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Question Number 57719 by Joel578 last updated on 10/Apr/19
Find all complex number z that satisfy  sinh z = i
$$\mathrm{Find}\:\mathrm{all}\:\mathrm{complex}\:\mathrm{number}\:{z}\:\mathrm{that}\:\mathrm{satisfy} \\ $$$$\mathrm{sinh}\:{z}\:=\:{i} \\ $$
Answered by mr W last updated on 10/Apr/19
sinh z=((e^z −e^(−z) )/2)=i  ⇒(e^z )^2 −2i(e^z )−1=0  ⇒(e^z −i)^2 =0  ⇒e^z =i=cos (2nπ+(π/2))+i sin (2nπ+(π/2))=e^((2nπ+(π/2))i)   ⇒z=(2nπ+(π/2))i
$$\mathrm{sinh}\:{z}=\frac{{e}^{{z}} −{e}^{−{z}} }{\mathrm{2}}={i} \\ $$$$\Rightarrow\left({e}^{{z}} \right)^{\mathrm{2}} −\mathrm{2}{i}\left({e}^{{z}} \right)−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\left({e}^{{z}} −{i}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{e}^{{z}} ={i}=\mathrm{cos}\:\left(\mathrm{2}{n}\pi+\frac{\pi}{\mathrm{2}}\right)+{i}\:\mathrm{sin}\:\left(\mathrm{2}{n}\pi+\frac{\pi}{\mathrm{2}}\right)={e}^{\left(\mathrm{2}{n}\pi+\frac{\pi}{\mathrm{2}}\right){i}} \\ $$$$\Rightarrow{z}=\left(\mathrm{2}{n}\pi+\frac{\pi}{\mathrm{2}}\right){i} \\ $$

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