IPA : [ ˈɟøkvonaːʃ ˈkomplɛks ˈsaːmboːl]
gyökvonás komplex számból
( matematika ) Minden nem nulla komplex számnak pontosan n darab n -edik gyöke van a komplex számok között, és ezek egy origó középpontú szabályos sokszög csúcsaiban helyezkednek el. A z szám n-edik gyökei azért alkotnak szabályos n-szöget, mert ez
w
0
{\displaystyle w_{0}}
z
=
r
(
cos
φ
+
i
sin
φ
)
,
n
∈
N
.
{\displaystyle z=r(\cos \varphi +i\sin \varphi ),n\in \mathbb {N} .}
Ekkor
z
n
=
r
n
(
cos
φ
+
2
k
π
n
+
i
sin
φ
+
2
k
π
n
)
,
k
=
0
,
1
,
…
,
n
−
1
{\displaystyle {\sqrt[{n}]{z}}={\sqrt[{n}]{r}}\left(\cos {\dfrac {\varphi +2k\pi }{n}}+i\sin {\dfrac {\varphi +2k\pi }{n}}\right),\quad k=0,1,\ldots ,n-1}
Egy
z
=
r
(
cos
φ
+
i
sin
φ
)
{\displaystyle z=r(\cos \varphi +i\sin \varphi )}
z
n
=
r
n
(
cos
(
φ
n
+
2
k
π
n
)
+
i
sin
(
φ
n
+
2
k
π
n
)
)
=
{\displaystyle {\sqrt[{n}]{z}}={\sqrt[{n}]{r}}\left(\cos \left({\dfrac {\varphi }{n}}+{\dfrac {2k\pi }{n}}\right)+i\sin \left({\dfrac {\varphi }{n}}+{\dfrac {2k\pi }{n}}\right)\right)=}
r
n
(
cos
φ
n
+
i
sin
φ
n
)
(
cos
2
k
π
n
+
i
sin
2
k
π
n
)
{\displaystyle {\sqrt[{n}]{r}}\left(\cos {\dfrac {\varphi }{n}}+i\sin {\dfrac {\varphi }{n}}\right)\left(\cos {\dfrac {2k\pi }{n}}+i\sin {\dfrac {2k\pi }{n}}\right)}
k
=
0
,
1
,
…
,
n
−
1.
{\displaystyle k=0,1,\ldots ,n-1.}
Azaz, ha
w
0
:=
r
n
(
cos
φ
n
+
i
sin
φ
n
)
{\displaystyle w_{0}:={\sqrt[{n}]{r}}\left(\cos {\dfrac {\varphi }{n}}+i\sin {\dfrac {\varphi }{n}}\right)}
w
0
,
w
0
ε
1
,
w
0
ε
1
2
,
…
,
w
0
ε
1
n
−
1
{\displaystyle w_{0},w_{0}\varepsilon _{1},w_{0}\varepsilon _{1}^{2},\dots ,w_{0}\varepsilon _{1}^{n-1}}
n
{\displaystyle n}
z
{\displaystyle z}
Számítsuk ki
−
8
4
{\displaystyle {\sqrt[{4}]{-8}}}
−
8
=
8
(
cos
π
+
i
sin
π
)
.
{\displaystyle -8=8(\cos \pi +i\sin \pi ).}
8
4
(
cos
π
4
+
i
sin
π
4
)
=
8
4
(
1
2
+
i
1
2
)
=
2
4
+
2
4
i
{\displaystyle {\sqrt[{4}]{8}}\left(\cos {\dfrac {\pi }{4}}+i\sin {\dfrac {\pi }{4}}\right)={\sqrt[{4}]{8}}\left({\dfrac {1}{\sqrt {2}}}+i{\dfrac {1}{\sqrt {2}}}\right)={\sqrt[{4}]{2}}+{\sqrt[{4}]{2}}i}
8
4
(
cos
(
π
4
+
2
π
4
)
+
i
sin
(
π
4
+
2
π
4
)
)
=
8
4
(
cos
3
π
4
+
i
sin
3
π
4
)
=
−
2
4
+
2
4
i
{\displaystyle {\sqrt[{4}]{8}}\left(\cos \left({\dfrac {\pi }{4}}+{\dfrac {2\pi }{4}}\right)+i\sin \left({\dfrac {\pi }{4}}+{\dfrac {2\pi }{4}}\right)\right)={\sqrt[{4}]{8}}\left(\cos {\dfrac {3\pi }{4}}+i\sin {\dfrac {3\pi }{4}}\right)=-{\sqrt[{4}]{2}}+{\sqrt[{4}]{2}}i}
8
4
(
cos
(
π
4
+
4
π
4
)
+
i
sin
(
π
4
+
4
π
4
)
)
=
8
4
(
cos
5
π
4
+
i
sin
5
π
4
)
=
−
2
4
−
2
4
i
{\displaystyle {\sqrt[{4}]{8}}\left(\cos \left({\dfrac {\pi }{4}}+{\dfrac {4\pi }{4}}\right)+i\sin \left({\dfrac {\pi }{4}}+{\dfrac {4\pi }{4}}\right)\right)={\sqrt[{4}]{8}}\left(\cos {\dfrac {5\pi }{4}}+i\sin {\dfrac {5\pi }{4}}\right)=-{\sqrt[{4}]{2}}-{\sqrt[{4}]{2}}i}
8
4
(
cos
(
π
4
+
6
π
4
)
+
i
sin
(
π
4
+
6
π
4
)
)
=
8
4
(
cos
7
π
4
+
i
sin
7
π
4
)
=
2
4
−
2
4
i
{\displaystyle {\sqrt[{4}]{8}}\left(\cos \left({\dfrac {\pi }{4}}+{\dfrac {6\pi }{4}}\right)+i\sin \left({\dfrac {\pi }{4}}+{\dfrac {6\pi }{4}}\right)\right)={\sqrt[{4}]{8}}\left(\cos {\dfrac {7\pi }{4}}+i\sin {\dfrac {7\pi }{4}}\right)={\sqrt[{4}]{2}}-{\sqrt[{4}]{2}}i}
![{\displaystyle {\sqrt[{n}]{z}}={\sqrt[{n}]{r}}\left(\cos {\dfrac {\varphi +2k\pi }{n}}+i\sin {\dfrac {\varphi +2k\pi }{n}}\right),\quad k=0,1,\ldots ,n-1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e3071156fc2f43547b082f4c1ddba42f34cd0e3b)
![{\displaystyle {\sqrt[{n}]{z}}={\sqrt[{n}]{r}}\left(\cos \left({\dfrac {\varphi }{n}}+{\dfrac {2k\pi }{n}}\right)+i\sin \left({\dfrac {\varphi }{n}}+{\dfrac {2k\pi }{n}}\right)\right)=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f7f4eaa4635693fc47af2c562cadfaa798b755b8)
![{\displaystyle {\sqrt[{n}]{r}}\left(\cos {\dfrac {\varphi }{n}}+i\sin {\dfrac {\varphi }{n}}\right)\left(\cos {\dfrac {2k\pi }{n}}+i\sin {\dfrac {2k\pi }{n}}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7094f156f33c1f8bef2864dbfa2feba81436fc1f)
![{\displaystyle w_{0}:={\sqrt[{n}]{r}}\left(\cos {\dfrac {\varphi }{n}}+i\sin {\dfrac {\varphi }{n}}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/992e76d9c5f93aba51cffd80dc7b6f2a933ac87e)
![{\displaystyle {\sqrt[{4}]{-8}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a807ef2d5465e58a487b69403841f5fcb0925860)
![{\displaystyle {\sqrt[{4}]{8}}\left(\cos {\dfrac {\pi }{4}}+i\sin {\dfrac {\pi }{4}}\right)={\sqrt[{4}]{8}}\left({\dfrac {1}{\sqrt {2}}}+i{\dfrac {1}{\sqrt {2}}}\right)={\sqrt[{4}]{2}}+{\sqrt[{4}]{2}}i}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c184e4b3ebf53133cfdf8960992787402dc33507)
![{\displaystyle {\sqrt[{4}]{8}}\left(\cos \left({\dfrac {\pi }{4}}+{\dfrac {2\pi }{4}}\right)+i\sin \left({\dfrac {\pi }{4}}+{\dfrac {2\pi }{4}}\right)\right)={\sqrt[{4}]{8}}\left(\cos {\dfrac {3\pi }{4}}+i\sin {\dfrac {3\pi }{4}}\right)=-{\sqrt[{4}]{2}}+{\sqrt[{4}]{2}}i}](https://wikimedia.org/api/rest_v1/media/math/render/svg/abccbabd12f99ca70bf93375c7bf400b29b7f8b7)
![{\displaystyle {\sqrt[{4}]{8}}\left(\cos \left({\dfrac {\pi }{4}}+{\dfrac {4\pi }{4}}\right)+i\sin \left({\dfrac {\pi }{4}}+{\dfrac {4\pi }{4}}\right)\right)={\sqrt[{4}]{8}}\left(\cos {\dfrac {5\pi }{4}}+i\sin {\dfrac {5\pi }{4}}\right)=-{\sqrt[{4}]{2}}-{\sqrt[{4}]{2}}i}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bcde2dbf02add92dbaa851cdfeaad3053c579ad5)
![{\displaystyle {\sqrt[{4}]{8}}\left(\cos \left({\dfrac {\pi }{4}}+{\dfrac {6\pi }{4}}\right)+i\sin \left({\dfrac {\pi }{4}}+{\dfrac {6\pi }{4}}\right)\right)={\sqrt[{4}]{8}}\left(\cos {\dfrac {7\pi }{4}}+i\sin {\dfrac {7\pi }{4}}\right)={\sqrt[{4}]{2}}-{\sqrt[{4}]{2}}i}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dee68cb2e73af937ea715f65db993b6a715dfb00)