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

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