Geometrische Darstellung einer komplexen Zahl. Kartesische Form
z
=
a
+
b
i
{\displaystyle z=a+b\mathrm {i} }
Polarform (trigonometrische Darstellung)
z
=
r
(
cos
φ
+
i
sin
φ
)
{\displaystyle z=r(\cos \varphi +\mathrm {i} \sin \varphi )}
Polarform (Exponentialdarstellung)
z
=
r
e
i
φ
{\displaystyle z=r\mathrm {e} ^{\mathrm {i} \varphi }}
Für alle
z
,
z
1
,
z
2
∈
C
{\displaystyle z,z_{1},z_{2}\in \mathbb {C} }
z
1
+
z
2
¯
=
z
¯
1
+
z
¯
2
{\displaystyle {\overline {z_{1}+z_{2}}}={\bar {z}}_{1}+{\bar {z}}_{2}}
z
1
−
z
2
¯
=
z
¯
1
−
z
¯
2
{\displaystyle {\overline {z_{1}-z_{2}}}={\bar {z}}_{1}-{\bar {z}}_{2}}
z
1
z
2
¯
=
z
¯
1
z
¯
2
{\displaystyle {\overline {z_{1}z_{2}}}={\bar {z}}_{1}\,{\bar {z}}_{2}}
z
2
≠
0
⟹
(
z
1
z
2
)
¯
=
z
¯
1
z
¯
2
{\displaystyle z_{2}\neq 0\implies {\overline {{\Big (}{\frac {z_{1}}{z_{2}}}{\Big )}}}={\frac {{\bar {z}}_{1}}{{\bar {z}}_{2}}}}
|
z
¯
|
=
|
z
|
{\displaystyle |{\bar {z}}|=|z|}
z
¯
¯
=
z
{\displaystyle {\bar {\bar {z}}}=z}
z
z
¯
=
|
z
|
2
{\displaystyle z{\bar {z}}=|z|^{2}}
Re
(
z
)
=
z
+
z
¯
2
{\displaystyle \operatorname {Re} (z)={\frac {z+{\bar {z}}}{2}}}
Im
(
z
)
=
z
−
z
¯
2
i
{\displaystyle \operatorname {Im} (z)={\frac {z-{\bar {z}}}{2i}}}
e
z
¯
=
e
z
¯
{\displaystyle {\overline {\mathrm {e} ^{z}}}=\mathrm {e} ^{\bar {z}}}
sin
(
z
)
¯
=
sin
(
z
¯
)
{\displaystyle {\overline {\sin(z)}}=\sin({\bar {z}})}
cos
(
z
)
¯
=
cos
(
z
¯
)
{\displaystyle {\overline {\cos(z)}}=\cos({\bar {z}})}
Für alle
z
∈
C
∖
{
x
∈
R
∣
x
≤
0
}
{\displaystyle z\in \mathbb {C} \setminus \{x\in \mathbb {R} \mid x\leq 0\}}
x
∈
C
{\displaystyle x\in \mathbb {C} }
z
x
¯
=
z
¯
x
¯
{\displaystyle {\overline {z^{x}}}={\bar {z}}^{\bar {x}}}
ln
(
z
)
¯
=
ln
(
z
¯
)
{\displaystyle {\overline {\ln(z)}}=\ln({\bar {z}})}
arg
(
z
¯
)
=
−
arg
(
z
)
{\displaystyle \arg({\bar {z}})=-\arg(z)}
Allgemeine Potenzfunktion
f
:
R
2
→
C
,
f
(
x
,
y
)
:=
x
y
{\displaystyle f\colon \mathbb {R} ^{2}\to \mathbb {C} ,\;f(x,y):=x^{y}}
Allgemeine Potenzfunktion
z
=
x
y
{\displaystyle z=x^{y}}
Definitionen:
e
z
:=
e
a
cos
(
b
)
+
i
e
a
sin
(
b
)
(
z
=
a
+
b
i
)
{\displaystyle \mathrm {e} ^{z}:=\mathrm {e} ^{a}\cos(b)+\mathrm {i} \,\mathrm {e} ^{a}\sin(b)\qquad (z=a+b\mathrm {i} )}
ln
(
z
)
:=
ln
(
|
z
|
)
+
arg
(
z
)
i
(
z
≠
0
)
{\displaystyle \ln(z):=\ln(|z|)+\operatorname {arg} (z)\,\mathrm {i} \qquad (z\neq 0)}
z
w
:=
e
w
ln
(
z
)
(
z
≠
0
)
{\displaystyle z^{w}:=\mathrm {e} ^{w\ln(z)}\qquad (z\neq 0)}
Für alle
z
,
z
1
,
z
2
∈
C
{\displaystyle z,z_{1},z_{2}\in \mathbb {C} }
e
z
1
+
z
2
=
e
z
1
e
z
2
{\displaystyle \mathrm {e} ^{z_{1}+z_{2}}=\mathrm {e} ^{z_{1}}\mathrm {e} ^{z_{2}}}
e
−
z
=
1
e
z
{\displaystyle \mathrm {e} ^{-z}={\frac {1}{\mathrm {e^{z}} }}}
e
z
≠
0
{\displaystyle \mathrm {e} ^{z}\neq 0}
e
i
z
=
cos
(
z
)
+
i
sin
(
z
)
{\displaystyle \mathrm {e} ^{\mathrm {i} z}=\cos(z)+\mathrm {i} \sin(z)}
∀
k
∈
Z
:
e
2
k
π
i
=
1
{\displaystyle \forall k\in \mathbb {Z} \colon \;\mathrm {e} ^{2k\pi \mathrm {i} }=1}
∀
k
∈
Z
:
e
(
2
k
+
1
)
π
i
+
1
=
0
{\displaystyle \forall k\in \mathbb {Z} \colon \;\mathrm {e} ^{(2k+1)\pi \mathrm {i} }+1=0}
Für alle
z
∈
C
∖
{
0
}
{\displaystyle z\in \mathbb {C} \setminus \{0\}}
x
,
y
∈
C
{\displaystyle x,y\in \mathbb {C} }
z
x
+
y
=
z
x
z
y
{\displaystyle z^{x+y}=z^{x}z^{y}}
z
x
−
y
=
z
x
z
y
{\displaystyle z^{x-y}={\frac {z^{x}}{z^{y}}}}
z
−
x
=
1
z
x
{\displaystyle z^{-x}={\frac {1}{z^{x}}}}
z
0
=
1
{\displaystyle z^{0}=1}
Für alle
r
>
0
{\displaystyle r>0}
z
∈
C
∖
{
0
}
{\displaystyle z\in \mathbb {C} \setminus \{0\}}
x
∈
C
{\displaystyle x\in \mathbb {C} }
(
r
z
)
x
=
r
x
z
x
{\displaystyle (rz)^{x}=r^{x}z^{x}}
(
z
r
)
x
=
z
x
r
x
{\displaystyle {\Big (}{\frac {z}{r}}{\Big )}^{x}={\frac {z^{x}}{r^{x}}}}
(
1
r
)
x
=
1
r
x
=
r
−
x
{\displaystyle {\Big (}{\frac {1}{r}}{\Big )}^{x}={\frac {1}{r^{x}}}=r^{-x}}
Für alle
r
>
0
{\displaystyle r>0}
z
∈
C
∖
{
x
∈
R
∣
x
≤
0
}
{\displaystyle z\in \mathbb {C} \setminus \{x\in \mathbb {R} \mid x\leq 0\}}
x
∈
C
{\displaystyle x\in \mathbb {C} }
(
r
z
)
x
=
r
x
z
x
{\displaystyle {\Big (}{\frac {r}{z}}{\Big )}^{x}={\frac {r^{x}}{z^{x}}}}
Graph der Funktion f (z ) = z 5 −1. Die Nullstellen von f heißen fünfte Einheitswurzeln . Die n -ten Wurzeln einer komplexen Zahl bilden immer ein regelmäßiges n -Eck, dessen Zentrum im Koordinatenursprung liegt. Sei
φ
:=
arg
(
z
)
{\displaystyle \varphi :=\arg(z)}
n
∈
N
{\displaystyle n\in \mathbb {N} }
z
=
w
n
⟺
w
=
|
z
|
n
exp
(
i
φ
+
2
k
π
i
n
)
,
k
∈
{
0
,
1
,
…
,
n
−
1
}
.
{\displaystyle z=w^{n}\iff w={\sqrt[{n}]{|z|}}\exp {\Big (}{\frac {\mathrm {i} \varphi +2k\pi \mathrm {i} }{n}}{\Big )},\;k\in \{0,1,\ldots ,n-1\}.}
Hauptwert:
z
n
=
|
z
|
n
exp
(
i
φ
n
)
.
{\displaystyle {\sqrt[{n}]{z}}={\sqrt[{n}]{|z|}}\exp {\Big (}{\frac {\mathrm {i} \varphi }{n}}{\Big )}.}
Hauptwert, allgemein für
z
,
x
∈
C
∖
{
0
}
{\displaystyle z,x\in \mathbb {C} \setminus \{0\}}
z
x
:=
exp
(
ln
(
z
)
x
)
.
{\displaystyle {\sqrt[{x}]{z}}:=\exp {\bigg (}{\frac {\ln(z)}{x}}{\bigg )}.}
Ist
α
{\displaystyle \alpha \,}
z
{\displaystyle z\,}
|
z
α
|
∈
Θ
(
|
z
|
Re
α
)
{\displaystyle \left|z^{\alpha }\right|\in \Theta \left(|z|^{{\text{Re}}\,\alpha }\right)}
|
z
|
→
∞
{\displaystyle |z|\to \infty \,}
Θ
{\displaystyle \Theta }
Sind
z
1
,
z
2
{\displaystyle z_{1},z_{2}\,}
α
{\displaystyle \alpha \,}
(
z
1
⋅
z
2
)
α
=
z
1
α
⋅
z
2
α
{\displaystyle (z_{1}\cdot z_{2})^{\alpha }=z_{1}^{\alpha }\cdot z_{2}^{\alpha }}
(
z
1
z
2
)
α
=
z
1
α
z
2
α
{\displaystyle \left({\frac {z_{1}}{z_{2}}}\right)^{\alpha }={\frac {z_{1}^{\alpha }}{z_{2}^{\alpha }}}}
Ist
z
{\displaystyle z\,}
0
z
=
{
0
Re
(
z
)
>
0
1
z
=
0
nicht definiert
sonst
{\displaystyle 0^{z}=\left\{{\begin{matrix}0&{\text{Re}}(z)>0\\1&z=0\\{\text{nicht definiert}}&{\text{sonst}}\end{matrix}}\right.}
(
a
2
+
b
2
)
(
c
2
+
d
2
)
=
(
a
c
−
b
d
)
2
+
(
a
d
+
b
c
)
2
{\displaystyle (a^{2}+b^{2})(c^{2}+d^{2})=(ac-bd)^{2}+(ad+bc)^{2}\,}
Beweis (Formel von Fibonacci)
Aus
(
a
+
i
b
)
(
c
+
i
d
)
=
(
a
c
−
b
d
)
+
i
(
a
d
+
b
c
)
{\displaystyle (a+ib)(c+id)=(ac-bd)+i(ad+bc)\,}
folgt
|
a
+
i
b
|
2
⋅
|
c
+
i
d
|
2
=
|
(
a
c
−
b
d
)
+
i
(
a
d
+
b
c
)
|
2
{\displaystyle |a+ib|^{2}\cdot |c+id|^{2}=|(ac-bd)+i(ad+bc)|^{2}}
a
+
i
b
=
a
2
+
b
2
+
a
2
+
i
Θ
(
b
)
a
2
+
b
2
−
a
2
{\displaystyle {\sqrt {a+ib}}={\sqrt {\frac {{\sqrt {a^{2}+b^{2}}}+a}{2}}}+i\,\Theta (b)\,{\sqrt {\frac {{\sqrt {a^{2}+b^{2}}}-a}{2}}}}
Θ
(
b
)
=
{
1
,
b
≥
0
−
1
,
b
<
0
{\displaystyle \Theta (b)=\left\{{\begin{matrix}1&,&b\geq 0\\\\-1&,&b<0\end{matrix}}\right.}
Beweis
Für jede von Null verschiedene komplexe Zahl
a
+
i
b
{\displaystyle a+ib\,}
a
+
i
b
{\displaystyle a+ib\,}
Mit
a
+
i
b
=
x
+
i
y
{\displaystyle {\sqrt {a+ib}}=x+iy}
x
≥
0
{\displaystyle x\geq 0}
x
=
0
{\displaystyle x=0\,}
y
≥
0
{\displaystyle y\geq 0}
Wenn
(
x
+
i
y
)
2
=
a
+
i
b
{\displaystyle (x+iy)^{2}=a+ib\,}
x
2
−
y
2
=
a
,
2
x
y
=
b
{\displaystyle x^{2}-y^{2}=a\,,\,2xy=b}
x
2
+
y
2
=
a
2
+
b
2
{\displaystyle x^{2}+y^{2}={\sqrt {a^{2}+b^{2}}}}
Daher ist
x
2
=
(
x
2
+
y
2
)
+
(
x
2
−
y
2
)
2
=
a
2
+
b
2
+
a
2
⇒
x
=
a
2
+
b
2
+
a
2
≥
0
{\displaystyle x^{2}={\frac {(x^{2}+y^{2})+(x^{2}-y^{2})}{2}}={\frac {{\sqrt {a^{2}+b^{2}}}+a}{2}}\Rightarrow x={\sqrt {\frac {{\sqrt {a^{2}+b^{2}}}+a}{2}}}\geq 0}
und
y
2
=
(
x
2
+
y
2
)
−
(
x
2
−
y
2
)
2
=
a
2
+
b
2
−
a
2
⇒
y
=
Θ
(
b
)
a
2
+
b
2
−
a
2
{\displaystyle y^{2}={\frac {(x^{2}+y^{2})-(x^{2}-y^{2})}{2}}={\frac {{\sqrt {a^{2}+b^{2}}}-a}{2}}\Rightarrow y=\Theta (b)\,{\sqrt {\frac {{\sqrt {a^{2}+b^{2}}}-a}{2}}}}
da im Fall
x
>
0
sgn
(
y
)
=
sgn
(
2
x
y
)
=
sgn
(
b
)
{\displaystyle x>0\quad {\text{sgn}}(y)={\text{sgn}}(2xy)={\text{sgn}}(b)}
x
=
0
{\displaystyle x=0\,}
b
=
0
{\displaystyle b=0\,}
y
≥
0
{\displaystyle y\geq 0}
Vergleich verschiedener Darstellungen zum Thema bei Wikibooks
Die komplexen Zahlen werden in folgenden Büchern von Wikibooks behandelt:
Einzelne Kapitel anderer Bücher richten sich an bestimmte Zielgruppen:
![{\displaystyle z=w^{n}\iff w={\sqrt[{n}]{|z|}}\exp {\Big (}{\frac {\mathrm {i} \varphi +2k\pi \mathrm {i} }{n}}{\Big )},\;k\in \{0,1,\ldots ,n-1\}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a3bea17a7d314afd793ddaf022d01dc78fe6ffb1)
![{\displaystyle {\sqrt[{n}]{z}}={\sqrt[{n}]{|z|}}\exp {\Big (}{\frac {\mathrm {i} \varphi }{n}}{\Big )}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/227dd560e1f2d20d6625adc5234e19d73047676e)
![{\displaystyle {\sqrt[{x}]{z}}:=\exp {\bigg (}{\frac {\ln(z)}{x}}{\bigg )}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/38a68ab832b5581a372562e9d462e18087cf1b2f)
