Question Number 100442 by Dwaipayan Shikari last updated on 26/Jun/20
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\:\:\:\:\mathrm{log}\left(−\mathrm{2}\right)\:\:\left\{\mathrm{imaginary}\right\} \\ $$
Commented by mathmax by abdo last updated on 26/Jun/20
$$\mathrm{you}\:\mathrm{are}\:\mathrm{welcome}. \\ $$
Commented by Dwaipayan Shikari last updated on 26/Jun/20
$$\mathrm{log}\left(−\mathrm{2}\right)=\mathrm{log}\left(−\mathrm{1}\right)+\mathrm{log}\left(\mathrm{2}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{log}\left(\mathrm{e}^{\mathrm{i}\pi} \right)+\mathrm{log}\left(\mathrm{2}\right)\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{log}\left(\mathrm{e}^{\mathrm{i}\left(\mathrm{2k}\pi+\pi\right)} \right)+\mathrm{log}\left(\mathrm{2}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{i}\pi\left(\mathrm{2k}+\mathrm{1}\right)+\mathrm{log}\left(\mathrm{2}\right)\:\:\:\:\left\{\mathrm{k}\in\mathbb{Z}\right\} \\ $$$$\mathrm{Is}\:\:\mathrm{this}\:\mathrm{a}\:\mathrm{solution}??????? \\ $$$$\mathrm{I}\:\mathrm{am}\:\mathrm{not}\:\mathrm{sure} \\ $$$$ \\ $$
Commented by mr W last updated on 26/Jun/20
$${i}\:{think}\:{it}'{s}\:{correct}. \\ $$
Commented by abdomsup last updated on 26/Jun/20
$${ln}\left(−\mathrm{2}\right)\:={ln}\left(−\mathrm{1}\right)+{ln}\left(\mathrm{2}\right) \\ $$$${ln}\left({e}^{{i}\left(\pi+\mathrm{2}{k}\pi\right)} \right)+{ln}\left(\mathrm{2}\right) \\ $$$$={ln}\mathrm{2}\:+{i}\left(\mathrm{2}{k}+\mathrm{1}\right)\pi\:\:\:\:\left({k}\in{Z}\right) \\ $$
Commented by Dwaipayan Shikari last updated on 26/Jun/20
$$\mathrm{Thanking}\:\mathrm{you}\:\mathrm{for}\:\mathrm{your}\:\mathrm{confirmation}\:\mathrm{Mr}.\mathrm{W}\:\mathrm{sir}\:\mathrm{and}\:\mathrm{Abdomsup}\:\mathrm{sir} \\ $$