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Question Number 171421 by mokys last updated on 14/Jun/22

Prove that arg(e^z )= Im(z) + 2kπ ,k=0,±1,±2,...(√)

$${Prove}\:{that}\:{arg}\left({e}^{{z}} \right)=\:{Im}\left({z}\right)\:+\:\mathrm{2}{k}\pi\:,{k}=\mathrm{0},\pm\mathrm{1},\pm\mathrm{2},...\sqrt{} \\ $$

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