Question Number 6923 by Tawakalitu. last updated on 03/Aug/16
$${Find}\:{the}\:{principal}\:{value}\:{of}\:\left(\mathrm{3}\:+\:\mathrm{4}{i}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$
Commented by Yozzii last updated on 03/Aug/16
$$\mathrm{3}+\mathrm{4}{i}=\sqrt{\mathrm{3}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} }{e}^{{itan}^{−\mathrm{1}} \left(\mathrm{4}/\mathrm{3}\right)} =\mathrm{5}{e}^{{itan}^{−\mathrm{1}} \frac{\mathrm{4}}{\mathrm{3}}} \\ $$$$\therefore\:{principal}\:{value}\:{of}\:\left(\mathrm{3}+\mathrm{4}{i}\right)^{\mathrm{1}/\mathrm{3}} \:{is} \\ $$$$\left(\mathrm{5}{e}^{{itan}^{−\mathrm{1}} \frac{\mathrm{4}}{\mathrm{3}}} \right)^{\mathrm{1}/\mathrm{3}} =\mathrm{5}^{\mathrm{1}/\mathrm{3}} {e}^{{i}\frac{\mathrm{1}}{\mathrm{3}}{tan}^{−\mathrm{1}} \frac{\mathrm{4}}{\mathrm{3}}} . \\ $$$$−−−−−−−−−−−−−−−−−−−−−− \\ $$$${z}={re}^{{i}\theta} ={re}^{{i}\left(\theta+\mathrm{2}\pi{k}\right)} \:\:\:\left({k}\in\mathbb{Z}\right) \\ $$$$\Rightarrow{z}^{\mathrm{1}/{n}} ={r}^{\mathrm{1}/{n}} {e}^{{i}\frac{\theta+\mathrm{2}\pi{k}}{{n}}} \:\:\left({k}=\mathrm{0},\mathrm{1},\mathrm{2},...,{n}−\mathrm{1}\right)\:\:\:\:\:\: \\ $$$${principal}\:{value}\:{of}\:{z}^{\mathrm{1}/{n}} \:{occurs}\:{at}\:{k}=\mathrm{0}. \\ $$$$ \\ $$
Commented by Tawakalitu. last updated on 03/Aug/16
$${I}\:{really}\:{appreciate} \\ $$