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Question Number 57719 by Joel578 last updated on 10/Apr/19

$$\mathrm{Find}\mathrm{all}\mathrm{complex}\mathrm{number}z\mathrm{that}\mathrm{satisfy}$$$$\sinh z=i$$

Answered by mr W last updated on 10/Apr/19

$$\sinh z=\frac{e^z-e^{-z}}{2}=i$$$$\Rightarrow {\left(e^z\right)}^2-2i\left(e^z\right)-1=0$$$$\Rightarrow {(e^z-i)}^2=0$$$$\Rightarrow e^z=i=\cos (2n\pi +\frac{\pi }{2})+i\sin (2n\pi +\frac{\pi }{2})=e^{(2n\pi +\frac{\pi }{2})i}$$$$\Rightarrow z=(2n\pi +\frac{\pi }{2})i$$

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