Question Number 114409 by bemath last updated on 19/Sep/20
$${find}\:{solution}\:{set}\:{of}\:{equation} \\ $$$$\:\mathrm{sin}\:{x}\:=\:\mathrm{2}\:,\:{x}\:\in\:\mathbb{C}\:. \\ $$
Answered by bobhans last updated on 19/Sep/20
$$\mathrm{sin}\:{x}\:=\:\frac{{e}^{{ix}} −{e}^{−{ix}} }{\mathrm{2}{i}}\:=\:\mathrm{2}\: \\ $$$${e}^{{ix}} −{e}^{−{ix}} \:=\:\mathrm{4}{i}\:,\:{let}\:{e}^{{ix}} \:=\:{z}\: \\ $$$${z}−{z}^{−\mathrm{1}} \:=\:\mathrm{4}{i}\:;\:{z}^{\mathrm{2}} −\mathrm{4}{iz}−\mathrm{1}\:=\:\mathrm{0} \\ $$$${z}\:=\:\frac{\mathrm{4}{i}\pm\sqrt{−\mathrm{16}+\mathrm{4}}}{\mathrm{2}}\:=\:\left(\mathrm{2}\pm\sqrt{\mathrm{3}}\right){i}\: \\ $$$${then}\:{e}^{{ix}} \:=\:\left(\mathrm{2}\pm\sqrt{\mathrm{3}}\right){i}\:=\:\left(\mathrm{2}\pm\sqrt{\mathrm{3}}\right){e}^{{i}.\frac{\pi}{\mathrm{2}}} \\ $$$${ix}\:=\:\mathrm{ln}\:\left(\mathrm{2}\pm\sqrt{\mathrm{3}}\right)\:+{i}\left(\frac{\pi}{\mathrm{2}}+\mathrm{2}{n}\pi\right);\:{n}\in\mathbb{Z} \\ $$$${x}=\left(\mathrm{4}{n}+\mathrm{1}\right)\frac{\pi}{\mathrm{2}}\pm{i}\:\mathrm{ln}\:\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)\:;\:{n}\in\mathbb{Z} \\ $$