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Suport teoretic

Vom rezolva mai intai ecuatia $\large z^{n}=1$ in cazul general. Astfel:

$z_{k}=\cos\frac{2k\pi }{n}+i\sin\frac{2k\pi }{n}, k=\overline{0,n-1}.$

Notam cu $U_{n}=\begin{Bmatrix}\cos\frac{2k\pi }{n}+i\sin\frac{2k\pi }{n}, k=\overline{0,n-1} \end{Bmatrix}$ multimea radacinilor de ordinul $n$ ale unitatii, iar $z_{k}\in U_{n}$ se numesc radacini de ordinul $n$ ale unitatii.

In continuare vom afla elementele multimii $U_{n}$ pentru cazurile particulare $n\in\begin{Bmatrix}2,3,4 \end{Bmatrix}.$

Pentru $n=2$ ecuatia $\large z^{n}=1$ devine $\large z^{2}=1$ $\Rightarrow$

$z_{k}=cos\frac{2k\pi }{2}+isin\frac{2k\pi }{2}=cos{k\pi }+isin{k\pi }, k=\overline{0,1}.$

$z_{0}=\cos 0+i\sin 0=1$

$z_{1}=\cos \pi +i\sin \pi =-1$

$U_{2}=\begin{Bmatrix}1,-1 \end{Bmatrix}$

Pentru $n=3$ ecuatia $z^{n}=1$ devine $z^{3}=1$ $\Rightarrow$

$z_{k}=cos\frac{2k\pi }{3}+isin\frac{2k\pi }{3}, k=\overline{0,2}.$

$z_{0}=\cos 0+i\sin 0=1$

$z_{1}=\cos \frac{2\pi }{3}+i\sin\frac{2\pi }{3}=-\frac{1}{2}+i\frac{\sqrt{3}}{2}$

$z_{2}=\cos \frac{4\pi }{3}+i\sin\frac{4\pi }{3}=\frac{1}{2}-i\frac{\sqrt{3}}{2}$

$U_{3}=\begin{Bmatrix}1,-\frac{1}{2}+i\frac{\sqrt{3}}{2},\frac{1}{2}-i\frac{\sqrt{3}}{2} \end{Bmatrix}$

Pentru $n=4$ ecuatia $z^{n}=1$ devine $z^{4}=1$ $\Rightarrow$

$z_{k}=cos\frac{2k\pi }{4}+isin\frac{2k\pi }{4}=cos\frac{k\pi }{2}+isin\frac{k\pi }{2}, k=\overline{0,3}.$

$z_{0}=\cos 0+i\sin 0=1$

$z_{1}=\cos \frac{\pi }{2}+i\sin\frac{\pi }{2}=i$

$z_{2}=\cos&amp{\pi}+i\sin{\pi}{3}=-1$

$z_{3}=\cos \frac{3\pi }{2}+i\sin\frac{3\pi }{2}=-i$

$U_{4}=\begin{Bmatrix}1,-1,i,-i \end{Bmatrix}$

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