7. Példa ${\root 6 \of{1-i\over{\sqrt 3+i}}}=?$

Megoldás.


\begin{displaymath}{1-i\over{\sqrt 3+i}}={\sqrt2(\cos(-{\pi\over4})+i\sin(-{\pi\... ...t(-{5\pi\over12}\right)+i\sin\left(-{5\pi\over12}\right)\right)\end{displaymath}


\begin{displaymath}\Rightarrow {\root 6 \of{1-i\over{\sqrt 3+i}}}=\root 6 \of{1\... ...left({-{5\pi\over12}+2k\pi\over6}\right)\right) (k=0,1,\dots,5)\end{displaymath}


\begin{displaymath}k=0:  \root 6 \of{1\over{\sqrt 2}}\left(\cos\left({-{5\pi\ove... ...t({-5\pi\over72}\right)+i\sin\left({-5\pi\over72}\right)\right)\end{displaymath}

$                  \approx0.922-0.204 i$


\begin{displaymath}k=1:  \root 6 \of{1\over{\sqrt 2}}\left(\cos\left({-{5\pi\ove... ...t({19\pi\over72}\right)+i\sin\left({19\pi\over72}\right)\right)\end{displaymath}


\begin{displaymath}k=2:  \root 6 \of{1\over{\sqrt 2}}\left(\cos\left({-{5\pi\ove... ...t({43\pi\over72}\right)+i\sin\left({43\pi\over72}\right)\right)\end{displaymath}


\begin{displaymath}k=3:  \root 6 \of{1\over{\sqrt 2}}\left(\cos\left({-{5\pi\ove... ...t({67\pi\over72}\right)+i\sin\left({67\pi\over72}\right)\right)\end{displaymath}


\begin{displaymath}k=4:  \root 6 \of{1\over{\sqrt 2}}\left(\cos\left({-{5\pi\ove... ...t({91\pi\over72}\right)+i\sin\left({91\pi\over72}\right)\right)\end{displaymath}


\begin{displaymath}k=5:  \root 6 \of{1\over{\sqrt 2}}\left(\cos\left({-{5\pi\ove... ...{115\pi\over72}\right)+i\sin\left({115\pi\over72}\right)\right)\end{displaymath}


Róbert Vajda 2003-02-21