Kiejtés Főnév n -edik egységgyök
( matematika ) Az 1 komplex szám
n
{\displaystyle n}
gyökét n-edik egységgyöknek nevezzük. Az n-edik egységgyök alakja
ε
k
:=
cos
2
k
π
n
+
i
sin
2
k
π
n
,
k
=
0
,
1
,
…
,
n
−
1
{\displaystyle \varepsilon _{k}:=\cos {\dfrac {2k\pi }{n}}+i\sin {\dfrac {2k\pi }{n}},\quad k=0,1,\ldots ,n-1}
ε
k
n
=
1
{\displaystyle \varepsilon _{k}^{n}=1}
k
=
0
,
1
,
…
,
n
−
1
{\displaystyle k=0,1,\ldots ,n-1}
ε
k
=
ε
1
k
,
k
=
0
,
1
,
…
,
n
−
1
{\displaystyle \varepsilon _{k}=\varepsilon _{1}^{k},\quad k=0,1,\ldots ,n-1}
{
ε
0
,
ε
1
,
ε
2
,
…
,
ε
n
−
1
}
=
{
1
,
ε
1
,
ε
1
2
,
…
,
ε
1
n
−
1
}
{\displaystyle \left\{\varepsilon _{0},\varepsilon _{1},\varepsilon _{2},\ldots ,\varepsilon _{n-1}\right\}=\left\{1,\varepsilon _{1},\varepsilon _{1}^{2},\ldots ,\varepsilon _{1}^{n-1}\right\}}
Az n-edik komplex egységgyökök halmaza zárt a szorzás műveletére vonatkozóan, azaz két n-edik egységgyök szorzata n-edik egységgyök.
Minden
1
≤
k
,
ℓ
≤
n
−
1
{\displaystyle 1\leq k,\ell \leq n-1}
ε
k
ε
ℓ
=
{
ε
k
+
ℓ
,
ha
k
+
ℓ
<
n
ε
k
+
ℓ
−
n
,
ha
k
+
ℓ
≥
n
{\displaystyle \varepsilon _{k}\varepsilon _{\ell }=\left\{{\begin{array}{l l }{\varepsilon _{k+\ell },}&{{\text{ ha }}k+\ell <n}\\{\varepsilon _{k+\ell -n},}&{{\text{ ha }}k+\ell \geq n}\end{array}}\right.}
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}
Azt is könnyű ellenőrizni, hogy ha
w
{\displaystyle w}
n
{\displaystyle n}
z
{\displaystyle z}
w
,
w
ε
1
,
w
ε
1
2
,
…
,
w
ε
1
n
−
1
{\displaystyle w,w\varepsilon _{1},w\varepsilon _{1}^{2},\ldots ,w\varepsilon _{1}^{n-1}}
n
{\displaystyle n}
z
{\displaystyle z}
ε
1
{\displaystyle \varepsilon _{1}}
2
π
n
{\displaystyle {\dfrac {2\pi }{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)=}](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)
