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Question Number 197794 by universe last updated on 28/Sep/23

$$\mathrm{if}\mathrm{x}=\log \tan (\frac{\pi }{4}+\frac{\mathrm{y}}{2}),\mathrm{prove}\mathrm{that}$$$$\mathrm{y}=-i\log \tan (\frac{ix}{2}+\frac{\pi }{4})\mathrm{here}i=\sqrt{-1}$$

Answered by Frix last updated on 29/Sep/23

$$x=\ln \tan \frac{2y+\pi }{4}\iff y=-\frac{\pi }{2}+{2\tan }^{-1}{\mathrm{e}}^x$$$$-\frac{\pi }{2}+{2\tan }^{-1}{\mathrm{e}}^x=-\mathrm{i}\ln \tan \frac{\pi +2\mathrm{i}x}{4}$$$$(-\frac{\pi }{2}+{2\tan }^{-1}{\mathrm{e}}^x)\mathrm{i}=\ln \tan \frac{\pi +2\mathrm{i}x}{4}$$$${\mathrm{e}}^{(-\frac{\pi }{2}+{2\tan }^{-1}{\mathrm{e}}^x)\mathrm{i}}=\tan \frac{\pi +2\mathrm{i}x}{4}$$$$\sin \left({2\tan }^{-1}{\mathrm{e}}^x\right)-\mathrm{i}\cos \left({2\tan }^{-1}{\mathrm{e}}^x\right)=\tan \frac{\pi +2\mathrm{i}x}{4}$$$$\frac{{2\mathrm{e}}^x}{{\mathrm{e}}^{2x}+1}+\frac{{\mathrm{e}}^{2x}-1}{{\mathrm{e}}^{2x}+1}\mathrm{i}=\tan \frac{\pi +2\mathrm{i}x}{4}$$$$\frac{1}{\cosh x}+\mathrm{i}\tanh x=\tan \frac{\pi +2\mathrm{i}x}{4}$$$$\tan (a+b\mathrm{i})=\frac{{2\mathrm{e}}^{2b}\sin 2a+({\mathrm{e}}^{4b}-1)\mathrm{i}}{{\mathrm{e}}^{4b}+{2\mathrm{e}}^{2b}\cos 2a+1}\stackrel{a=\frac{\pi }{4}}{=}$$$$=\frac{1}{\cosh 2b}+\mathrm{i}\tanh 2b\stackrel{b=\frac{x}{2}}{=}$$$$=\frac{1}{\cosh x}+\mathrm{i}\tanh x$$

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