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Question Number 100442 by Dwaipayan Shikari last updated on 26/Jun/20

$$\mathrm{Find}\mathrm{the}\mathrm{value}\mathrm{of}\log (-2)\left\{\mathrm{imaginary}\right\}$$

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

$$\log (-2)=\log (-1)+\log \left(2\right)$$$$=\log \left({\mathrm{e}}^{\mathrm{i}\pi }\right)+\log \left(2\right)$$$$=\log \left({\mathrm{e}}^{\mathrm{i}(2\mathrm{k}\pi +\pi )}\right)+\log \left(2\right)$$$$=\mathrm{i}\pi (2\mathrm{k}+1)+\log \left(2\right)\{\mathrm{k}\in \mathbb{Z}\}$$$$\mathrm{Is}\mathrm{this}\mathrm{a}\mathrm{solution}???????$$$$\mathrm{I}\mathrm{am}\mathrm{not}\mathrm{sure}$$$$$$

Commented by mr W last updated on 26/Jun/20

$$ithink{it}^{\prime }scorrect.$$

Commented by abdomsup last updated on 26/Jun/20

$$ln(-2)=ln(-1)+ln\left(2\right)$$$$ln\left(e^{i(\pi +2k\pi )}\right)+ln\left(2\right)$$$$=ln2+i(2k+1)\pi (k\in Z)$$

Commented by Dwaipayan Shikari last updated on 26/Jun/20

$$\mathrm{Thanking}\mathrm{you}\mathrm{for}\mathrm{your}\mathrm{confirmation}\mathrm{Mr}.\mathrm{W}\mathrm{sir}\mathrm{and}\mathrm{Abdomsup}\mathrm{sir}$$

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