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Question Number 68207 by mr W last updated on 07/Sep/19

$$solveforx\in \mathbb{C}$$$$\sin x=z(z=a+bi={re}^{i\theta })$$

Commented by mathmax by abdo last updated on 07/Sep/19

$$sinx=z\iff \frac{e^{ix}-e^{-ix}}{2i}=z\iff e^{ix}-e^{-ix}=2ir{e}^{i\theta }$$$$lett=e^{ix}\Rightarrow ix=ln\left(t\right)+i2k\pi withk\in Z$$$$\left(e\right)\Rightarrow t-t^{-1}=2ir{e}^{i\theta }\Rightarrow t^2-1=2ir{e}^{i\theta }t\Rightarrow$$$$t^2-2{ire}^{i\theta }t-1=0\to {\mathrm{\Delta }}^{{}^{\prime }}={\left({ire}^{i\theta }\right)}^2+1=-r^2{e}^{2i\theta }+1$$$$t_1=ir{e}^{i\theta }+\sqrt{1-r^2{e}^{2i\theta }}and{t}_2=ir{e}^{i\theta }-\sqrt{1-r^2{e}^{2i\theta }}$$$$ix=ln\left(t_1\right)+i2k\pi \Rightarrow -x={ilnt}_1-2k\pi \Rightarrow x=-{ilnt}_1+2k\pi$$$$=-iln(r{e}^{i\theta }-i\sqrt{1-r^2{e}^{2i\theta }})+2k\pi$$$$ix={lnt}_2+i2k\pi \Rightarrow -x={ilnt}_2-2k\pi \Rightarrow x=-{ilnt}_2+2k\pi$$$$=-iln(ir{e}^{i\theta }-\sqrt{1-r^2{e}^{2i\theta }})+2k\pi$$

Answered by Smail last updated on 07/Sep/19

$$\frac{e^{ix}-e^{-ix}}{2}{e}^{-i\pi /2}={re}^{i\theta }$$$$e^{2ix}-1-2{re}^{ix}\times e^{i(\theta +\pi /2)}=0$$$${(e^{ix}-{re}^{i(\theta +\pi /2)})}^2=1+r^2{e}^{2i(\theta +\pi /2)}$$$$e^{ix}=\underset{-}{+}{(1+r^2{e}^{2i(\theta +\pi /2)})}^{1/2}+{re}^{i(\theta +\pi /2)}$$$$x=-iln\left({re}^{i(\theta +\pi /2)}\underset{-}{+}\sqrt{1+r^2{e}^{2i(\theta +\pi /2)}}\right)$$$$x=-{isinh}^{-1}\left({re}^{i(\theta +\pi /2)}\right)$$

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