U skupu realnih brojeva
R
{\displaystyle \mathbb {R} }
jednačina
x
2
−
1
=
0
{\displaystyle x^{2}-1=0}
x
2
−
1
=
0
⇒
x
=
±
1
{\displaystyle x^{2}-1=0\Rightarrow x=\pm 1}
Slična jednačina
x
2
+
1
=
0
{\displaystyle x^{2}+1=0}
R
{\displaystyle \mathbb {R} }
i
{\displaystyle i}
i
2
=
−
1
{\displaystyle i^{2}=-1}
x
2
+
1
=
0
⇒
x
=
±
i
{\displaystyle x^{2}+1=0\quad \Rightarrow \quad x=\pm i}
i
2
=
−
1
,
i
3
=
−
i
,
i
4
=
−
i
⋅
i
=
−
(
−
1
)
=
1
,
i
5
=
i
,
i
6
=
−
1
,
…
{\displaystyle \displaystyle i^{2}=-1,\quad i^{3}=-i,\quad i^{4}=-i\cdot i=-(-1)=1,\quad i^{5}=i,\quad i^{6}=-1,\ldots }
[2]
Na ovaj način dobili smo skup kompleksnih brojeva.
Kompleksan broj je broj oblika
x
+
y
i
{\displaystyle x+yi\,}
gdje su x i y realni brojevi , a i se naziva imaginarna jedinica i ima osobinu i2 = -1.
Realni broj
x
{\displaystyle x}
Re
(
z
)
{\displaystyle \operatorname {Re} (z)}
y
{\displaystyle y}
Im
(
z
)
{\displaystyle \operatorname {Im} (z)}
Skup kompleksnih brojeva možemo smatrati proširenjem skupa realnih brojeva, odnosno svaki realni broj x možemo posmatrati kao kompleksni, uzimajući u prethodnoj notaciji da je y = 0, tj.
x
=
x
+
0
i
{\displaystyle x=x+0i}
Povremeno se moze naići na definiciju
i
=
−
1
{\displaystyle i={\sqrt {-1}}}
−
1
⋅
−
1
=
1
=
1
{\displaystyle {\sqrt {-1}}\cdot {\sqrt {-1}}={\sqrt {1}}=1}
i
=
−
1
{\displaystyle i={\sqrt {-1}}}
i
⋅
i
=
−
1
{\displaystyle i\cdot i=-1}
Kompleksni brojevi se mogu formalno definisati kao dvodimenzionalni vektori ili uređeni parovi
(
x
,
y
)
{\displaystyle (x,y)}
Definiciju kompleksnih brojeva kao uređenih parova dao je William R. Hamilton , irski matematičar (1805– 1865.) Ta se definicija temelji samo na osobini realnih brojeva, čime se izbjegava donekle nerazjašnjeni pojam broja
−
1
{\displaystyle {\sqrt {-1}}}
S druge strane, zapis oblika
z
=
x
+
y
i
{\displaystyle z=x+yi}
Oba oblika kompleksnog broja
z
=
x
+
y
i
{\displaystyle z=x+yi}
z
=
(
x
,
y
)
{\displaystyle z=(x,y)}
Skup kompleksnih brojeva
C
{\displaystyle \mathbb {C} }
z
=
x
+
i
y
{\displaystyle z=x+iy}
x
,
y
∈
R
{\displaystyle x,y\in \mathbb {R} }
Posebno je
0
=
0
+
i
0
{\displaystyle 0=0+i0}
x
=
Re
z
{\displaystyle x=\operatorname {Re} z}
z
{\displaystyle z}
y
=
Im
z
{\displaystyle y=\operatorname {Im} z}
z
{\displaystyle z}
Algebarski oblik kompleksnog broja je
z
=
x
+
i
y
{\displaystyle z=x+iy}
x
,
y
∈
R
{\displaystyle x,y\in \mathbb {R} }
Trigonometrijski oblik kompleksnog broja je
z
=
r
(
cos
θ
+
i
sin
θ
)
,
r
≥
0
,
θ
∈
R
{\displaystyle z=r(\cos \theta +i\sin \theta ),r\geq 0,\theta \in \mathbb {R} }
pri čemu je
r
=
|
z
|
{\displaystyle r=|z|}
θ
=
Arg
z
{\displaystyle \theta =\operatorname {Arg} z}
Eksponencijalni oblik kompleksnog broja je
z
=
r
i
θ
{\displaystyle z=r^{i\theta }}
r
≥
0
,
θ
∈
R
{\displaystyle r\geq 0,\theta \in \mathbb {R} }
pri čemu je
r
=
|
z
|
{\displaystyle r=|z|}
θ
=
Arg
z
{\displaystyle \theta =\operatorname {Arg} z}
Dva kompleksna broja su jednaka ako su im jednaki realni i imaginarni dijelovi.
z
1
=
z
2
⇔
(
Re
(
z
1
)
=
Re
(
z
2
)
∧
Im
(
z
1
)
=
Im
(
z
2
)
)
.
{\displaystyle z_{1}=z_{2}\quad \Leftrightarrow \quad (\operatorname {Re} (z_{1})=\operatorname {Re} (z_{2})\,\land \,\operatorname {Im} (z_{1})=\operatorname {Im} (z_{2})).}
Konjugirano kompleksni broj broja
z
=
x
+
i
y
{\displaystyle z=x+iy}
z
¯
=
x
−
i
y
{\displaystyle {\bar {z}}=x-iy}
Modul ili apsolutna vrijednost kompleksnog broja
z
{\displaystyle z}
r
=
|
z
|
=
x
2
+
y
2
{\displaystyle r=\vert z\vert ={\sqrt {x^{2}+y^{2}}}}
malo U skupu kompleksnih brojeva definisano je sabiranje .
Neka su
z
1
=
x
1
+
i
y
1
{\displaystyle z_{1}=x_{1}+iy_{1}}
z
2
=
x
2
+
i
y
2
{\displaystyle z_{2}=x_{2}+iy_{2}}
z
1
+
z
2
=
x
1
+
x
2
+
i
(
y
1
+
y
2
)
{\displaystyle \displaystyle z_{1}+z_{2}\displaystyle =x_{1}+x_{2}+i(y_{1}+y_{2})}
[2]
i oduzimanje
z
1
−
z
2
=
x
1
−
x
2
+
i
(
y
1
−
y
2
)
{\displaystyle \displaystyle z_{1}-z_{2}\displaystyle =x_{1}-x_{2}+i(y_{1}-y_{2})}
Osobine sabiranja kompleksnih brojeva
z
1
+
z
2
=
z
2
+
z
1
{\displaystyle z_{1}+z_{2}=z_{2}+z_{1}}
∀
z
1
,
z
2
∈
C
{\displaystyle \forall z_{1},z_{2}\in \mathbb {C} }
komutativnost sabiranja
z
1
+
(
z
2
+
z
3
)
=
(
z
1
+
z
2
)
+
z
3
,
{\displaystyle z_{1}+(z_{2}+z_{3})=(z_{1}+z_{2})+z_{3},}
∀
z
1
,
z
2
,
z
3
∈
C
{\displaystyle \forall z_{1},z_{2},z_{3}\in \mathbb {C} }
asocijativnost sabiranja
∃
0
∈
C
z
+
0
=
z
{\displaystyle \exists 0\in \mathbb {C} z+0=z}
∀
z
∈
C
{\displaystyle \forall z\in \mathbb {C} }
0(nula) za sabiranje
Kompleksni broj
0
=
(
0
,
0
)
=
0
+
0
i
{\displaystyle 0=(0,0)=0+0i}
(
∀
z
∈
C
)
(
∃
(
−
z
)
∈
C
z
+
(
−
z
)
=
0
{\displaystyle (\forall z\in \mathbb {C} )(\exists (-z)\in \mathbb {C} z+(-z)=0}
Kompleksni broj
−
z
=
(
−
x
,
−
y
)
=
−
x
−
y
i
{\displaystyle -z=(-x,-y)=-x-yi}
[3]
Neka su
z
1
=
x
1
+
i
y
1
{\displaystyle z_{1}=x_{1}+iy_{1}}
z
2
=
x
2
+
i
y
2
{\displaystyle z_{2}=x_{2}+iy_{2}}
U skupu kompleksnih brojeva definisano je množenje
z
1
⋅
z
2
=
(
x
1
+
i
y
1
)
(
x
2
+
i
y
2
)
=
x
1
x
2
+
i
y
1
x
2
+
i
x
1
y
2
+
i
2
y
1
y
2
=
x
1
x
2
−
y
1
y
2
+
i
(
x
1
y
2
+
x
2
y
1
)
{\displaystyle \displaystyle z_{1}\cdot z_{2}\displaystyle =(x_{1}+iy_{1})(x_{2}+iy_{2})=x_{1}x_{2}+iy_{1}x_{2}+ix_{1}y_{2}+i^{2}y_{1}y_{2}\displaystyle =x_{1}x_{2}-y_{1}y_{2}+i(x_{1}y_{2}+x_{2}y_{1})}
z
1
⋅
z
2
=
z
2
⋅
z
1
{\displaystyle z_{1}\cdot z_{2}=z_{2}\cdot z_{1}}
∀
z
1
,
z
2
∈
C
{\displaystyle \forall z_{1},z_{2}\in \mathbb {C} }
z
1
⋅
(
z
2
⋅
z
3
)
=
(
z
1
⋅
z
2
)
⋅
z
3
{\textstyle z_{1}\cdot (z_{2}\cdot z_{3})=(z_{1}\cdot z_{2})\cdot z_{3}}
∀
z
1
,
z
2
,
z
3
∈
C
{\displaystyle \forall z_{1},z_{2},z_{3}\in \mathbb {C} }
∃
1
∈
C
z
⋅
1
=
z
{\displaystyle \exists 1\in \mathbb {C} z\cdot 1=z}
∀
z
∈
C
{\displaystyle \forall z\in \mathbb {C} }
1
{\displaystyle 1}
(
∀
z
∈
C
)
(
z
≠
0
)
(
∃
z
′
∈
C
)
z
⋅
(
−
z
)
=
1
{\displaystyle (\forall z\in \mathbb {C} )(z\neq 0)(\exists z'\in \mathbb {C} )z\cdot (-z)=1}
z
1
⋅
(
z
2
+
z
3
)
=
z
1
⋅
z
2
+
z
1
⋅
z
3
{\displaystyle z_{1}\cdot (z_{2}+z_{3})=z_{1}\cdot z_{2}+z_{1}\cdot z_{3}}
∀
z
1
,
z
2
,
z
3
∈
C
{\displaystyle \forall z_{1},z_{2},z_{3}\in \mathbb {C} }
[3]
Realan proizvod dva kompleksna broja
U skupu kompleksnih brojeva skalarnom proizvodu vektora odgovara pojam realnog proizvoda kompleksnih brojeva koji je skalarni proizvod vektora koji su određeni kompleksnim brojevima koji se množe.
Definicija
Realan proizvod kompleksnih brojeva
a
{\displaystyle a}
b
{\displaystyle b}
a
∘
b
{\displaystyle a\circ b}
a
∘
b
=
1
2
(
a
¯
b
+
a
b
¯
)
{\displaystyle a\circ b={\frac {1}{2}}({\overline {a}}b+a{\overline {b}})}
Neka su A i B tačke određene kompleksnim brojevima
a
=
|
a
|
(
cos
φ
+
i
sin
φ
)
{\displaystyle a=|a|(\cos \varphi +i\sin \varphi )}
b
=
|
b
|
(
cos
ψ
+
i
sin
ψ
)
{\displaystyle b=|b|(\cos \psi +i\sin \psi )}
a
∘
b
=
|
a
|
|
b
|
(
cos
φ
+
i
sin
ψ
)
=
|
O
A
|
|
A
B
|
cos
A
O
B
^
{\displaystyle a\circ b=|a||b|(\cos \varphi +i\sin \psi )=|OA||AB|\cos {\widehat {AOB}}}
Osobine realnog proizvoda dva kompleksna broja
a
∘
a
=
|
a
|
2
{\displaystyle a\circ a=|a|^{2}}
a
∘
b
=
b
∘
a
{\displaystyle a\circ b=b\circ a}
a
∘
b
¯
=
a
∘
b
{\displaystyle {\overline {a\circ b}}=a\circ b}
(
α
a
)
∘
b
=
α
(
a
∘
b
)
=
a
∘
(
α
b
)
{\displaystyle (\alpha a)\circ b=\alpha (a\circ b)=a\circ (\alpha b)}
(
a
z
)
(
b
z
)
=
|
z
|
2
(
a
∘
b
)
{\displaystyle (az)(bz)=|z|^{2}(a\circ b)}
a
∘
b
=
0
⇔
O
A
⊥
O
B
{\displaystyle a\circ b=0\quad \Leftrightarrow \quad OA\perp OB}
a
{\displaystyle a}
b
{\displaystyle b}
Realan proizvod kompleksnih brojeva
a
{\displaystyle a}
b
{\displaystyle b}
O
{\displaystyle O}
A
B
{\displaystyle AB}
A
{\displaystyle A}
B
{\displaystyle B}
a
{\displaystyle a}
b
{\displaystyle b}
Tačka
M
{\displaystyle M}
a
+
b
2
{\displaystyle {\frac {a+b}{2}}}
O
{\displaystyle O}
M
{\displaystyle M}
r
=
a
−
b
2
=
|
a
−
b
|
2
{\displaystyle r={\frac {a-b}{2}}={\frac {|a-b|}{2}}}
O
M
2
−
r
2
=
|
a
+
b
2
|
−
|
a
−
b
2
|
=
(
a
+
b
)
(
a
¯
+
b
¯
)
4
−
(
a
−
b
)
(
a
¯
−
b
¯
)
4
=
a
∘
b
{\displaystyle OM^{2}-r^{2}=\left|{\frac {a+b}{2}}\right|-\left|{\frac {a-b}{2}}\right|={\frac {(a+b)({\overline {a}}+{\overline {b}})}{4}}-{\frac {(a-b)({\overline {a}}-{\overline {b}})}{4}}=a\circ b}
Neka su tačke
A
{\displaystyle A}
B
{\displaystyle B}
C
{\displaystyle C}
D
{\displaystyle D}
a
{\displaystyle a}
b
{\displaystyle b}
c
{\displaystyle c}
d
{\displaystyle d}
A
B
⊥
C
D
{\displaystyle AB\perp CD}
(
a
+
b
)
∘
(
c
+
d
)
=
0
{\displaystyle (a+b)\circ (c+d)=0}
b
−
a
d
−
c
∈
i
R
{
0
}
{\displaystyle {\frac {b-a}{d-c}}\in i\mathbb {R} \ {\begin{Bmatrix}0\end{Bmatrix}}}
Re
(
b
−
a
d
−
c
)
=
0
{\displaystyle \operatorname {Re} \left({\frac {b-a}{d-c}}\right)=0}
Središte kružnice opisane oko trougla
A
B
C
{\displaystyle ABC}
A
{\displaystyle A}
B
{\displaystyle B}
C
{\displaystyle C}
A
B
C
{\displaystyle ABC}
a
{\displaystyle a}
b
{\displaystyle b}
c
{\displaystyle c}
H
{\displaystyle H}
h
=
a
+
b
+
c
{\displaystyle h=a+b+c}
Kompleksan proizvod dva kompleksna broja
Kompleksan proizvod dva kompleksna broja je analogan vektorskom proizvodu vektora.
Definicija Kompleksan broj
a
×
b
=
a
¯
b
−
a
b
¯
2
{\displaystyle a\times b={\frac {{\overline {a}}b-a{\overline {b}}}{2}}}
a
{\displaystyle a}
b
{\displaystyle b}
Neka su
A
{\displaystyle A}
B
{\displaystyle B}
a
=
|
a
|
(
cos
φ
+
i
sin
φ
)
{\displaystyle a=|a|(\cos \varphi +i\sin \varphi )}
a
=
|
b
|
(
cos
ψ
+
i
sin
ψ
)
{\displaystyle a=|b|(\cos \psi +i\sin \psi )}
|
a
×
b
|
=
|
a
|
|
b
|
sin
(
φ
−
ψ
)
=
|
O
A
|
|
A
B
|
sin
A
O
B
^
=
2
P
A
O
B
{\displaystyle |a\times b|=|a||b|\sin(\varphi -\psi )=|OA||AB|\sin {\widehat {AOB}}=2P_{AOB}}
Neka su
a
{\displaystyle a}
b
{\displaystyle b}
c
{\displaystyle c}
a
×
b
¯
=
−
a
×
b
{\displaystyle {\overline {a\times b}}=-a\times b}
a
×
b
=
0
⇔
a
=
0
∨
b
=
0
∨
a
=
λ
b
{\displaystyle a\times b=0\quad \Leftrightarrow \quad a=0\lor b=0\lor a=\lambda b}
λ
∈
R
{
0
}
{\displaystyle \lambda \in \mathbb {R} \ {\begin{Bmatrix}0\end{Bmatrix}}}
a
×
b
=
−
b
×
a
{\displaystyle a\times b=-b\times a}
α
(
a
×
b
)
=
(
α
a
)
×
b
=
a
×
(
α
b
)
{\displaystyle \alpha (a\times b)=(\alpha a)\times b=a\times (\alpha b)}
∀
α
∈
R
{\displaystyle \forall \alpha \in \mathbb {R} }
Ako su
A
(
a
)
{\displaystyle A(a)}
B
(
b
)
{\displaystyle B(b)}
O
(
0
)
{\displaystyle O(0)}
a
×
b
=
0
{\displaystyle a\times b=0}
O
{\displaystyle O}
A
{\displaystyle A}
B
{\displaystyle B}
Neka su
A
(
a
{\displaystyle A(a}
B
(
b
{\displaystyle B(b}
a
{\displaystyle a}
b
{\displaystyle b}
a
×
b
=
{
2
i
P
A
O
B
za trougao
A
O
B
pozitivno orijentisan
−
2
i
P
A
O
B
za trougao
A
O
B
negativno orijentisan
{\displaystyle a\times b={\begin{cases}2iP_{AOB}{\text{ za trougao }}AOB{\text{ pozitivno orijentisan}}\\-2iP_{AOB}{\text{ za trougao }}AOB{\text{ negativno orijentisan}}\end{cases}}}
A
(
a
)
{\displaystyle A(a)}
B
(
b
)
{\displaystyle B(b)}
C
(
c
)
{\displaystyle C(c)}
P
A
B
C
=
{
1
2
(
a
×
b
+
b
×
c
+
c
×
a
)
ako je
A
B
C
pozitivno orjentisan
1
2
(
a
×
b
+
b
×
c
+
c
×
a
)
ako je
A
B
C
negativno orjentisan
{\displaystyle P_{ABC}={\begin{cases}{\frac {1}{2}}(a\times b+b\times c+c\times a){\text{ ako je }}ABC{\text{ pozitivno orjentisan}}\\{\frac {1}{2}}(a\times b+b\times c+c\times a){\text{ ako je }}ABC{\text{ negativno orjentisan}}\end{cases}}}
Neka su
A
(
a
)
{\displaystyle A(a)}
B
(
b
)
{\displaystyle B(b)}
C
(
c
)
{\displaystyle C(c)}
Tačke
A
{\displaystyle A}
B
{\displaystyle B}
C
{\displaystyle C}
(
b
−
a
)
×
(
c
−
a
)
=
0
{\displaystyle (b-a)\times (c-a)=0}
a
×
b
+
b
×
c
+
c
×
a
=
0
{\displaystyle a\times b+b\times c+c\times a=0}
Neka su
A
(
a
)
{\displaystyle A(a)}
B
(
b
)
{\displaystyle B(b)}
C
(
c
)
{\displaystyle C(c)}
D
(
d
)
{\displaystyle D(d)}
A
B
∥
C
D
{\displaystyle AB\parallel CD}
(
b
−
a
)
×
(
d
−
c
)
=
0
{\displaystyle (b-a)\times (d-c)=0}
Dijeljenje kompleksnih brojeva
z
1
z
2
=
x
1
+
i
y
1
x
2
+
i
y
2
⋅
x
2
−
i
y
2
x
2
−
i
y
2
=
x
1
x
2
+
y
1
y
2
x
2
2
+
y
2
2
+
i
y
1
x
2
−
x
1
y
2
x
2
2
+
y
2
2
,
za
z
2
≠
0
{\displaystyle \displaystyle {\frac {z_{1}}{z_{2}}}\displaystyle ={\frac {x_{1}+iy_{1}}{x_{2}+iy_{2}}}\cdot {\frac {x_{2}-iy_{2}}{x_{2}-iy_{2}}}\displaystyle ={\frac {x_{1}x_{2}+y_{1}y_{2}}{x_{2}^{2}+y_{2}^{2}}}+i{\frac {y_{1}x_{2}-x_{1}y_{2}}{x_{2}^{2}+y_{2}^{2}}},\quad {\text{za}}\quad z_{2}\neq 0}
U svakom skupu brojeva dijeljenje se definiše kao množenje inverznim elementom. Uvjerimo se da za
∀
z
∈
C
)
(
z
≠
0
)
∃
z
′
∈
C
{\displaystyle \forall z\in \mathbb {C} )(z\neq 0)\exists z'\in \mathbb {C} }
Neka je
z
=
x
+
y
i
≠
0
{\displaystyle z=x+yi\neq 0}
x
2
+
y
2
≠
0
{\displaystyle x^{2}+y^{2}\neq 0}
z
′
=
x
x
2
+
y
2
+
−
y
x
2
+
y
2
i
{\displaystyle z'={\frac {x}{x^{2}+y^{2}}}+{\frac {-y}{x^{2}+y^{2}}}i}
1
z
=
z
¯
z
z
¯
=
z
¯
x
2
+
y
2
=
x
x
2
+
y
2
−
y
x
2
+
y
2
i
.
{\displaystyle {\frac {1}{z}}={\frac {\bar {z}}{z{\bar {z}}}}={\frac {\bar {z}}{x^{2}+y^{2}}}={\frac {x}{x^{2}+y^{2}}}-{\frac {y}{x^{2}+y^{2}}}i.}
imamo
z
′
⋅
z
=
z
⋅
z
′
=
1
{\displaystyle z'\cdot z=z\cdot z'=1}
z
′
=
z
−
1
=
1
z
{\displaystyle z'=z^{-1}={\frac {1}{z}}}
z
n
=
r
n
(
cos
n
θ
+
i
sin
n
θ
)
=
r
n
e
i
n
θ
{\displaystyle z^{n}=r^{n}(\cos n\theta +i\sin n\theta )=r^{n}e^{in\theta }}
n
∈
N
{\displaystyle n\in N}
z
m
z
n
=
z
m
+
n
{\displaystyle z^{m}z^{n}=z^{m+n}}
(
z
1
z
2
)
n
=
(
z
1
z
2
)
n
{\displaystyle (z_{1}z_{2})^{n}=(z_{1}z_{2})^{n}}
(
z
m
)
n
=
z
m
n
{\displaystyle (z^{m})^{n}=z^{mn}}
i
4
k
=
1
{\displaystyle i^{4k}=1}
i
4
k
+
1
=
i
{\displaystyle i^{4k+1}=i}
i
4
k
+
2
=
−
1
{\displaystyle i^{4k+2}=-1}
i
4
k
+
3
=
−
i
{\displaystyle i^{4k+3}=-i}
[5]
z
n
=
{
u
0
,
u
1
.
.
.
u
n
}
{\displaystyle {\sqrt[{n}]{z}}={\begin{Bmatrix}u_{0},u_{1}...u_{n}\end{Bmatrix}}}
n
∈
N
{\displaystyle n\in N}
gdje je
u
k
=
r
n
(
cos
r
n
n
+
i
sin
θ
+
2
k
π
n
)
{\displaystyle u_{k}={\sqrt[{n}]{r}}(\cos {\frac {\sqrt[{n}]{r}}{n}}+i\sin {\frac {\theta +2k\pi }{n}})}
k
=
0
,
1
,
…
,
(
n
−
1
)
{\displaystyle k=0,1,\dots ,(n-1)}
u
k
=
r
n
e
i
(
θ
+
2
k
π
)
/
2
{\displaystyle u_{k}={\sqrt[{n}]{r}}e^{i(\theta +2k\pi )/2}}
k
=
0
,
1
,
…
,
(
n
−
1
)
{\displaystyle k=0,1,\dots ,(n-1)}
i
=
1
2
2
+
i
1
2
2
=
2
2
(
1
+
i
)
.
{\displaystyle {\sqrt {i}}={\frac {1}{2}}{\sqrt {2}}+i{\frac {1}{2}}{\sqrt {2}}={\frac {\sqrt {2}}{2}}(1+i).}
Ovaj rezultat možemo dobiti na sljedeći način
i
=
(
a
+
b
i
)
2
{\displaystyle i=(a+bi)^{2}\!}
i
=
a
2
+
2
a
b
i
−
b
2
.
{\displaystyle i=a^{2}+2abi-b^{2}.\!}
Dobijamo dvije jednačine
{
2
a
b
=
1
a
2
−
b
2
=
0
{\displaystyle {\begin{cases}2ab=1\!\\a^{2}-b^{2}=0\!\end{cases}}}
čija su rješenja
a
=
b
=
±
1
2
.
{\displaystyle a=b=\pm {\frac {1}{\sqrt {2}}}.}
Izbor glavnog korjena daje
a
=
b
=
1
2
.
{\displaystyle a=b={\frac {1}{\sqrt {2}}}.}
Rezultat možemo dobiti pomoću Moavrova formule
i
=
cos
(
π
2
)
+
i
sin
(
π
2
)
{\displaystyle i=\cos \left({\frac {\pi }{2}}\right)+i\sin \left({\frac {\pi }{2}}\right)}
i
=
(
cos
(
π
2
)
+
i
sin
(
π
2
)
)
1
2
=
cos
(
π
4
)
+
i
sin
(
π
4
)
=
1
2
+
i
(
1
2
)
=
1
2
(
1
+
i
)
.
{\displaystyle {\begin{aligned}{\sqrt {i}}&=\left(\cos \left({\frac {\pi }{2}}\right)+i\sin \left({\frac {\pi }{2}}\right)\right)^{\frac {1}{2}}\\&=\cos \left({\frac {\pi }{4}}\right)+i\sin \left({\frac {\pi }{4}}\right)\\&={\frac {1}{\sqrt {2}}}+i\left({\frac {1}{\sqrt {2}}}\right)={\frac {1}{\sqrt {2}}}(1+i).\\\end{aligned}}}
Iz trigonometrijskih identiteta
cos
(
a
)
cos
(
b
)
−
sin
(
a
)
sin
(
b
)
=
cos
(
a
+
b
)
{\displaystyle \cos(a)\cos(b)-\sin(a)\sin(b)=\cos(a+b)}
cos
(
a
)
sin
(
b
)
+
sin
(
a
)
cos
(
b
)
=
sin
(
a
+
b
)
{\displaystyle \cos(a)\sin(b)+\sin(a)\cos(b)=\sin(a+b)}
imamo
z
1
z
2
=
r
1
r
2
(
cos
(
φ
1
+
φ
2
)
+
i
sin
(
φ
1
+
φ
2
)
)
.
{\displaystyle z_{1}z_{2}=r_{1}r_{2}(\cos(\varphi _{1}+\varphi _{2})+i\sin(\varphi _{1}+\varphi _{2})).\,}
Primjer
(
2
+
i
)
(
3
+
i
)
=
5
+
5
i
.
{\displaystyle (2+i)(3+i)=5+5i.\,}
π
4
=
arctan
1
2
+
arctan
1
3
{\displaystyle {\frac {\pi }{4}}=\arctan {\frac {1}{2}}+\arctan {\frac {1}{3}}}
Dijeljenje
z
1
z
2
=
r
1
r
2
(
cos
(
φ
1
−
φ
2
)
+
i
sin
(
φ
1
−
φ
2
)
)
.
{\displaystyle {\frac {z_{1}}{z_{2}}}={\frac {r_{1}}{r_{2}}}\left(\cos(\varphi _{1}-\varphi _{2})+i\sin(\varphi _{1}-\varphi _{2})\right).}
Ponekad je kompleksne brojeve potrebno pisati u trigonometrijskom obliku
a
+
b
i
=
ρ
(
cos
ϕ
+
i
sin
ϕ
)
{\displaystyle a+bi=\rho (\cos \phi +i\sin \phi )\,}
ρ
=
a
2
+
b
2
,
ϕ
=
arctan
b
a
{\displaystyle \rho ={\sqrt {a^{2}+b^{2}}},\ \phi =\arctan {\frac {b}{a}}}
a
>
0
{\displaystyle a>0}
ϕ
=
π
+
arctan
b
a
{\displaystyle \phi =\pi +\arctan {\frac {b}{a}}}
a
<
0
{\displaystyle a<0}
a
=
0
{\displaystyle a=0}
ϕ
=
π
2
{\displaystyle \phi ={\frac {\pi }{2}}}
b
>
0
{\displaystyle b>0}
ϕ
=
−
π
2
{\displaystyle \phi =-{\frac {\pi }{2}}}
b
<
0
{\displaystyle b<0}
Broj
ρ
{\displaystyle \rho }
ϕ
{\displaystyle \phi }
kompleksnog broja
Množenje Množenje kompleksnih brojeva u trigonometrijskom obliku je slično množenju kompleksnih brojeva u standardnom obliku.
Neka su zadani kompleksni brojevi
z
1
=
r
1
(
cos
φ
1
+
i
sin
φ
1
)
{\displaystyle z_{1}=r_{1}(\cos \varphi _{1}+i\sin \varphi _{1})}
z
2
=
r
2
(
cos
φ
2
+
i
sin
φ
2
)
{\displaystyle z_{2}=r_{2}(\cos \varphi _{2}+i\sin \varphi _{2})}
onda je [6]
Sada, kada smo odredili brojeve, mogu se pomnožiti:
z
1
z
2
=
r
1
(
cos
φ
1
+
i
sin
φ
1
)
r
2
(
cos
φ
2
+
i
sin
φ
2
)
=
{\displaystyle z_{1}z_{2}=r_{1}(\cos \varphi _{1}+i\sin \varphi _{1})r_{2}(\cos \varphi _{2}+i\sin \varphi _{2})=}
r
1
r
2
(
cos
φ
1
cos
φ
2
+
i
cos
φ
1
sin
φ
2
+
i
cos
φ
2
sin
φ
1
+
i
2
sin
φ
1
sin
φ
2
)
=
{\displaystyle r_{1}r_{2}(\cos \varphi _{1}\cos \varphi _{2}+i\cos \varphi _{1}\sin \varphi _{2}+i\cos \varphi _{2}\sin \varphi _{1}+i^{2}\sin \varphi _{1}\sin \varphi _{2})=}
r
1
r
2
(
(
cos
φ
1
cos
φ
2
−
sin
φ
1
sin
φ
2
)
+
i
(
cos
φ
1
sin
φ
2
cos
φ
2
sin
φ
1
)
=
{\displaystyle r_{1}r_{2}((\cos \varphi _{1}\cos \varphi _{2}-\sin \varphi _{1}\sin \varphi _{2})+i(\cos \varphi _{1}\sin \varphi _{2}\cos \varphi _{2}\sin \varphi _{1})=}
r
1
r
2
(
cos
(
φ
1
+
φ
2
)
+
i
sin
(
φ
1
+
φ
2
)
{\displaystyle r_{1}r_{2}(\cos(\varphi _{1}+\varphi _{2})+i\sin(\varphi _{1}+\varphi _{2})}
Dijeljenje Neka su zadani kompleksni brojevi
z
1
=
r
1
(
cos
φ
1
+
i
sin
φ
1
)
{\displaystyle z_{1}=r_{1}(\cos \varphi _{1}+i\sin \varphi _{1})}
z
2
=
r
2
(
cos
φ
2
+
i
sin
φ
2
)
{\displaystyle z_{2}=r_{2}(\cos \varphi _{2}+i\sin \varphi _{2})}
z
1
z
2
=
r
1
(
cos
φ
1
+
i
sin
φ
1
)
r
2
(
cos
φ
2
+
i
sin
φ
2
)
=
r
1
(
cos
φ
1
+
i
sin
φ
1
)
r
2
(
cos
φ
2
+
i
sin
φ
2
)
⋅
r
1
(
cos
φ
2
−
i
sin
φ
2
)
r
2
(
cos
φ
2
−
i
sin
φ
2
)
{\displaystyle {\frac {z_{1}}{z_{2}}}={\frac {r_{1}(\cos \varphi _{1}+i\sin \varphi _{1})}{r_{2}(\cos \varphi _{2}+i\sin \varphi _{2})}}={\frac {r_{1}(\cos \varphi _{1}+i\sin \varphi _{1})}{r_{2}(\cos \varphi _{2}+i\sin \varphi _{2})}}\cdot {\frac {r_{1}(\cos \varphi _{2}-i\sin \varphi _{2})}{r_{2}(\cos \varphi _{2}-i\sin \varphi _{2})}}}
[6]
u općem slučaju važi
r
1
r
2
⋅
cos
φ
1
cos
φ
2
−
i
cos
φ
1
sin
φ
2
+
i
cos
φ
2
sin
φ
1
−
i
2
sin
φ
1
sin
φ
2
cos
2
φ
2
+
sin
2
φ
2
{\displaystyle {\frac {r_{1}}{r_{2}}}\cdot {\frac {\cos \varphi _{1}\cos \varphi _{2}-i\cos \varphi _{1}\sin \varphi _{2}+i\cos \varphi _{2}\sin \varphi _{1}-i^{2}\sin \varphi _{1}\sin \varphi _{2}}{\cos ^{2}\varphi _{2}+\sin ^{2}\varphi _{2}}}}
r
1
r
2
(
cos
(
φ
1
−
φ
2
)
+
i
sin
(
φ
1
−
φ
2
)
)
=
r
1
r
2
(
cis
(
φ
1
−
φ
2
)
)
{\displaystyle {\frac {r_{1}}{r_{2}}}(\cos(\varphi _{1}-\varphi _{2})+i\sin(\varphi _{1}-\varphi _{2}))={\frac {r_{1}}{r_{2}}}(\operatorname {cis} (\varphi _{1}-\varphi _{2}))}
Neka je
z
=
r
(
cos
θ
+
i
sin
θ
)
=
r
cis
θ
{\displaystyle z=r(\cos \theta +i\sin \theta )=r\operatorname {cis} \theta }
z
2
=
z
⋅
z
{\displaystyle z^{2}=z\cdot z}
z
2
=
r
cis
θ
⋅
r
cis
θ
=
r
2
cis
(
θ
+
θ
)
=
r
2
cis
2
θ
{\displaystyle z^{2}=r\operatorname {cis} \theta \cdot r\operatorname {cis} \theta =r^{2}\operatorname {cis} (\theta +\theta )=r^{2}\operatorname {cis} 2\theta }
z
3
=
r
2
cis
2
θ
⋅
r
cis
θ
=
r
3
cis
(
2
θ
+
θ
)
=
r
3
cis
3
θ
{\displaystyle z^{3}=r^{2}\operatorname {cis} 2\theta \cdot r\operatorname {cis} \theta =r^{3}\operatorname {cis} (2\theta +\theta )=r^{3}\operatorname {cis} 3\theta }
z
4
=
r
3
cis
3
θ
⋅
r
cis
θ
=
r
4
cis
(
3
θ
+
θ
)
=
r
4
cis
4
θ
{\displaystyle z^{4}=r^{3}\operatorname {cis} 3\theta \cdot r\operatorname {cis} \theta =r^{4}\operatorname {cis} (3\theta +\theta )=r^{4}\operatorname {cis} 4\theta }
Na ovaj način dobijamo opšti oblik De Moavrova teorema koji ima važnu ulogu u elektrotehnici
(
cos
ϕ
+
i
sin
ϕ
)
n
=
cos
n
ϕ
+
i
sin
n
ϕ
{\displaystyle (\cos \phi +i\sin \phi )^{n}=\cos n\phi +i\sin n\phi \,}
[7]
Helmuth Gericke (1970). Geschichte des Zahlbegriffs . Mannheim: Bibliographisches Institut. str. 57–67.
![{\displaystyle {\sqrt[{n}]{z}}={\begin{Bmatrix}u_{0},u_{1}...u_{n}\end{Bmatrix}}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/7cf0e60c4d3392df53c167b88044feef6328bd51)
![{\displaystyle u_{k}={\sqrt[{n}]{r}}(\cos {\frac {\sqrt[{n}]{r}}{n}}+i\sin {\frac {\theta +2k\pi }{n}})}](http://wikimedia.org/api/rest_v1/media/math/render/svg/5a69be16f09046f52c9dc440a8a5940b07440fea)
![{\displaystyle u_{k}={\sqrt[{n}]{r}}e^{i(\theta +2k\pi )/2}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/133d35aa2d3bede7963901515ab63b4928ba11f2)
