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Question Number 50849 by peter frank last updated on 21/Dec/18

$$\mathrm{solve}forzintheformx+iy$$$$iftanz=0.5$$

Answered by ajfour last updated on 21/Dec/18

$$e^{iz}=\cos z+i\sin z$$$$e^{-iz}=\cos z-i\sin z$$$$\tan z=\frac{e^{iz}-e^{-iz}}{i(e^{iz}+e^{-iz})}$$$$let{e}^{iz}=e^{-y+ix}=e^{-y}(\cos x+i\sin x)$$$$\Rightarrow e^{-iz}=e^{y-ix}=e^y(\cos x-i\sin x)$$$$\Rightarrow \tan z=\frac{(e^{-y}-e^y)\cos x+i(e^{-y}+e^y)\sin x}{i[(e^{-y}+e^y)\cos x+i(e^{-y}-e^y)\sin x]}$$$$let{e}^{-y}+e^y=u,e^{-y}-e^y=v$$$$\tan z=\frac{(-iv\cos x+u\sin x)(u\cos x-iv\sin x)}{u^2\cos {}^{2}x+v^2\sin {}^{2}x}$$$$\tan z=\frac{-iuv+(u^2-v^2)\sin 2x}{2(u^2\cos {}^{2}x+v^2\sin {}^{2}x)}$$$$if\tan z=\frac{1}{2},\Rightarrow$$$$uv=0or{e}^{-2y}=e^{2y}$$$$\Rightarrow y=0$$$$\Rightarrow u=2,v=0$$$$\frac{(u^2-v^2)\sin 2x}{2(u^2\cos {}^{2}x+v^2\sin {}^{2}x)}=\frac{1}{2}$$$$\Rightarrow \tan x=\frac{1}{2}$$$$\Rightarrow \mathit{z}=x={\tan }^{-1}\frac{1}{2}+n\pi .$$

Commented by peter frank last updated on 21/Dec/18

$$ans[z=\frac{\pi \mathrm{n}}{2}+26.56+0.00000119i]$$

Commented by MJS last updated on 21/Dec/18

$$\mathrm{sorry}\mathrm{but}\mathrm{this}\mathrm{is}\mathrm{wrong}.\mathrm{possibly}\mathrm{the}\mathrm{calculator}$$$$\mathrm{approximates}\mathrm{in}\mathrm{a}\mathrm{crazy}\mathrm{way}.$$$$\tan (a+b\mathrm{i})=\frac{1}{2}$$$$\mathrm{has}\mathrm{only}\mathrm{real}\mathrm{solutions}\Rightarrow b=0$$

Commented by peter frank last updated on 22/Dec/18

$$thankyoubothsirs$$

Answered by MJS last updated on 21/Dec/18

$$\tan (x+\mathrm{i}y)=\frac{1}{2};\left(\begin{array}{c}x\\ y\end{array}\right)\in {\mathbb{R}}^2$$$$x+\mathrm{i}y=n\pi +\arctan \frac{1}{2};n\in \mathbb{Z}$$$$\Rightarrow y=0;x=n\pi +\arctan \frac{1}{2};n\in \mathbb{Z}$$

Answered by peter frank last updated on 21/Dec/18

$$\tanh \left(iz\right)=itanz$$$$\frac{1}{2}=\frac{\tanh \left(iz\right)}{i}$$$$\frac{1}{2}=\frac{e^{iz}-e^{-iz}}{i(e^{iz}+e^{-iz})}$$$$multiplyby{e}^{iz}andsimplify$$$$e^{2iz}=\frac{3}{5}+\frac{4i}{5}$$$$z=x+iy$$$$e^{2ix}.e^{-2y}=\frac{3}{5}+\frac{4i}{5}$$$$from{Eurel}^{\prime }sformof$$$$complexno$$$$e^{2xi}=\cos 2x+i\sin 2x$$$$e^{-2y}(\cos 2x+i\sin 2x)=\frac{3}{4}+\frac{4i}{5}$$$$comparerealandImmarginary$$$$e^{-2y}\cos 2x=\frac{3}{5}....\left(i\right)$$$$e^{-2y}\sin 2x=\frac{4}{5}....\left(ii\right)$$$$divideiibyi$$$$tan2x=\frac{4}{3}$$$$tan2x=\tan (53.13010235)\approx \tan 53.13$$$$2x=\pi n+53.{13}^{°}$$$$x=\frac{\pi n}{2}+26.565$$$$n=0,1,2,3$$$$....$$$$from$$$$e^{-2y}\cos 2x=\frac{3}{5}....\left(i\right)$$$$e^{-2y}=\frac{0.6}{\cos 2x}$$$$-2y=ln\left(\frac{0.6}{\cos 2x}\right)$$$$y=\frac{1}{2}ln\left(\frac{0.6}{\cos 2x}\right)$$$$z=x+iy$$$$z=\frac{\pi n}{2}+26.565-\frac{1}{2}ln\left(\frac{0.6}{\cos 2x}\right)$$$$......$$

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