6. Példa $\root 3 \of{-8}=?$

Megoldás.

$-8=8\left(\cos({\pi})+i\sin(\pi)\right)~\Rightarrow$


\begin{displaymath}\root 3 \of{-8}=\root 3 \of 8\left(\cos\left({\pi+2k\pi\over3}\right)+i\sin\left({\pi+2k\pi\over3}\right)\right) k=0,1,2. \end{displaymath}


\begin{displaymath}k=0:  \root 3 \of 8\left(\cos\left({\pi+2\cdot0\pi\over3}\rig... ...s\left({\pi\over3}\right)+i\sin\left({\pi\over3}\right)\right)=\end{displaymath}

$                           =2\cdot\left({1\over2}+{\sqrt3\over2}i\right)=1+\sqrt3 i$


\begin{displaymath}k=1:  \root 3 \of 8\left(\cos\left({\pi+2\cdot1\pi\over3}\rig... ...left({3\pi\over3}\right)+i\sin\left({3\pi\over3}\right)\right)=\end{displaymath}

$                           =2\cdot(-1+0i)=-2$


\begin{displaymath}k=2:  \root 3 \of 8\left(\cos\left({\pi+2\cdot2\pi\over3}\rig... ...left({5\pi\over3}\right)+i\sin\left({5\pi\over3}\right)\right)=\end{displaymath}

$                           =2\cdot\left({1\over2}-{\sqrt3\over2}i\right)=1-\sqrt3 i$


Róbert Vajda 2003-02-21