Question Number 198302 by sonukgindia last updated on 17/Oct/23
Answered by mr W last updated on 17/Oct/23
$${z}_{\mathrm{1}} =\mathrm{1}+{i}\sqrt{\mathrm{3}}=\mathrm{2}{e}^{{i}\left(\frac{\pi}{\mathrm{3}}\right)} \\ $$$${z}_{\mathrm{2}} =\mathrm{2}{e}^{{i}\left(\frac{\pi}{\mathrm{3}}+\frac{\mathrm{2}\pi}{\mathrm{3}}\right)} =\mathrm{2}{e}^{{i}\pi} =−\mathrm{2} \\ $$$${z}_{\mathrm{2}} =\mathrm{2}{e}^{{i}\left(\frac{\pi}{\mathrm{3}}−\frac{\mathrm{2}\pi}{\mathrm{3}}\right)} =\mathrm{2}{e}^{{i}\left(−\frac{\pi}{\mathrm{3}}\right)} =\mathrm{1}−{i}\sqrt{\mathrm{3}} \\ $$