Question Number 130619 by mohammad17 last updated on 27/Jan/21
$${solve}\:{the}\:{equation}\:{sinz}={i}\:? \\ $$
Answered by Dwaipayan Shikari last updated on 27/Jan/21
$$\frac{{e}^{{iz}} −{e}^{−{iz}} }{\mathrm{2}{i}}={i}\Rightarrow{e}^{\mathrm{2}{iz}} +\mathrm{2}{e}^{{iz}} −\mathrm{1}=\mathrm{0}\Rightarrow{e}^{{iz}} =\frac{−\mathrm{2}\pm\sqrt{\mathrm{8}}}{\mathrm{2}} \\ $$$${iz}={log}\left(−\mathrm{1}\pm\sqrt{\mathrm{2}}\right)\Rightarrow{z}=−{ilog}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right) \\ $$$${iz}={log}\left(−\mathrm{1}\right)+{log}\left(\sqrt{\mathrm{2}}+\mathrm{1}\right) \\ $$$${z}=\frac{\mathrm{4}{k}+\mathrm{1}}{\mathrm{2}}\pi−{ilog}\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)\:\:\left({k}\in\mathbb{Z}\right) \\ $$