Question Number 130742 by mohammad17 last updated on 28/Jan/21
Answered by Dwaipayan Shikari last updated on 28/Jan/21
$${z}=\mathrm{9}+\mathrm{3}{i}\:\:=\sqrt{\mathrm{9}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} }\:{e}^{{itan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{3}}} =\mathrm{3}\sqrt{\mathrm{10}}{e}^{{itan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{3}}} \\ $$$${z}=\mathrm{7}+\mathrm{2}{i}=\sqrt{\mathrm{53}}\:{e}^{{itan}^{−\mathrm{1}} \frac{\mathrm{2}}{\mathrm{7}}} \\ $$$${z}=\mathrm{2}+\mathrm{6}{i}=\mathrm{2}\sqrt{\mathrm{10}}\:{e}^{{itan}^{−\mathrm{1}} \mathrm{3}} \:\: \\ $$