Question Number 167119 by mathlove last updated on 07/Mar/22
$$\left({z}+\mathrm{3}{i}\right)^{\mathrm{20}} =? \\ $$
Answered by mr W last updated on 07/Mar/22
$${z}={a}+{bi} \\ $$$$\left({z}+\mathrm{3}{i}\right)^{\mathrm{20}} \\ $$$$=\left({a}+\left({b}+\mathrm{3}\right){i}\right)^{\mathrm{20}} \\ $$$$=\left({a}^{\mathrm{2}} +\left({b}+\mathrm{3}\right)^{\mathrm{2}} \right)^{\mathrm{10}} {e}^{{i}\left(\mathrm{20}\:\mathrm{tan}^{−\mathrm{1}} \frac{{b}+\mathrm{3}}{{a}}\right)} \\ $$