
{\displaystyle \sin 3\varphi =3\cos ^{2}\varphi \sin \varphi -\sin ^{3}\varphi =3(1-\sin ^{2}\varphi )\sin \varphi -\sin ^{3}\varphi =3\sin \varphi -4\sin ^{3}\varphi .}
Kompleksinių skaičių laukas
keisti
Formaliai kompleksinis skaičius gali būti apibrėžtas kaip išrikiuota dviejų realių skaičių (a , b ) pora su įvestomis operacijomis:
(
a
,
b
)
+
(
c
,
d
)
=
(
a
+
c
,
b
+
d
)
{\displaystyle (a,b)+(c,d)=(a+c,b+d)\,}
(
a
,
b
)
⋅
(
c
,
d
)
=
(
a
c
−
b
d
,
b
c
+
a
d
)
.
{\displaystyle (a,b)\cdot (c,d)=(ac-bd,bc+ad).\,}
Taip apibrėžti kompleksiniai skaičiai sudaro lauką , kompleksinių skaičių lauką, žymimą C (laukas matematikoje yra algebrinė struktūra, kurioje apibrėžtos sudėties, atimties, daugybos ir dalybos operacijos, turinčios tam tikras algebrines savybes. Pvz., realieji skaičiai yra laukas).
Realusis skaičius a yra sutapatinamas su kompleksiniu skaičiumi (a , 0), ir tuo būdu realiųjų skaičių laukas R tampa C dalimi. Menamasis vienetas i apibrėžiamas kaip kompleksinis skaičius (0, 1), kuris tenkina:
(
a
,
b
)
=
a
⋅
(
1
,
0
)
+
b
⋅
(
0
,
1
)
=
a
+
b
i
ir
i
2
=
(
0
,
1
)
⋅
(
0
,
1
)
=
(
−
1
,
0
)
=
−
1.
{\displaystyle (a,b)=a\cdot (1,0)+b\cdot (0,1)=a+bi\quad {\text{ir}}\quad i^{2}=(0,1)\cdot (0,1)=(-1,0)=-1.}
Lauke C mes turime:
vienetinį elementą sudėčiai („nulį“): (0, 0)
vienetinį elementą daugybai („vienetą“): (1, 0)
atvirkštinį elementą sudėties operacijai (a ,b ): (−a , −b )
atvirkštinį elementą sandaugos operacijai nenuliniam (a , b ):
(
a
a
2
+
b
2
,
−
b
a
2
+
b
2
)
.
{\displaystyle \left({a \over a^{2}+b^{2}},{-b \over a^{2}+b^{2}}\right).}
Kompleksinių skaičių plokštuma
keisti
Kiekvienam kompleksiniam skaičiui z = a + bi galima vienareikšmiškai priskirti plokštumos, kurioje yra Dekarto koordinačių sistema, tašką (a; b). Pagrindiniai kompleksinių skaičių veiksmai gali būti interpretuojami geometriškai: kompleksiniai skaičiai a + ib ir c + id gali būti sumuojami kaip dvimačiai vektoriai (a; b) ir (c; d).
Kompleksiniai skaičiai trigonometrijoje.
Greta algebrinės formos (
z
=
(
a
,
b
)
=
a
+
b
⋅
i
{\displaystyle z=(a,b)=a+b\cdot i}
) dar yra trigonometrinė kompleksinių skaičių užrašymo forma:
z
=
r
(
cos
φ
+
i
sin
φ
)
=
r
e
i
φ
{\displaystyle z=r(\cos \varphi \ +i\sin \varphi \ )=re^{i\varphi }}
,
Čia
r
=
a
2
+
b
2
{\displaystyle r={\sqrt {a^{2}+b^{2}}}}
,
cos
φ
=
a
r
,
{\displaystyle \cos \varphi \ ={\frac {a}{r}},}
sin
φ
=
b
r
,
{\displaystyle \sin \varphi \ ={\frac {b}{r}},}
.
Formulė kai
r
=
1
{\displaystyle r=1}
yra vadinama Oilerio formule :
e
i
φ
=
cos
φ
+
i
sin
φ
{\displaystyle e^{i\varphi }=\cos \varphi +i\sin \varphi }
.
Šiuo atveju kompleksinis skaičius
(
a
,
b
)
{\displaystyle (a,b)}
turi paprastą geometrinę interpretaciją. a yra atkarpos ilgis x ašimi, o b - y ašimi. Kampas
ϕ
{\displaystyle \phi }
yra kampas tarp x ašies ir tiesės jungiančios koordinačių pradžią (0,0) ir tašką (a, b).
r
{\displaystyle r}
yra atkarpos ilgis nuo koordinačių pradžios (0, 0) iki taško (a, b).
Dviejų kompleksinių skaičių daugyba atrodys taip:
z
=
z
1
z
2
=
r
1
e
i
φ
1
⋅
r
2
e
i
φ
2
=
r
1
r
2
e
i
(
φ
1
+
φ
2
)
{\displaystyle z=z_{1}z_{2}=r_{1}\,e^{i\varphi _{1}}\cdot r_{2}\,e^{i\varphi _{2}}=r_{1}\,r_{2}\,e^{i(\varphi _{1}+\varphi _{2})}\,}
dalyba:
z
=
z
1
z
2
=
r
1
e
i
φ
1
r
2
e
i
φ
2
=
r
1
r
2
e
i
(
φ
1
−
φ
2
)
.
{\displaystyle z={\frac {z_{1}}{z_{2}}}={\frac {r_{1}\,e^{i\varphi _{1}}}{r_{2}\,e^{i\varphi _{2}}}}={\frac {r_{1}}{r_{2}}}\,e^{i(\varphi _{1}-\varphi _{2})}.\,}
Kėlimui laipsniu yra naudojama Muavro formulė :
z
n
=
(
r
e
i
φ
)
n
=
r
n
e
i
n
φ
=
r
n
(
cos
n
φ
+
i
sin
n
φ
)
{\displaystyle z^{n}={\big (}r\,e^{i\varphi }{\big )}^{n}=r^{n}\,e^{in\varphi }=r^{n}(\cos {n\varphi }+i\sin {n\varphi })\,}
Šaknies traukimo operacija:
ω
=
z
n
,
{\displaystyle \omega ={\sqrt[{n}]{z}},}
ω
k
=
r
n
(
cos
φ
+
2
π
k
n
+
i
sin
φ
+
2
π
k
n
)
{\displaystyle \omega _{k}={\sqrt[{n}]{r}}\left(\cos {\frac {\varphi \ +2\pi \ k}{n}}+i\sin {\frac {\varphi \ +2\pi \ k}{n}}\right)}
- egzistuoja lygiai n skirtingų šaknų. Kai k kinta nuo 0 iki (n-1) visos gaunamos reikšmės yra skirtingos. Kai
k
≥
n
,
{\displaystyle k\geq n,}
gaunamos reikšmės kartojasi.
Šaknies traukimo iš kompleksinio skaičiaus formulė:
α
n
=
β
k
=
r
n
(
cos
ϕ
+
2
k
π
n
+
i
sin
ϕ
+
2
k
π
n
)
,
(
k
=
0
,
1
,
2
,
.
.
.
,
n
−
1
)
.
{\displaystyle {\sqrt[{n}]{\alpha }}=\beta _{k}={\sqrt[{n}]{r}}(\cos {\frac {\phi +2k\pi }{n}}+i\sin {\frac {\phi +2k\pi }{n}}),\quad (k=0,\;1,\;2,\;...,n-1).}
Pavyzdis . Išskaičiuosime
4
4
.
{\displaystyle {\sqrt[{4}]{4}}.}
Čia
ϕ
=
0
{\displaystyle \phi =0}
ir
r
=
4
{\displaystyle r=4}
. Iš bendrosios šaknies formulės
β
k
=
2
(
cos
2
k
π
4
+
i
sin
2
k
π
4
)
(
k
=
0
,
1
,
2
,
3
)
.
{\displaystyle \beta _{k}={\sqrt {2}}(\cos {\frac {2k\pi }{4}}+i\sin {\frac {2k\pi }{4}})\quad (k=0,\;1,\;2,\;3).}
Todėl
β
0
=
2
(
cos
2
⋅
0
⋅
π
4
+
i
sin
2
⋅
0
⋅
π
4
)
=
2
(
cos
0
+
i
sin
0
)
=
2
(
1
+
i
⋅
0
)
=
2
,
{\displaystyle \beta _{0}={\sqrt {2}}(\cos {\frac {2\cdot 0\cdot \pi }{4}}+i\sin {\frac {2\cdot 0\cdot \pi }{4}})={\sqrt {2}}(\cos 0+i\sin 0)={\sqrt {2}}(1+i\cdot 0)={\sqrt {2}},}
β
1
=
2
(
cos
2
⋅
1
⋅
π
4
+
i
sin
2
⋅
1
⋅
π
4
)
=
2
(
cos
π
2
+
i
sin
π
2
)
=
2
(
0
+
i
⋅
1
)
=
i
2
,
{\displaystyle \beta _{1}={\sqrt {2}}(\cos {\frac {2\cdot 1\cdot \pi }{4}}+i\sin {\frac {2\cdot 1\cdot \pi }{4}})={\sqrt {2}}(\cos {\frac {\pi }{2}}+i\sin {\frac {\pi }{2}})={\sqrt {2}}(0+i\cdot 1)=i{\sqrt {2}},}
β
2
=
2
(
cos
2
⋅
2
⋅
π
4
+
i
sin
2
⋅
2
⋅
π
4
)
=
2
(
cos
π
+
i
sin
π
)
=
2
(
−
1
+
i
⋅
0
)
=
−
2
,
{\displaystyle \beta _{2}={\sqrt {2}}(\cos {\frac {2\cdot 2\cdot \pi }{4}}+i\sin {\frac {2\cdot 2\cdot \pi }{4}})={\sqrt {2}}(\cos \pi +i\sin \pi )={\sqrt {2}}(-1+i\cdot 0)=-{\sqrt {2}},}
β
3
=
2
(
cos
2
⋅
3
⋅
π
4
+
i
sin
2
⋅
3
⋅
π
4
)
=
2
(
cos
3
π
2
+
i
sin
3
π
2
)
=
2
(
0
+
i
⋅
(
−
1
)
)
=
−
i
2
.
{\displaystyle \beta _{3}={\sqrt {2}}(\cos {\frac {2\cdot 3\cdot \pi }{4}}+i\sin {\frac {2\cdot 3\cdot \pi }{4}})={\sqrt {2}}(\cos {\frac {3\pi }{2}}+i\sin {\frac {3\pi }{2}})={\sqrt {2}}(0+i\cdot (-1))=-i{\sqrt {2}}.}
Nesunku matyti, kad
(
β
k
)
4
=
4
{\displaystyle (\beta _{k})^{4}=4}
visiems
k
=
0
,
1
,
2
,
3.
{\displaystyle k=0,\;1,\;2,\;3.}
(
β
0
)
4
=
(
2
)
4
=
2
2
=
4.
{\displaystyle (\beta _{0})^{4}=\left({\sqrt {2}}\right)^{4}=2^{2}=4.}
(
β
1
)
4
=
(
i
2
)
4
=
i
4
(
2
)
4
=
i
⋅
i
⋅
i
⋅
i
(
2
)
4
=
−
1
⋅
i
⋅
i
(
2
)
4
=
(
−
1
)
⋅
(
−
1
)
⋅
2
2
=
4.
{\displaystyle (\beta _{1})^{4}=\left(i{\sqrt {2}}\right)^{4}=i^{4}\left({\sqrt {2}}\right)^{4}=i\cdot i\cdot i\cdot i\left({\sqrt {2}}\right)^{4}=-1\cdot i\cdot i\left({\sqrt {2}}\right)^{4}=(-1)\cdot (-1)\cdot 2^{2}=4.}
(
β
2
)
4
=
(
−
2
)
4
=
(
−
1
)
4
⋅
2
2
=
4.
{\displaystyle (\beta _{2})^{4}=\left(-{\sqrt {2}}\right)^{4}=(-1)^{4}\cdot 2^{2}=4.}
(
β
3
)
4
=
(
−
i
2
)
4
=
i
4
(
−
2
)
4
=
i
⋅
i
⋅
i
⋅
i
(
2
)
4
=
−
1
⋅
i
⋅
i
(
−
2
)
4
=
(
−
1
)
⋅
(
−
1
)
⋅
(
−
2
)
4
=
(
−
2
)
4
=
2
2
=
4.
{\displaystyle (\beta _{3})^{4}=\left(-i{\sqrt {2}}\right)^{4}=i^{4}\left(-{\sqrt {2}}\right)^{4}=i\cdot i\cdot i\cdot i\left({\sqrt {2}}\right)^{4}=-1\cdot i\cdot i\left(-{\sqrt {2}}\right)^{4}=(-1)\cdot (-1)\cdot \left(-{\sqrt {2}}\right)^{4}=\left(-{\sqrt {2}}\right)^{4}=2^{2}=4.}
Tuo patikriname šaknies traukimą.
Ištraukti penkto laipsnio šaknį iš kompleksinio skaičiaus
α
=
−
16
+
i
16
3
.
{\displaystyle \alpha =-16+i16{\sqrt {3}}.}
Sprendimas . Randame spindulį
r
=
16
2
+
(
16
3
)
2
=
256
+
256
⋅
3
=
1024
=
32.
{\displaystyle r={\sqrt {16^{2}+(16{\sqrt {3}})^{2}}}={\sqrt {256+256\cdot 3}}={\sqrt {1024}}=32.}
Toliau taikydami formule galime užrašyti
α
5
=
β
k
=
r
5
(
cos
ϕ
+
2
k
π
5
+
i
sin
ϕ
+
2
k
π
5
)
(
k
=
0
,
1
,
2
,
3
,
4
)
.
{\displaystyle {\sqrt[{5}]{\alpha }}=\beta _{k}={\sqrt[{5}]{r}}(\cos {\frac {\phi +2k\pi }{5}}+i\sin {\frac {\phi +2k\pi }{5}})\quad (k=0,\;1,\;2,\;3,\;4).}
Pažymėkime:
ϵ
=
α
r
=
−
16
+
i
16
3
32
=
−
1
2
+
i
3
2
.
{\displaystyle \epsilon ={\frac {\alpha }{r}}={\frac {-16+i16{\sqrt {3}}}{32}}=-{\frac {1}{2}}+i{\frac {\sqrt {3}}{2}}.}
Dabar žinome, kad
ϕ
=
120
∘
{\displaystyle \phi =120^{\circ }}
arba
ϕ
=
2
π
3
,
{\displaystyle \phi ={\frac {2\pi }{3}},}
nes
cos
2
π
3
=
−
1
2
{\displaystyle \cos {\frac {2\pi }{3}}=-{\frac {1}{2}}}
ir
sin
2
π
3
=
3
2
.
{\displaystyle \sin {\frac {2\pi }{3}}={\frac {\sqrt {3}}{2}}.}
Arba
arccos
(
−
1
2
)
=
2
π
3
{\displaystyle \arccos \left(-{\frac {1}{2}}\right)={\frac {2\pi }{3}}}
ir
arcsin
3
2
=
π
3
.
{\displaystyle \arcsin {\frac {\sqrt {3}}{2}}={\frac {\pi }{3}}.}
Bet kadangi realioji dalis neigiama tai
ϕ
=
120
∘
,
{\displaystyle \phi =120^{\circ },}
o ne
ϕ
=
60
∘
.
{\displaystyle \phi =60^{\circ }.}
Dabar galime rasti visas šaknis:
β
0
=
32
5
(
cos
2
π
3
+
2
⋅
0
⋅
π
5
+
i
sin
2
π
3
+
2
⋅
0
⋅
π
5
)
=
{\displaystyle \beta _{0}={\sqrt[{5}]{32}}(\cos {\frac {{\frac {2\pi }{3}}+2\cdot 0\cdot \pi }{5}}+i\sin {\frac {{\frac {2\pi }{3}}+2\cdot 0\cdot \pi }{5}})=}
=
2
(
cos
2
π
3
5
+
i
sin
2
π
3
5
)
=
2
(
cos
2
π
15
+
i
sin
2
π
15
)
=
2
(
cos
(
24
∘
)
+
i
sin
(
24
∘
)
=
{\displaystyle =2(\cos {\frac {\frac {2\pi }{3}}{5}}+i\sin {\frac {\frac {2\pi }{3}}{5}})=2(\cos {\frac {2\pi }{15}}+i\sin {\frac {2\pi }{15}})=2(\cos(24^{\circ })+i\sin(24^{\circ })=}
=
2
(
0.913545457
+
i
0.406736643
)
=
1.827090915
+
i
0.813473286
;
{\displaystyle =2(0.913545457+i0.406736643)=1.827090915+i0.813473286;}
β
1
=
32
5
(
cos
2
π
3
+
2
⋅
1
⋅
π
5
+
i
sin
2
π
3
+
2
⋅
1
⋅
π
5
)
=
{\displaystyle \beta _{1}={\sqrt[{5}]{32}}\left(\cos {\frac {{\frac {2\pi }{3}}+2\cdot 1\cdot \pi }{5}}+i\sin {\frac {{\frac {2\pi }{3}}+2\cdot 1\cdot \pi }{5}}\right)=}
=
2
(
cos
(
2
π
3
⋅
5
+
2
π
5
)
+
i
sin
(
2
π
3
⋅
5
+
2
π
5
)
)
=
{\displaystyle =2\left(\cos \left({\frac {2\pi }{3\cdot 5}}+{\frac {2\pi }{5}}\right)+i\sin \left({\frac {2\pi }{3\cdot 5}}+{\frac {2\pi }{5}}\right)\right)=}
=
2
(
cos
2
π
+
6
π
3
⋅
5
+
i
sin
2
π
+
6
π
3
⋅
5
)
=
2
(
cos
8
π
15
+
i
sin
8
π
15
)
=
{\displaystyle =2\left(\cos {\frac {2\pi +6\pi }{3\cdot 5}}+i\sin {\frac {2\pi +6\pi }{3\cdot 5}}\right)=2\left(\cos {\frac {8\pi }{15}}+i\sin {\frac {8\pi }{15}}\right)=}
=
2
(
cos
(
96
∘
)
+
i
sin
(
96
∘
)
)
=
2
(
−
0.104528463
+
i
0.994521895
)
=
−
0.209056926
+
i
1.989043791
;
{\displaystyle =2(\cos(96^{\circ })+i\sin(96^{\circ }))=2(-0.104528463+i0.994521895)=-0.209056926+i1.989043791;}
β
2
=
32
5
(
cos
2
π
3
+
2
⋅
2
⋅
π
5
+
i
sin
2
π
3
+
2
⋅
2
⋅
π
5
)
=
{\displaystyle \beta _{2}={\sqrt[{5}]{32}}\left(\cos {\frac {{\frac {2\pi }{3}}+2\cdot 2\cdot \pi }{5}}+i\sin {\frac {{\frac {2\pi }{3}}+2\cdot 2\cdot \pi }{5}}\right)=}
=
2
(
cos
(
2
π
3
⋅
5
+
4
π
5
)
+
i
sin
(
2
π
3
⋅
5
+
4
π
5
)
)
=
{\displaystyle =2\left(\cos \left({\frac {2\pi }{3\cdot 5}}+{\frac {4\pi }{5}}\right)+i\sin \left({\frac {2\pi }{3\cdot 5}}+{\frac {4\pi }{5}}\right)\right)=}
=
2
(
cos
2
π
+
12
π
3
⋅
5
+
i
sin
2
π
+
12
π
3
⋅
5
)
=
2
(
cos
14
π
15
+
i
sin
14
π
15
)
=
{\displaystyle =2\left(\cos {\frac {2\pi +12\pi }{3\cdot 5}}+i\sin {\frac {2\pi +12\pi }{3\cdot 5}}\right)=2\left(\cos {\frac {14\pi }{15}}+i\sin {\frac {14\pi }{15}}\right)=}
=
2
(
cos
(
168
∘
)
+
i
sin
(
168
∘
)
)
=
2
(
−
0.9781476
+
i
0.20791169
)
=
−
1.956295201
+
i
0.415823381
;
{\displaystyle =2(\cos(168^{\circ })+i\sin(168^{\circ }))=2(-0.9781476+i0.20791169)=-1.956295201+i0.415823381;}
β
3
=
32
5
(
cos
2
π
3
+
2
⋅
3
⋅
π
5
+
i
sin
2
π
3
+
2
⋅
3
⋅
π
5
)
=
{\displaystyle \beta _{3}={\sqrt[{5}]{32}}\left(\cos {\frac {{\frac {2\pi }{3}}+2\cdot 3\cdot \pi }{5}}+i\sin {\frac {{\frac {2\pi }{3}}+2\cdot 3\cdot \pi }{5}}\right)=}
=
2
(
cos
(
2
π
3
⋅
5
+
6
π
5
)
+
i
sin
(
2
π
3
⋅
5
+
6
π
5
)
)
=
{\displaystyle =2\left(\cos \left({\frac {2\pi }{3\cdot 5}}+{\frac {6\pi }{5}}\right)+i\sin \left({\frac {2\pi }{3\cdot 5}}+{\frac {6\pi }{5}}\right)\right)=}
=
2
(
cos
2
π
+
18
π
3
⋅
5
+
i
sin
2
π
+
18
π
3
⋅
5
)
=
2
(
cos
20
π
15
+
i
sin
20
π
15
)
=
{\displaystyle =2\left(\cos {\frac {2\pi +18\pi }{3\cdot 5}}+i\sin {\frac {2\pi +18\pi }{3\cdot 5}}\right)=2\left(\cos {\frac {20\pi }{15}}+i\sin {\frac {20\pi }{15}}\right)=}
=
2
(
cos
(
240
∘
)
+
i
sin
(
240
∘
)
)
=
2
(
−
0.5
+
i
3
2
)
=
−
1
+
i
3
;
{\displaystyle =2(\cos(240^{\circ })+i\sin(240^{\circ }))=2(-0.5+i{{\sqrt {3}} \over 2})=-1+i{\sqrt {3}};}
β
4
=
32
5
(
cos
2
π
3
+
2
⋅
4
⋅
π
5
+
i
sin
2
π
3
+
2
⋅
4
⋅
π
5
)
=
{\displaystyle \beta _{4}={\sqrt[{5}]{32}}\left(\cos {\frac {{\frac {2\pi }{3}}+2\cdot 4\cdot \pi }{5}}+i\sin {\frac {{\frac {2\pi }{3}}+2\cdot 4\cdot \pi }{5}}\right)=}
=
2
(
cos
(
2
π
3
⋅
5
+
8
π
5
)
+
i
sin
(
2
π
3
⋅
5
+
8
π
5
)
)
=
{\displaystyle =2\left(\cos \left({\frac {2\pi }{3\cdot 5}}+{\frac {8\pi }{5}}\right)+i\sin \left({\frac {2\pi }{3\cdot 5}}+{\frac {8\pi }{5}}\right)\right)=}
=
2
(
cos
2
π
+
24
π
3
⋅
5
+
i
sin
2
π
+
24
π
3
⋅
5
)
=
2
(
cos
26
π
15
+
i
sin
26
π
15
)
=
{\displaystyle =2\left(\cos {\frac {2\pi +24\pi }{3\cdot 5}}+i\sin {\frac {2\pi +24\pi }{3\cdot 5}}\right)=2\left(\cos {\frac {26\pi }{15}}+i\sin {\frac {26\pi }{15}}\right)=}
=
2
(
cos
(
312
∘
)
+
i
sin
(
312
∘
)
)
=
2
(
0.669130606
−
i
0.743144825
)
=
1.338261213
−
i
1.486289651.
{\displaystyle =2(\cos(312^{\circ })+i\sin(312^{\circ }))=2(0.669130606-i0.743144825)=1.338261213-i1.486289651.}
Ištraukti kubinę šaknį iš
a
=
−
q
2
+
q
2
4
+
p
3
27
{\displaystyle a=-{\frac {q}{2}}+{\sqrt {{\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}}}}
ir
b
=
−
q
2
−
q
2
4
+
p
3
27
,
{\displaystyle b=-{\frac {q}{2}}-{\sqrt {{\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}}},}
kai
p
=
−
19
{\displaystyle p=-19}
ir
q
=
30.
{\displaystyle q=30.}
Gauname
a
=
−
q
2
+
q
2
4
+
p
3
27
=
−
30
2
+
30
2
4
+
(
−
19
)
3
27
=
−
15
+
900
4
+
−
6859
27
=
−
15
+
225
+
−
6859
27
=
−
15
+
225
⋅
27
−
6859
27
=
{\displaystyle a=-{\frac {q}{2}}+{\sqrt {{\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}}}=-{\frac {30}{2}}+{\sqrt {{\frac {30^{2}}{4}}+{\frac {(-19)^{3}}{27}}}}=-15+{\sqrt {{\frac {900}{4}}+{\frac {-6859}{27}}}}=-15+{\sqrt {225+{\frac {-6859}{27}}}}=-15+{\sqrt {\frac {225\cdot 27-6859}{27}}}=}
=
−
15
+
6075
−
6859
27
=
−
15
+
−
784
27
=
−
15
+
28
−
1
27
=
−
15
+
i
28
27
=
−
15
+
5.3886025
i
;
{\displaystyle =-15+{\sqrt {\frac {6075-6859}{27}}}=-15+{\sqrt {-{\frac {784}{27}}}}=-15+28{\sqrt {-{\frac {1}{27}}}}=-15+i{\frac {28}{\sqrt {27}}}=-15+5.3886025i;}
b
=
−
15
−
6075
−
6859
27
=
−
15
−
−
784
27
=
−
15
−
28
−
1
27
=
−
15
−
i
28
27
=
−
15
−
5.3886025
i
.
{\displaystyle b=-15-{\sqrt {\frac {6075-6859}{27}}}=-15-{\sqrt {-{\frac {784}{27}}}}=-15-28{\sqrt {-{\frac {1}{27}}}}=-15-i{\frac {28}{\sqrt {27}}}=-15-5.3886025i.}
Surandame vektoriaus ilgį r (tiek a tiek b ilgis r vienodas):
r
=
(
−
15
)
2
+
(
28
1
27
)
2
=
225
+
784
27
=
225
⋅
27
+
784
27
=
6075
+
784
27
=
6859
27
≈
254.037037037
≈
15.938539.
{\displaystyle r={\sqrt {(-15)^{2}+(28{\sqrt {\frac {1}{27}}})^{2}}}={\sqrt {225+{\frac {784}{27}}}}={\sqrt {\frac {225\cdot 27+784}{27}}}={\sqrt {\frac {6075+784}{27}}}={\sqrt {\frac {6859}{27}}}\approx {\sqrt {254.037037037}}\approx 15.938539.}
Padalinę a ir b iš r gausime normalizuotus a ir b [tipo vektorius]:
a
n
=
a
r
=
−
15
+
5.3886025
i
15.938539
=
−
0.941115117
+
0.33808635
i
;
{\displaystyle a_{n}={\frac {a}{r}}={\frac {-15+5.3886025i}{15.938539}}=-0.941115117+0.33808635i;}
b
n
=
a
r
=
−
15
−
5.3886025
i
15.938539
=
−
0.941115117
−
0.33808635
i
.
{\displaystyle b_{n}={\frac {a}{r}}={\frac {-15-5.3886025i}{15.938539}}=-0.941115117-0.33808635i.}
Atėjo laikas išsiaiškinti kokiuose vienetinio apskritimo ketvirčiuose guli vektoriai
a
n
{\displaystyle a_{n}}
ir
b
n
{\displaystyle b_{n}}
(kiek laipsnių vektoriai
a
n
{\displaystyle a_{n}}
ir
b
n
{\displaystyle b_{n}}
pasisukę prieš laikrodžio rodyklę nuo ašies Ox ).
Vektoriaus
a
n
{\displaystyle a_{n}}
realioji dalis yra su minuso ženklu, o menamoji dalis su pliuso ženklu. Todėl vektoriaus
a
n
{\displaystyle a_{n}}
realioji dalis yra
cos
ϕ
1
=
−
0.941115117
,
ϕ
1
=
arccos
(
−
0.941115117
)
=
2.79670995
{\displaystyle \cos \phi _{1}=-0.941115117,\;\;\phi _{1}=\arccos(-0.941115117)=2.79670995}
ir iš to galima pasakyti, kad vektorius
a
n
{\displaystyle a_{n}}
yra arba antrame arba trečiame ketvirtyje. Bet menamoji vektoriaus
a
n
{\displaystyle a_{n}}
dalis yra teigiama, todėl vektorius
a
n
{\displaystyle a_{n}}
guli antrame ketvirtyje.
sin
ϕ
1
=
0.33808635
,
ϕ
1
=
arcsin
(
0.33808635
)
=
0.344882767.
{\displaystyle \sin \phi _{1}=0.33808635,\;\;\phi _{1}=\arcsin(0.33808635)=0.344882767.}
2.79670995
π
⋅
180
=
160.239677
{\displaystyle {\frac {2.79670995}{\pi }}\cdot 180=160.239677}
laipsnių.
0.344882767
π
⋅
180
=
19.760326976
{\displaystyle {\frac {0.344882767}{\pi }}\cdot 180=19.760326976}
laipsnių.
Vektorius
a
n
=
−
0.941115117
+
0.33808635
i
{\displaystyle a_{n}=-0.941115117+0.33808635i}
yra antrame ketvirtyje, todėl
ϕ
1
=
160.239677
{\displaystyle \phi _{1}=160.239677}
laipsnių.
Vektorius
b
n
=
−
0.941115117
−
0.33808635
i
{\displaystyle b_{n}=-0.941115117-0.33808635i}
yra trečiame ketvirtyje, nes realioji dalis ir menamoji dalis yra neigiamos. Trečias ketvirtis yra nuo 180 laipsnių iki 270 laipsnių, todėl
ϕ
2
{\displaystyle \phi _{2}}
negali būti
160.239677
{\displaystyle 160.239677}
laipsnių. Tačiau
sin
ϕ
2
=
−
0.33808635
,
ϕ
2
=
arcsin
(
−
0.33808635
)
=
−
0.344882767.
{\displaystyle \sin \phi _{2}=-0.33808635,\;\;\phi _{2}=\arcsin(-0.33808635)=-0.344882767.}
−
0.344882767
π
⋅
180
=
−
19.760326976
{\displaystyle {\frac {-0.344882767}{\pi }}\cdot 180=-19.760326976}
laipsnių (arba 360-19.760326976=340.239673024 laipsnių). Kadangi vektorius
b
n
{\displaystyle b_{n}}
guli trečiame ketvirtyje, tai
ϕ
2
=
180
+
19.760326976
=
199.760326976
{\displaystyle \phi _{2}=180+19.760326976=199.760326976}
laipsnių (arba
ϕ
2
=
π
+
0.344882767
=
3.48647542
{\displaystyle \phi _{2}=\pi +0.344882767=3.48647542}
radianų).
Pažymėkime
α
=
a
3
=
−
q
2
+
q
2
4
+
p
3
27
3
=
r
a
n
3
=
15.938539
3
−
0.941115117
+
0.33808635
i
3
;
{\displaystyle \alpha ={\sqrt[{3}]{a}}={\sqrt[{3}]{-{\frac {q}{2}}+{\sqrt {{\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}}}}}={\sqrt[{3}]{ra_{n}}}={\sqrt[{3}]{15.938539}}{\sqrt[{3}]{-0.941115117+0.33808635i}};}
β
=
b
3
=
−
q
2
−
q
2
4
+
p
3
27
3
=
r
b
n
3
=
15.938539
3
−
0.941115117
−
0.33808635
i
3
.
{\displaystyle \beta ={\sqrt[{3}]{b}}={\sqrt[{3}]{-{\frac {q}{2}}-{\sqrt {{\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}}}}}={\sqrt[{3}]{rb_{n}}}={\sqrt[{3}]{15.938539}}{\sqrt[{3}]{-0.941115117-0.33808635i}}.}
Tada žinodami, kad
k
=
3
{\displaystyle k=3}
(0, 1, 2) ir
n
=
3
{\displaystyle n=3}
gauname
α
1
=
r
n
(
cos
ϕ
+
2
k
π
n
+
i
sin
ϕ
+
2
k
π
n
)
=
15.938539
3
(
cos
ϕ
1
+
2
⋅
0
π
3
+
i
sin
ϕ
1
+
2
⋅
0
π
3
)
=
2.516611459
(
cos
ϕ
1
3
+
i
sin
ϕ
1
3
)
=
{\displaystyle \alpha _{1}={\sqrt[{n}]{r}}(\cos {\frac {\phi +2k\pi }{n}}+i\sin {\frac {\phi +2k\pi }{n}})={\sqrt[{3}]{15.938539}}(\cos {\frac {\phi _{1}+2\cdot 0\pi }{3}}+i\sin {\frac {\phi _{1}+2\cdot 0\pi }{3}})=2.516611459(\cos {\frac {\phi _{1}}{3}}+i\sin {\frac {\phi _{1}}{3}})=}
=
2.516611459
(
cos
160.239677
3
+
i
sin
160.239677
3
)
=
2.516611459
(
cos
(
53.41322567
)
+
i
sin
(
53.41322567
)
)
=
2.516611459
(
0.5960395
+
i
0.80295508
)
=
1.4999998357
+
2.02072596
i
;
{\displaystyle =2.516611459(\cos {\frac {160.239677}{3}}+i\sin {\frac {160.239677}{3}})=2.516611459(\cos(53.41322567)+i\sin(53.41322567))=2.516611459(0.5960395+i0.80295508)=1.4999998357+2.02072596i;}
α
2
=
15.938539
3
(
cos
ϕ
1
+
2
⋅
1
π
3
+
i
sin
ϕ
1
+
2
⋅
1
π
3
)
=
2.516611459
(
cos
ϕ
1
+
2
π
3
+
i
sin
ϕ
1
+
2
π
3
)
=
2.516611459
(
cos
160.239677
+
360
3
+
i
sin
160.239677
+
360
3
)
=
{\displaystyle \alpha _{2}={\sqrt[{3}]{15.938539}}(\cos {\frac {\phi _{1}+2\cdot 1\pi }{3}}+i\sin {\frac {\phi _{1}+2\cdot 1\pi }{3}})=2.516611459(\cos {\frac {\phi _{1}+2\pi }{3}}+i\sin {\frac {\phi _{1}+2\pi }{3}})=2.516611459(\cos {\frac {160.239677+360}{3}}+i\sin {\frac {160.239677+360}{3}})=}
=
2.516611459
(
cos
520.239677
3
+
i
sin
520.239677
3
)
=
2.516611459
(
cos
(
173.41322567
)
+
i
sin
(
173.41322567
)
)
=
2.516611459
(
−
0.99339927
+
0.114707846
i
)
=
−
2.499999986
+
0.288675079
i
;
{\displaystyle =2.516611459(\cos {\frac {520.239677}{3}}+i\sin {\frac {520.239677}{3}})=2.516611459(\cos(173.41322567)+i\sin(173.41322567))=2.516611459(-0.99339927+0.114707846i)=-2.499999986+0.288675079i;}
α
3
=
15.938539
3
(
cos
ϕ
1
+
2
⋅
2
π
3
+
i
sin
ϕ
1
+
2
⋅
2
π
3
)
=
2.516611459
(
cos
ϕ
1
+
4
π
3
+
i
sin
ϕ
1
+
4
π
3
)
=
2.516611459
(
cos
160.239677
+
720
3
+
i
sin
160.239677
+
720
3
)
=
{\displaystyle \alpha _{3}={\sqrt[{3}]{15.938539}}(\cos {\frac {\phi _{1}+2\cdot 2\pi }{3}}+i\sin {\frac {\phi _{1}+2\cdot 2\pi }{3}})=2.516611459(\cos {\frac {\phi _{1}+4\pi }{3}}+i\sin {\frac {\phi _{1}+4\pi }{3}})=2.516611459(\cos {\frac {160.239677+720}{3}}+i\sin {\frac {160.239677+720}{3}})=}
=
2.516611459
(
cos
880.239677
3
+
i
sin
880.239677
3
)
=
2.516611459
(
cos
(
293.41322567
)
+
i
sin
(
293.41322567
)
)
=
2.516611459
(
0.397359727
−
0.917662927
i
)
=
1.0000000423
−
2.309401038
i
.
{\displaystyle =2.516611459(\cos {\frac {880.239677}{3}}+i\sin {\frac {880.239677}{3}})=2.516611459(\cos(293.41322567)+i\sin(293.41322567))=2.516611459(0.397359727-0.917662927i)=1.0000000423-2.309401038i.}
Radome trys a šaknis. Analogiškai randame trys b šaknis:
β
1
=
r
n
(
cos
ϕ
+
2
k
π
n
+
i
sin
ϕ
+
2
k
π
n
)
=
15.938539
3
(
cos
ϕ
2
+
2
⋅
0
π
3
+
i
sin
ϕ
2
+
2
⋅
0
π
3
)
=
2.516611459
(
cos
ϕ
2
3
+
i
sin
ϕ
2
3
)
=
{\displaystyle \beta _{1}={\sqrt[{n}]{r}}(\cos {\frac {\phi +2k\pi }{n}}+i\sin {\frac {\phi +2k\pi }{n}})={\sqrt[{3}]{15.938539}}(\cos {\frac {\phi _{2}+2\cdot 0\pi }{3}}+i\sin {\frac {\phi _{2}+2\cdot 0\pi }{3}})=2.516611459(\cos {\frac {\phi _{2}}{3}}+i\sin {\frac {\phi _{2}}{3}})=}
=
2.516611459
(
cos
199.760326976
3
+
i
sin
199.760326976
3
)
=
2.516611459
(
cos
(
66.5867756587
)
+
i
sin
(
66.5867756587
)
)
=
2.516611459
(
0.3973597
+
i
0.917662936
)
=
0.99999997
+
2.30940106
i
;
{\displaystyle =2.516611459(\cos {\frac {199.760326976}{3}}+i\sin {\frac {199.760326976}{3}})=2.516611459(\cos(66.5867756587)+i\sin(66.5867756587))=2.516611459(0.3973597+i0.917662936)=0.99999997+2.30940106i;}
β
2
=
15.938539
3
(
cos
ϕ
2
+
2
⋅
1
π
3
+
i
sin
ϕ
2
+
2
⋅
1
π
3
)
=
2.516611459
(
cos
ϕ
2
+
2
π
3
+
i
sin
ϕ
2
+
2
π
3
)
=
2.516611459
(
cos
199.760326976
+
360
3
+
i
sin
199.760326976
+
360
3
)
=
{\displaystyle \beta _{2}={\sqrt[{3}]{15.938539}}(\cos {\frac {\phi _{2}+2\cdot 1\pi }{3}}+i\sin {\frac {\phi _{2}+2\cdot 1\pi }{3}})=2.516611459(\cos {\frac {\phi _{2}+2\pi }{3}}+i\sin {\frac {\phi _{2}+2\pi }{3}})=2.516611459(\cos {\frac {199.760326976+360}{3}}+i\sin {\frac {199.760326976+360}{3}})=}
=
2.516611459
(
cos
559.760326976
3
+
i
sin
559.760326976
3
)
=
2.516611459
(
cos
(
186.58677566
)
+
i
sin
(
186.58677566
)
)
=
2.516611459
(
−
0.9933992676
−
i
0.1147078688
)
=
−
2.49999998
−
0.2886751369
i
;
{\displaystyle =2.516611459(\cos {\frac {559.760326976}{3}}+i\sin {\frac {559.760326976}{3}})=2.516611459(\cos(186.58677566)+i\sin(186.58677566))=2.516611459(-0.9933992676-i0.1147078688)=-2.49999998-0.2886751369i;}
β
3
=
15.938539
3
(
cos
ϕ
2
+
2
⋅
2
π
3
+
i
sin
ϕ
2
+
2
⋅
2
π
3
)
=
2.516611459
(
cos
ϕ
2
+
4
π
3
+
i
sin
ϕ
2
+
4
π
3
)
=
2.516611459
(
cos
199.760326976
+
720
3
+
i
sin
199.760326976
+
720
3
)
=
{\displaystyle \beta _{3}={\sqrt[{3}]{15.938539}}(\cos {\frac {\phi _{2}+2\cdot 2\pi }{3}}+i\sin {\frac {\phi _{2}+2\cdot 2\pi }{3}})=2.516611459(\cos {\frac {\phi _{2}+4\pi }{3}}+i\sin {\frac {\phi _{2}+4\pi }{3}})=2.516611459(\cos {\frac {199.760326976+720}{3}}+i\sin {\frac {199.760326976+720}{3}})=}
=
2.516611459
(
cos
919.760326976
3
+
i
sin
919.760326976
3
)
=
2.516611459
(
cos
(
306.58677566
)
+
i
sin
(
306.58677566
)
)
=
2.516611459
(
0.59603956
−
i
0.80295507
)
=
1.499999987
−
2.02072592
i
.
{\displaystyle =2.516611459(\cos {\frac {919.760326976}{3}}+i\sin {\frac {919.760326976}{3}})=2.516611459(\cos(306.58677566)+i\sin(306.58677566))=2.516611459(0.59603956-i0.80295507)=1.499999987-2.02072592i.}
Kubinės lygties
x
3
−
19
x
+
30
=
0
{\displaystyle x^{3}-19x+30=0}
sprendinys turėtų būti
x
0
=
α
+
β
{\displaystyle x_{0}=\alpha +\beta }
, bet kadangi
q
2
4
+
p
3
27
=
30
2
4
+
(
−
19
)
3
27
=
900
4
+
−
6859
27
=
225
−
254.037037
=
−
29.037037037
<
0
,
{\displaystyle {\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}={\frac {30^{2}}{4}}+{\frac {(-19)^{3}}{27}}={\frac {900}{4}}+{\frac {-6859}{27}}=225-254.037037=-29.037037037<0,}
tai mes turime sudėti rastas
α
m
{\displaystyle \alpha _{m}}
ir
β
m
{\displaystyle \beta _{m}}
reikšmes taip, kad menamosios dalys pasinaikintų ir liktų tik 3 realieji sprendiniai. Tokiu budu gauname trys realiuosius lygties
x
3
−
19
x
+
30
=
0
{\displaystyle x^{3}-19x+30=0}
sprendinius:
x
1
=
α
1
+
β
3
=
3
;
{\displaystyle x_{1}=\alpha _{1}+\beta _{3}=3;}
x
2
=
α
2
+
β
2
=
−
5
;
{\displaystyle x_{2}=\alpha _{2}+\beta _{2}=-5;}
x
3
=
α
3
+
β
1
=
2.
{\displaystyle x_{3}=\alpha _{3}+\beta _{1}=2.}
Arba tiesiog realiąsias dalis padauginti iš 2, o menamąsias dalis ignoruoti. Nes, pavyzdžiui,
α
1
{\displaystyle \alpha _{1}}
ir
β
3
{\displaystyle \beta _{3}}
yra jungtiniai kompleksiniai skaičiai, kaip ir kitos dvi poros.
Daugiau apie kubinės lygties sprendimą žiūrėti čia https://lt.wikibooks.org/wiki/Diskriminantas#Kubinės_lygties_sprendimas_Kordano_metodu
Jei bet kuris kompleksinis skaičius, pakeltas n-tu laipsniu, lygus 1, tai jis yra to laipsnio vieneto šaknis.
ϵ
k
⋅
ϵ
j
=
(
cos
ϕ
+
2
k
π
n
+
i
sin
ϕ
+
2
k
π
n
)
⋅
(
cos
ϕ
+
2
j
π
n
+
i
sin
ϕ
+
2
j
π
n
)
=
cos
2
ϕ
+
2
(
k
+
j
)
π
n
+
i
sin
2
ϕ
+
2
(
k
+
j
)
π
n
.
{\displaystyle \epsilon _{k}\cdot \epsilon _{j}=(\cos {\frac {\phi +2k\pi }{n}}+i\sin {\frac {\phi +2k\pi }{n}})\cdot (\cos {\frac {\phi +2j\pi }{n}}+i\sin {\frac {\phi +2j\pi }{n}})=\cos {\frac {2\phi +2(k+j)\pi }{n}}+i\sin {\frac {2\phi +2(k+j)\pi }{n}}.}
Pavyzdis . Rasime šaknis ištrauktas iš
1
3
.
{\displaystyle {\sqrt[{3}]{1}}.}
ϵ
k
=
1
3
=
cos
2
k
π
3
+
i
sin
2
k
π
3
,
(
k
=
0
,
1
,
2
)
.
{\displaystyle \epsilon _{k}={\sqrt[{3}]{1}}=\cos {\frac {2k\pi }{3}}+i\sin {\frac {2k\pi }{3}},\quad (k=0,\;1,\;2).}
ϵ
0
=
1
3
=
cos
2
⋅
0
⋅
π
3
+
i
sin
2
⋅
0
⋅
π
3
=
cos
0
+
i
sin
0
=
1.
{\displaystyle \epsilon _{0}={\sqrt[{3}]{1}}=\cos {\frac {2\cdot 0\cdot \pi }{3}}+i\sin {\frac {2\cdot 0\cdot \pi }{3}}=\cos 0+i\sin 0=1.}
ϵ
1
=
1
3
=
cos
2
⋅
1
⋅
π
3
+
i
sin
2
⋅
1
⋅
π
3
=
cos
(
2.094395102
)
+
i
sin
(
2.094395102
)
=
−
1
2
+
i
3
2
.
{\displaystyle \epsilon _{1}={\sqrt[{3}]{1}}=\cos {\frac {2\cdot 1\cdot \pi }{3}}+i\sin {\frac {2\cdot 1\cdot \pi }{3}}=\cos(2.094395102)+i\sin(2.094395102)=-{\frac {1}{2}}+i{\frac {\sqrt {3}}{2}}.}
ϵ
2
=
1
3
=
cos
2
⋅
2
⋅
π
3
+
i
sin
2
⋅
2
⋅
π
3
=
cos
4
π
3
+
i
sin
4
π
3
=
cos
(
4.188790205
)
+
i
sin
(
4.188790205
)
=
−
1
2
−
i
3
2
.
{\displaystyle \epsilon _{2}={\sqrt[{3}]{1}}=\cos {\frac {2\cdot 2\cdot \pi }{3}}+i\sin {\frac {2\cdot 2\cdot \pi }{3}}=\cos {\frac {4\pi }{3}}+i\sin {\frac {4\pi }{3}}=\cos(4.188790205)+i\sin(4.188790205)=-{\frac {1}{2}}-i{\frac {\sqrt {3}}{2}}.}
Šaknies traukimas iš kompleksinio skaičiaus
keisti
a
1
+
a
2
i
=
±
(
a
1
+
a
1
2
+
a
2
2
2
+
i
−
a
1
+
a
1
2
+
a
2
2
2
)
,
k
a
i
a
2
>
0.
{\displaystyle {\sqrt {a_{1}+a_{2}i}}=\pm ({\sqrt {\frac {a_{1}+{\sqrt {a_{1}^{2}+a_{2}^{2}}}}{2}}}+i{\sqrt {\frac {-a_{1}+{\sqrt {a_{1}^{2}+a_{2}^{2}}}}{2}}}),\quad kai\quad a_{2}>0.}
a
1
+
a
2
i
=
±
(
a
1
+
a
1
2
+
a
2
2
2
−
i
−
a
1
+
a
1
2
+
a
2
2
2
)
,
k
a
i
a
2
<
0.
{\displaystyle {\sqrt {a_{1}+a_{2}i}}=\pm ({\sqrt {\frac {a_{1}+{\sqrt {a_{1}^{2}+a_{2}^{2}}}}{2}}}-i{\sqrt {\frac {-a_{1}+{\sqrt {a_{1}^{2}+a_{2}^{2}}}}{2}}}),\quad kai\;\;\;\;a_{2}<0.}
a
1
+
0
i
=
±
a
1
+
a
1
2
+
a
2
2
2
,
j
e
i
a
1
>
0.
{\displaystyle {\sqrt {a_{1}+0i}}=\pm {\sqrt {\frac {a_{1}+{\sqrt {a_{1}^{2}+a_{2}^{2}}}}{2}}},\quad jei\quad a_{1}>0.}
a
1
+
0
i
=
±
i
−
a
1
+
a
1
2
+
a
2
2
2
,
j
e
i
a
1
<
0.
{\displaystyle {\sqrt {a_{1}+0i}}=\pm i{\sqrt {\frac {-a_{1}+{\sqrt {a_{1}^{2}+a_{2}^{2}}}}{2}}},\quad jei\quad a_{1}<0.}
0
+
i
0
=
0.
{\displaystyle {\sqrt {0+i0}}=0.}
3
+
4
i
=
±
(
3
+
3
2
+
4
2
2
+
i
−
3
+
3
2
+
4
2
2
)
=
±
(
3
+
25
2
+
i
−
3
+
25
2
)
=
±
(
8
2
+
i
2
2
)
=
±
(
4
+
i
1
)
=
±
(
2
+
i
)
.
{\displaystyle {\sqrt {3+4i}}=\pm ({\sqrt {\frac {3+{\sqrt {3^{2}+4^{2}}}}{2}}}+i{\sqrt {\frac {-3+{\sqrt {3^{2}+4^{2}}}}{2}}})=\pm ({\sqrt {\frac {3+{\sqrt {25}}}{2}}}+i{\sqrt {\frac {-3+{\sqrt {25}}}{2}}})=\pm ({\sqrt {\frac {8}{2}}}+i{\sqrt {\frac {2}{2}}})=\pm ({\sqrt {4}}+i{\sqrt {1}})=\pm (2+i).}
Patikriname:
(
2
+
i
)
2
=
4
−
1
+
4
i
=
3
+
4
i
,
{\displaystyle (2+i)^{2}=4-1+4i=3+4i,}
(
−
2
−
i
)
2
=
(
−
2
)
2
+
2
(
−
2
)
(
−
i
)
+
(
−
i
)
2
=
4
+
4
i
−
1
=
3
+
4
i
.
{\displaystyle (-2-i)^{2}=(-2)^{2}+2(-2)(-i)+(-i)^{2}=4+4i-1=3+4i.}
3
−
4
i
=
±
(
3
+
3
2
+
4
2
2
−
i
−
3
+
3
2
+
4
2
2
)
=
±
(
3
+
25
2
−
i
−
3
+
25
2
)
=
±
(
8
2
−
i
2
2
)
=
±
(
4
−
i
1
)
=
±
(
2
−
i
)
.
{\displaystyle {\sqrt {3-4i}}=\pm ({\sqrt {\frac {3+{\sqrt {3^{2}+4^{2}}}}{2}}}-i{\sqrt {\frac {-3+{\sqrt {3^{2}+4^{2}}}}{2}}})=\pm ({\sqrt {\frac {3+{\sqrt {25}}}{2}}}-i{\sqrt {\frac {-3+{\sqrt {25}}}{2}}})=\pm ({\sqrt {\frac {8}{2}}}-i{\sqrt {\frac {2}{2}}})=\pm ({\sqrt {4}}-i{\sqrt {1}})=\pm (2-i).}
(
−
2
+
i
)
2
=
4
−
1
−
4
i
=
3
−
4
i
.
{\displaystyle (-2+i)^{2}=4-1-4i=3-4i.}
−
25
=
i
−
(
−
25
)
+
(
−
25
)
2
+
0
2
2
=
i
25
+
625
2
=
i
25
+
25
2
=
i
50
2
=
i
25
=
±
5
i
.
{\displaystyle {\sqrt {-25}}=i{\sqrt {\frac {-(-25)+{\sqrt {(-25)^{2}+0^{2}}}}{2}}}=i{\sqrt {\frac {25+{\sqrt {625}}}{2}}}=i{\sqrt {\frac {25+25}{2}}}=i{\sqrt {\frac {50}{2}}}=i{\sqrt {25}}=\pm 5i.}
−
25
=
i
25
+
0
=
±
5
i
.
{\displaystyle {\sqrt {-25}}=i{\sqrt {25+0}}=\pm 5i.}
21
−
20
i
=
±
(
1
2
⋅
(
21
+
21
2
+
(
−
20
)
2
)
−
i
1
2
⋅
(
−
21
+
21
2
+
(
−
20
)
2
)
)
=
±
(
1
2
⋅
(
21
+
441
+
400
)
−
i
1
2
⋅
(
−
21
+
441
+
400
)
)
=
{\displaystyle {\sqrt {21-20i}}=\pm \left({\sqrt {{\frac {1}{2}}\cdot (21+{\sqrt {21^{2}+(-20)^{2}}})}}-i{\sqrt {{\frac {1}{2}}\cdot (-21+{\sqrt {21^{2}+(-20)^{2}}})}}\right)=\pm \left({\sqrt {{\frac {1}{2}}\cdot (21+{\sqrt {441+400}})}}-i{\sqrt {{\frac {1}{2}}\cdot (-21+{\sqrt {441+400}})}}\right)=}
=
±
(
1
2
⋅
(
21
+
841
)
−
i
1
2
⋅
(
−
21
+
841
)
)
=
±
(
1
2
⋅
(
21
+
29
)
−
i
1
2
⋅
(
−
21
+
29
)
)
=
±
(
1
2
⋅
50
−
i
1
2
⋅
8
)
=
±
(
25
−
i
4
)
=
±
(
5
−
i
2
)
.
{\displaystyle =\pm \left({\sqrt {{\frac {1}{2}}\cdot (21+{\sqrt {841}})}}-i{\sqrt {{\frac {1}{2}}\cdot (-21+{\sqrt {841}})}}\right)=\pm \left({\sqrt {{\frac {1}{2}}\cdot (21+29)}}-i{\sqrt {{\frac {1}{2}}\cdot (-21+29)}}\right)=\pm \left({\sqrt {{\frac {1}{2}}\cdot 50}}-i{\sqrt {{\frac {1}{2}}\cdot 8}}\right)=\pm ({\sqrt {25}}-i{\sqrt {4}})=\pm (5-i2).}
Turime kompleksinio skaičiaus koordinates
0.8
+
0.6
i
{\displaystyle 0.8+0.6i}
. Šis taškas yra ant apskritimo (kurio spindulys r=1) linijos. Žinome, kad
cos
ϕ
1
=
0.8
{\displaystyle \cos \phi _{1}=0.8}
it
sin
ϕ
1
=
0.6
{\displaystyle \sin \phi _{1}=0.6}
. Tada
ϕ
1
=
arccos
(
0.8
)
=
0.643501108
{\displaystyle \phi _{1}=\arccos(0.8)=0.643501108}
radiano arba 36.86989765 laipsnių,
ϕ
1
=
arcsin
(
0.6
)
=
0.643501108
{\displaystyle \phi _{1}=\arcsin(0.6)=0.643501108}
radiano arba 36.86989765 laipsnių.
Rasime šaknį kompleksinio skaičiaus
0.8
+
0.6
i
{\displaystyle 0.8+0.6i}
. Taigi:
0.8
+
0.6
i
=
±
(
0.8
+
0.8
2
+
0.6
2
2
+
i
−
0.8
+
0.8
2
+
0.6
2
2
)
=
±
(
0.8
+
0.64
+
0.36
2
+
i
−
0.8
+
0.64
+
0.36
2
)
=
{\displaystyle {\sqrt {0.8+0.6i}}=\pm ({\sqrt {\frac {0.8+{\sqrt {0.8^{2}+0.6^{2}}}}{2}}}+i{\sqrt {\frac {-0.8+{\sqrt {0.8^{2}+0.6^{2}}}}{2}}})=\pm ({\sqrt {\frac {0.8+{\sqrt {0.64+0.36}}}{2}}}+i{\sqrt {\frac {-0.8+{\sqrt {0.64+0.36}}}{2}}})=}
=
±
(
0.8
+
1
2
+
i
−
0.8
+
1
2
)
=
±
(
1.8
2
+
i
0.2
2
)
=
±
(
0.9
+
i
0.1
)
=
±
(
0.948683298
+
0.316227766
i
)
.
{\displaystyle =\pm ({\sqrt {\frac {0.8+1}{2}}}+i{\sqrt {\frac {-0.8+1}{2}}})=\pm ({\sqrt {\frac {1.8}{2}}}+i{\sqrt {\frac {0.2}{2}}})=\pm ({\sqrt {0.9}}+i{\sqrt {0.1}})=\pm (0.948683298+0.316227766i).}
Patikriname kokį kampą su ašmi Ox sudaro gautos koordinatės esančios ant apskritimo lanko kurio spindulys r=1.
ϕ
2
=
arccos
(
0.948683298
)
=
0.321750554
{\displaystyle \phi _{2}=\arccos(0.948683298)=0.321750554}
radiano arba 18.43494883 laipsnio,
ϕ
2
=
arcsin
(
0.316227766
)
=
0.321750554
{\displaystyle \phi _{2}=\arcsin(0.316227766)=0.321750554}
radiano arba 18.43494882 laipsnio.
Naujai gautas kampas yra pusė pradinio kampo, nes
18
,
43494882
+
18
,
43494882
=
36
,
86989765
{\displaystyle 18,43494882+18,43494882=36,86989765}
laipsnio,
2
⋅
0.321750554
=
0.643501108.
{\displaystyle 2\cdot 0.321750554=0.643501108.}
Šaknies traukimo iš kompleksinio skaičiaus įrodymas
keisti
Dabar įrodysime, kad kvadratinė šaknis iš kompleksinio skaičiaus visada yra kompleksinis skaičius.
Tegu iš kompleksinio skaičiaus
a
1
+
a
2
i
{\displaystyle a_{1}+a_{2}i}
turime ištraukti kvadratinę šaknį. Jei ta šaknis yra kompleksinis skaičius, tai, pažymėję jį
u
1
+
u
2
i
{\displaystyle u_{1}+u_{2}i}
, turėsime:
a
1
+
i
a
2
=
u
1
+
u
2
i
,
{\displaystyle {\sqrt {a_{1}+ia_{2}}}=u_{1}+u_{2}i,}
a
1
+
a
2
i
=
(
u
1
+
u
2
i
)
2
,
{\displaystyle a_{1}+a_{2}i=(u_{1}+u_{2}i)^{2},}
a
1
+
a
2
i
=
u
1
2
−
u
2
2
+
2
u
1
u
2
i
.
{\displaystyle a_{1}+a_{2}i=u_{1}^{2}-u_{2}^{2}+2u_{1}u_{2}i.}
Iš čia gauname lygčių sistemą:
{
u
1
2
−
u
2
2
=
a
1
,
{\displaystyle u_{1}^{2}-u_{2}^{2}=a_{1},}
{
2
u
1
u
2
=
a
2
.
{\displaystyle 2u_{1}u_{2}=a_{2}.}
Abi puses pakeliame kvadratu:
{
u
1
4
−
2
u
1
2
u
2
2
+
u
2
4
=
a
1
2
,
{\displaystyle u_{1}^{4}-2u_{1}^{2}u_{2}^{2}+u_{2}^{4}=a_{1}^{2},}
{
4
u
1
2
u
2
2
=
a
2
2
.
{\displaystyle 4u_{1}^{2}u_{2}^{2}=a_{2}^{2}.}
Sudedame dvi lygtis ir gauname:
u
1
4
−
2
u
1
2
u
2
2
+
u
2
4
+
4
u
1
2
u
2
2
=
a
1
2
+
a
2
2
,
{\displaystyle u_{1}^{4}-2u_{1}^{2}u_{2}^{2}+u_{2}^{4}+4u_{1}^{2}u_{2}^{2}=a_{1}^{2}+a_{2}^{2},}
u
1
4
+
2
u
1
2
u
2
2
+
u
2
4
=
a
1
2
+
a
2
2
,
{\displaystyle u_{1}^{4}+2u_{1}^{2}u_{2}^{2}+u_{2}^{4}=a_{1}^{2}+a_{2}^{2},}
(
u
1
2
+
u
2
2
)
2
=
a
1
2
+
a
2
2
,
{\displaystyle (u_{1}^{2}+u_{2}^{2})^{2}=a_{1}^{2}+a_{2}^{2},}
u
1
2
+
u
2
2
=
a
1
2
+
a
2
2
.
{\displaystyle u_{1}^{2}+u_{2}^{2}={\sqrt {a_{1}^{2}+a_{2}^{2}}}.}
Pridėję prie paskutinės lygties lygtį
u
1
2
−
u
2
2
=
a
1
,
{\displaystyle u_{1}^{2}-u_{2}^{2}=a_{1},}
gauname:
u
1
2
+
u
2
2
+
u
1
2
−
u
2
2
=
a
1
2
+
a
2
2
+
a
1
,
{\displaystyle u_{1}^{2}+u_{2}^{2}+u_{1}^{2}-u_{2}^{2}={\sqrt {a_{1}^{2}+a_{2}^{2}}}+a_{1},}
2
u
1
2
=
a
1
2
+
a
2
2
+
a
1
;
{\displaystyle 2u_{1}^{2}={\sqrt {a_{1}^{2}+a_{2}^{2}}}+a_{1};}
iš lygties
u
1
2
+
u
2
2
=
a
1
2
+
a
2
2
{\displaystyle u_{1}^{2}+u_{2}^{2}={\sqrt {a_{1}^{2}+a_{2}^{2}}}}
atėmę lygtį
u
1
2
−
u
2
2
=
a
1
,
{\displaystyle u_{1}^{2}-u_{2}^{2}=a_{1},}
gauname:
u
1
2
+
u
2
2
−
(
u
1
2
−
u
2
2
)
=
a
1
2
+
a
2
2
−
a
1
,
{\displaystyle u_{1}^{2}+u_{2}^{2}-(u_{1}^{2}-u_{2}^{2})={\sqrt {a_{1}^{2}+a_{2}^{2}}}-a_{1},}
2
u
2
2
=
a
1
2
+
a
2
2
−
a
1
.
{\displaystyle 2u_{2}^{2}={\sqrt {a_{1}^{2}+a_{2}^{2}}}-a_{1}.}
Iš to gauname:
u
1
=
±
a
1
+
a
1
2
+
a
2
2
2
,
u
2
=
±
−
a
1
+
a
1
2
+
a
2
2
2
.
{\displaystyle u_{1}=\pm {\sqrt {\frac {a_{1}+{\sqrt {a_{1}^{2}+a_{2}^{2}}}}{2}}},\quad u_{2}=\pm {\sqrt {\frac {-a_{1}+{\sqrt {a_{1}^{2}+a_{2}^{2}}}}{2}}}.}
Šios gautos
u
1
{\displaystyle u_{1}}
ir
u
2
{\displaystyle u_{2}}
reikšmės turi tenkinti pirmos sistemos antrą lygtį
2
u
1
u
2
=
a
2
.
{\displaystyle 2u_{1}u_{2}=a_{2}.}
Todėl jei
a
2
{\displaystyle a_{2}}
teigiamas skaičius, tai
u
1
{\displaystyle u_{1}}
ir
u
2
{\displaystyle u_{2}}
reikia imti su tais pačiais ženklais, o jeigu
a
2
{\displaystyle a_{2}}
yra neigiamas skaičius, tai
u
1
{\displaystyle u_{1}}
ir
u
2
{\displaystyle u_{2}}
reikia imti su priešingais ženklais.
Kompleksinio skaičiaus argumentas
keisti
Kompleksinio skaičiaus z argumentas
ϕ
{\displaystyle \phi }
nusako kiek laipsnių (arba radianų) kompleksinio skaičiaus taškas (arba vektorius) pasisukęs nuo teigiamos Ox ašies prieš laikrodžio rodyklę ir žymimas
arg
z
=
ϕ
.
{\displaystyle \arg z=\phi .}
Pavyzdžiui, jei
z
=
i
,
{\displaystyle z=i,}
tai
arg
z
=
π
2
.
{\displaystyle \arg z={\frac {\pi }{2}}.}
Jei
z
=
x
+
y
i
,
{\displaystyle z=x+yi,}
tai
arg
z
=
arctan
y
x
,
x
>
0
;
{\displaystyle \arg z=\arctan {\frac {y}{x}},\quad x>0;}
arg
z
=
arctan
y
x
+
π
,
x
<
0
,
y
≥
0
;
{\displaystyle \arg z=\arctan {\frac {y}{x}}+\pi ,\quad x<0,\;\;y\geq 0;}
arg
z
=
arctan
y
x
−
π
,
x
<
0
,
y
<
0.
{\displaystyle \arg z=\arctan {\frac {y}{x}}-\pi ,\quad x<0,\;\;y<0.}
Jei duoti du kompleksiniai skaičiai
z
1
{\displaystyle z_{1}}
ir
z
2
,
{\displaystyle z_{2},}
tai
arg
(
z
1
⋅
z
2
)
=
arg
z
1
+
arg
z
2
,
{\displaystyle \arg(z_{1}\cdot z_{2})=\arg z_{1}+\arg z_{2},}
arg
z
1
z
2
=
arg
z
1
−
arg
z
2
.
{\displaystyle \arg {\frac {z_{1}}{z_{2}}}=\arg z_{1}-\arg z_{2}.}
Duoti du normalizuoti kompleksiniai skaičiai:
z
1
=
−
0.941115117
+
0.33808635
i
;
{\displaystyle z_{1}=-0.941115117+0.33808635i;}
z
2
=
−
0.941115117
−
0.33808635
i
.
{\displaystyle z_{2}=-0.941115117-0.33808635i.}
(normalizuoti reiškia, kad jų ilgiai lygūs
|
z
1
|
=
|
z
2
|
=
1
{\displaystyle |z_{1}|=|z_{2}|=1}
; gal reikėtų sakyti, normuoti , arba vienetinio ilgio ).
Rasime
arg
z
1
{\displaystyle \arg z_{1}}
ir
arg
z
2
.
{\displaystyle \arg z_{2}.}
arg
z
1
=
arctan
0.33808635
−
0.941115117
+
π
=
arctan
(
−
0.35924
)
+
π
=
−
19.7603
+
180
=
160.2397
;
{\displaystyle \arg z_{1}=\arctan {\frac {0.33808635}{-0.941115117}}+\pi =\arctan(-0.35924)+\pi =-19.7603+180=160.2397;}
arg
z
2
=
arctan
−
0.33808635
−
0.941115117
−
π
=
arctan
(
0.35924
)
−
π
=
19.7603
−
180
=
−
160.2397
=
360
−
160.2397
=
199.7603.
{\displaystyle \arg z_{2}=\arctan {\frac {-0.33808635}{-0.941115117}}-\pi =\arctan(0.35924)-\pi =19.7603-180=-160.2397=360-160.2397=199.7603.}
Taigi, gavome
arg
z
1
=
ϕ
1
=
160.2397
{\displaystyle \arg z_{1}=\phi _{1}=160.2397}
laipsnių ir
arg
z
2
=
ϕ
2
=
199.7603
{\displaystyle \arg z_{2}=\phi _{2}=199.7603}
laipsniai.
Apskaičiuoti
z
=
−
16
+
16
3
i
{\displaystyle z=-16+16{\sqrt {3}}i}
argumentą
arg
z
.
{\displaystyle \arg z.}
Sprendimas.
arg
z
=
arctan
y
x
+
π
,
nes
x
<
0
,
y
≥
0
,
{\displaystyle \arg z=\arctan {\frac {y}{x}}+\pi ,\quad {\text{nes}}\quad x<0,\;\;y\geq 0,}
arg
z
=
arctan
16
3
−
16
+
π
=
arctan
(
−
3
)
+
π
=
−
60
∘
+
180
∘
=
120
∘
.
{\displaystyle \arg z=\arctan {\frac {16{\sqrt {3}}}{-16}}+\pi =\arctan(-{\sqrt {3}})+\pi =-60^{\circ }+180^{\circ }=120^{\circ }.}
Jungtiniai kompleksiniai skaičiai
keisti
Kompleksinio skaičiaus
z
=
(
x
,
y
)
=
x
+
i
y
{\displaystyle z=(x,\;y)=x+iy}
jungtinis kompleksinis skaičius yra
z
¯
=
(
x
,
−
y
)
=
x
−
i
y
.
{\displaystyle {\overline {z}}=(x,\;-y)=x-iy.}
Savaime aišku, kompleksinis skaičius lygus nuliui tada ir tik tada, kai jo jungtinis skaičius lygus nuliui .
Pavyzdžiai
Kompleksiniai skaičiai
z
=
2
+
5
i
{\displaystyle z=2+5i}
ir
z
¯
=
2
−
5
i
{\displaystyle {\overline {z}}=2-5i}
yra jungtiniai (vienas kito atžvilgiu).
Kompleksinio skaičiaus
z
=
−
6
−
4
i
{\displaystyle z=-6-4i}
jungtinis kompleksinis skaičius yra
z
¯
=
−
6
+
4
i
.
{\displaystyle {\overline {z}}=-6+4i.}
Suma ir sandauga jungtinių kompleksinių skaičių yra realieji skaičiai. Iš tiesu,
α
+
α
¯
=
(
a
+
b
i
)
+
(
a
−
b
i
)
=
2
a
,
{\displaystyle \alpha +{\overline {\alpha }}=(a+bi)+(a-bi)=2a,}
α
α
¯
=
(
a
+
b
i
)
(
a
−
b
i
)
=
a
2
−
(
b
i
)
2
=
a
2
+
b
2
=
|
α
|
2
.
{\displaystyle \alpha {\overline {\alpha }}=(a+bi)(a-bi)=a^{2}-(bi)^{2}=a^{2}+b^{2}=|\alpha |^{2}.}
Jungtinių kompleksinių skaičių savybės
1)
α
¯
+
β
¯
=
α
+
β
¯
,
{\displaystyle {\overline {\alpha }}+{\overline {\beta }}={\overline {\alpha +\beta }},}
nes
(
a
+
b
i
)
+
(
c
+
d
i
)
=
(
a
+
c
)
+
(
b
+
d
)
i
,
{\displaystyle (a+bi)+(c+di)=(a+c)+(b+d)i,}
(
a
−
b
i
)
+
(
c
−
d
i
)
=
(
a
+
c
)
−
(
b
+
d
)
i
.
{\displaystyle (a-bi)+(c-di)=(a+c)-(b+d)i.}
2)
α
¯
−
β
¯
=
α
−
β
¯
,
{\displaystyle {\overline {\alpha }}-{\overline {\beta }}={\overline {\alpha -\beta }},}
nes
(
a
+
b
i
)
−
(
c
+
d
i
)
=
(
a
−
c
)
+
(
b
−
d
)
i
,
{\displaystyle (a+bi)-(c+di)=(a-c)+(b-d)i,}
(
a
−
b
i
)
−
(
c
−
d
i
)
=
(
a
−
c
)
−
(
b
−
d
)
i
.
{\displaystyle (a-bi)-(c-di)=(a-c)-(b-d)i.}
3)
α
¯
⋅
β
¯
=
α
β
¯
,
{\displaystyle {\overline {\alpha }}\cdot {\overline {\beta }}={\overline {\alpha \beta }},}
nes
(
a
+
b
i
)
(
c
+
d
i
)
=
(
a
c
−
b
d
)
+
(
a
d
+
b
c
)
i
,
{\displaystyle (a+bi)(c+di)=(ac-bd)+(ad+bc)i,}
(
a
−
b
i
)
(
c
−
d
i
)
=
(
a
c
−
b
d
)
−
(
a
d
+
b
c
)
i
.
{\displaystyle (a-bi)(c-di)=(ac-bd)-(ad+bc)i.}
4)
α
¯
β
¯
=
(
α
β
)
¯
,
{\displaystyle {\frac {\overline {\alpha }}{\overline {\beta }}}={\overline {{\Big (}{\frac {\alpha }{\beta }}{\Big )}}},}
nes
a
+
b
i
c
+
d
i
=
(
a
+
b
i
)
(
c
−
d
i
)
(
c
+
d
i
)
(
c
−
d
i
)
=
(
a
c
+
b
d
)
+
(
b
c
−
a
d
)
i
c
2
+
d
2
=
a
c
+
b
d
c
2
+
d
2
+
i
b
c
−
a
d
c
2
+
d
2
,
{\displaystyle {\frac {a+bi}{c+di}}={\frac {(a+bi)(c-di)}{(c+di)(c-di)}}={\frac {(ac+bd)+(bc-ad)i}{c^{2}+d^{2}}}={\frac {ac+bd}{c^{2}+d^{2}}}+i{\frac {bc-ad}{c^{2}+d^{2}}},}
a
−
b
i
c
−
d
i
=
(
a
−
b
i
)
(
c
+
d
i
)
(
c
−
d
i
)
(
c
+
d
i
)
=
(
a
c
+
b
d
)
−
(
b
c
−
a
d
)
i
c
2
+
d
2
=
a
c
+
b
d
c
2
+
d
2
−
i
b
c
−
a
d
c
2
+
d
2
.
{\displaystyle {\frac {a-bi}{c-di}}={\frac {(a-bi)(c+di)}{(c-di)(c+di)}}={\frac {(ac+bd)-(bc-ad)i}{c^{2}+d^{2}}}={\frac {ac+bd}{c^{2}+d^{2}}}-i{\frac {bc-ad}{c^{2}+d^{2}}}.}
Jeigu skaičius
α
{\displaystyle \alpha }
kokiu nors budu išreikštas per kompleksinius skaičius
β
1
,
β
2
,
.
.
.
,
β
n
{\displaystyle \beta _{1},\;\beta _{2},\;...,\;\beta _{n}}
taikant sudėtį, daugybą, atimtį ir dalybą, tai pakeičiant visus skaičius
β
k
{\displaystyle \beta _{k}}
į jiems jungtinius (pakeičiant
β
k
{\displaystyle \beta _{k}}
į
β
k
¯
{\displaystyle {\overline {\beta _{k}}}}
), mes gausime skaičių jungtinį skaičiui
α
{\displaystyle \alpha }
(gausime
α
¯
{\displaystyle {\overline {\alpha }}}
).
Kompleksinio skaičiaus
α
=
a
+
b
i
{\displaystyle \alpha =a+bi}
modulis yra jo (vektoriaus) ilgis ir lygus
|
α
|
=
a
2
+
b
2
.
{\displaystyle |\alpha |={\sqrt {a^{2}+b^{2}}}.}
Dviejų kompleksinių skaičių sandaugos modulis lygus tų kompleksinių skaičių modulių sandaugai. T. y., jei duoti du kompleksiniai skaičiai
α
{\displaystyle \alpha }
ir
β
,
{\displaystyle \beta ,}
tai
|
α
β
|
=
|
α
|
⋅
|
β
|
.
{\displaystyle |\alpha \beta |=|\alpha |\cdot |\beta |.}
Ši lygybė išplaukia iš kompleksinių skaičių daugybos trigonometrinėje formoje. Iš tikro, jei duoti du kompleksiniai skaičiai
z
1
=
r
1
(
cos
θ
1
+
i
sin
θ
1
)
{\displaystyle z_{1}=r_{1}(\cos \theta _{1}+i\sin \theta _{1})}
ir
z
2
=
r
2
(
cos
θ
2
+
i
sin
θ
2
)
,
{\displaystyle z_{2}=r_{2}(\cos \theta _{2}+i\sin \theta _{2}),}
tai
|
z
1
|
=
r
1
,
{\displaystyle |z_{1}|=r_{1},}
|
z
2
|
=
r
2
{\displaystyle |z_{2}|=r_{2}}
ir
|
z
1
z
2
|
=
r
1
r
2
,
{\displaystyle |z_{1}z_{2}|=r_{1}r_{2},}
nes
z
1
z
2
=
r
1
(
cos
θ
1
+
i
sin
θ
1
)
⋅
r
2
(
cos
θ
2
+
i
sin
θ
2
)
=
r
1
r
2
(
cos
(
θ
1
+
θ
2
)
+
i
sin
(
θ
1
+
θ
2
)
)
.
{\displaystyle z_{1}z_{2}=r_{1}(\cos \theta _{1}+i\sin \theta _{1})\cdot r_{2}(\cos \theta _{2}+i\sin \theta _{2})=r_{1}r_{2}(\cos(\theta _{1}+\theta _{2})+i\sin(\theta _{1}+\theta _{2})).}
Iš to, pavzdžiui, seka tokios lygybės:
|
a
z
n
|
=
|
a
|
⋅
|
z
n
|
=
|
a
|
⋅
|
z
|
n
{\displaystyle |az^{n}|=|a|\cdot |z^{n}|=|a|\cdot |z|^{n}\;}
(čia a ir z yra kompleksiniai skaičiai).
1 pav. OB < OA + AB.
Kompleksinių skaičių sumos moduliui yra tokios svarbios nelygybės:
|
α
|
−
|
β
|
≤
|
α
+
β
|
≤
|
α
|
+
|
β
|
,
(
11
)
{\displaystyle |\alpha |-|\beta |\leq |\alpha +\beta |\leq |\alpha |+|\beta |,\quad (11)}
t. y. sumos modulis dviejų kompleksinių skaičių mažesnis arba lygus sumai modulių dėmenų, bet didesnis arba lygus skirtumui šitų modulių . (11) nelygybė išplaukia iš žinomos teoremos elementariosios geometrijos apie trikampio kraštines žinant, kad
|
α
+
β
|
{\displaystyle |\alpha +\beta |}
lygus lygiagretainio įsitrižainei su kraštinėmis
|
α
|
{\displaystyle |\alpha |}
ir
|
β
|
.
{\displaystyle |\beta |.}
Iš 1 paveikslėlio matyti, kad
α
{\displaystyle \alpha }
ir
β
{\displaystyle \beta }
vektorių ilgių suma yra didesnė už vektoriaus
γ
=
α
+
β
{\displaystyle \gamma =\alpha +\beta }
ilgį (t. y.
|
α
+
β
|
≤
|
α
|
+
|
β
|
{\displaystyle |\alpha +\beta |\leq |\alpha |+|\beta |}
arba OB < OA + AB).
(11) nelygybėse lygybės ženklas gali būti gautas tik kai
α
,
{\displaystyle \alpha ,}
β
{\displaystyle \beta }
ir 0 guli ant vienos tiesės.
Iš (11), atsižvelgiant į tai, kad
α
−
β
=
α
+
(
−
β
)
{\displaystyle \alpha -\beta =\alpha +(-\beta )}
ir
|
−
β
|
=
|
β
|
,
(
12
)
{\displaystyle |-\beta |=|\beta |,\quad (12)}
išplaukia taip pat nelygybės
|
α
|
−
|
β
|
≤
|
α
−
β
|
≤
|
α
|
+
|
β
|
,
(
13
)
{\displaystyle |\alpha |-|\beta |\leq |\alpha -\beta |\leq |\alpha |+|\beta |,\quad (13)}
t. y. skirtumo moduliui tinka tokios pat nelygybės kaip ir sumos moduliui.
Nelygybes (11) galima buvo gauti taip pat štai tokiu budu. Tegu
α
=
r
(
cos
φ
+
i
sin
φ
)
,
{\displaystyle \alpha =r(\cos \varphi +i\sin \varphi ),}
β
=
r
′
(
cos
φ
′
+
i
sin
φ
′
)
{\displaystyle \beta =r'(\cos \varphi '+i\sin \varphi ')}
ir tegu trigonometrinė forma skaičiaus
α
+
β
{\displaystyle \alpha +\beta }
yra
α
+
β
=
R
(
cos
ψ
+
i
sin
ψ
)
.
{\displaystyle \alpha +\beta =R(\cos \psi +i\sin \psi ).}
Sudėdami atskirai realiąsias ir atskirai menamąsias dalis, gauname:
r
cos
φ
+
r
′
cos
φ
′
=
R
cos
ψ
,
{\displaystyle r\cos \varphi +r'\cos \varphi '=R\cos \psi ,}
r
sin
φ
+
r
′
sin
φ
′
=
R
sin
ψ
;
{\displaystyle r\sin \varphi +r'\sin \varphi '=R\sin \psi ;}
padaugindami abi dalis pirmos lygybės iš
cos
ψ
,
{\displaystyle \cos \psi ,}
abi dalis antros lygybės — iš
sin
ψ
{\displaystyle \sin \psi }
ir sudedami, gauname:
r
(
cos
φ
cos
ψ
+
sin
φ
sin
ψ
)
+
r
′
(
cos
φ
′
cos
ψ
+
sin
φ
′
sin
ψ
)
=
R
(
cos
2
ψ
+
sin
2
ψ
)
,
{\displaystyle r(\cos \varphi \cos \psi +\sin \varphi \sin \psi )+r'(\cos \varphi '\cos \psi +\sin \varphi '\sin \psi )=R(\cos ^{2}\psi +\sin ^{2}\psi ),}
t. y.
r
cos
(
φ
−
ψ
)
+
r
′
cos
(
φ
′
−
ψ
)
=
R
.
{\displaystyle r\cos(\varphi -\psi )+r'\cos(\varphi '-\psi )=R.}
Iš čia, kadangi kosinusas niekada nebūna daugiau už vientetą, seka nelygybė
r
+
r
′
≥
R
,
{\displaystyle r+r'\geq R,}
t. y.
|
α
|
+
|
β
|
≥
|
α
+
β
|
.
{\displaystyle |\alpha |+|\beta |\geq |\alpha +\beta |.}
Iš kitos pusės,
α
=
(
α
+
β
)
−
β
=
(
α
+
β
)
+
(
−
β
)
.
{\displaystyle \alpha =(\alpha +\beta )-\beta =(\alpha +\beta )+(-\beta ).}
Iš čia, pagal įrodyta ir iš (12),
[
|
(
α
+
β
)
+
(
−
β
)
|
≤
|
(
α
+
β
)
|
+
|
(
−
β
)
|
{\displaystyle |(\alpha +\beta )+(-\beta )|\leq |(\alpha +\beta )|+|(-\beta )|}
]
|
α
|
≤
|
α
+
β
|
+
|
−
β
|
=
|
α
+
β
|
+
|
β
|
,
{\displaystyle |\alpha |\leq |\alpha +\beta |+|-\beta |=|\alpha +\beta |+|\beta |,}
(
β
{\displaystyle \beta }
modulyje
|
α
+
β
|
{\displaystyle |\alpha +\beta |}
mažiau gali sumažint [
|
α
+
β
|
{\displaystyle |\alpha +\beta |}
] nei pridėti pats
β
{\displaystyle \beta }
modulis
|
β
|
{\displaystyle |\beta |}
)
iš kur
|
α
|
−
|
β
|
≤
|
α
+
β
|
.
{\displaystyle |\alpha |-|\beta |\leq |\alpha +\beta |.}
Tenka pastebėti, kad kompleksiniams skaičiams supratimai "daugiau" ir "mažiau" negali būti protingai nustatyti, nes šitie skaičiai, skirtingai nei realieji skaičiai, yra ne ant tiesios linijos, kurios taškai aiškiu budu sutvarkyti, o ant plokštumos. Todėl pačius kompleksinius skaičius (o ne jų modulius) niekada negalima sujunginėti nelygybės ženklais.
Šios nelygybės reikalingos įrodinėjant pagrindinę algebros teoremą . Pagrindinė algebros teorema kompleksinių skaičių yra tokia:
Bet koks polinomas su bet kokiais skaitiniais koeficientais, laipsnis kurio ne mažesnis už vienetą, turi bent vieną šaknį, bendru atveju kompleksinę .