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Question Number 197794 by universe last updated on 28/Sep/23
if x = log tan((π/4)+(y/2)), prove that y = −ilog tan(((ix)/2) + (π/4)) here i = (√(−1))
$$\:\:\:\:\mathrm{if}\:\mathrm{x}\:\:\:=\:\:\:\mathrm{log}\:\mathrm{tan}\left(\frac{\pi}{\mathrm{4}}+\frac{\mathrm{y}}{\mathrm{2}}\right),\:\:\mathrm{prove}\:\mathrm{that} \\ $$$$\:\:\:\:\:\mathrm{y}\:\:\:\:=\:\:\:−{i}\mathrm{log}\:\mathrm{tan}\left(\frac{{ix}}{\mathrm{2}}\:+\:\frac{\pi}{\mathrm{4}}\right)\:\:\:\:\:\mathrm{here}\:{i}\:\:=\:\sqrt{−\mathrm{1}} \\ $$
Answered by Frix last updated on 29/Sep/23
x=ln tan ((2y+π)/4) ⇔ y=−(π/2)+2tan^(−1) e^x −(π/2)+2tan^(−1) e^x =−i ln tan ((π+2ix)/4) (−(π/2)+2tan^(−1) e^x )i=ln tan ((π+2ix)/4) e^((−(π/2)+2tan^(−1) e^x )i) =tan ((π+2ix)/4) sin (2tan^(−1) e^x ) −i cos (2tan^(−1) e^x ) =tan ((π+2ix)/4) ((2e^x )/(e^(2x) +1))+((e^(2x) −1)/(e^(2x) +1))i=tan ((π+2ix)/4) (1/(cosh x))+i tanh x =tan ((π+2ix)/4) tan (a+bi) =((2e^(2b) sin 2a +(e^(4b) −1)i)/(e^(4b) +2e^(2b) cos 2a +1)) =^(a=(π/4)) =(1/(cosh 2b))+i tanh 2b =^(b=(x/2)) =(1/(cosh x))+i tanh x
$${x}=\mathrm{ln}\:\mathrm{tan}\:\frac{\mathrm{2}{y}+\pi}{\mathrm{4}}\:\Leftrightarrow\:{y}=−\frac{\pi}{\mathrm{2}}+\mathrm{2tan}^{−\mathrm{1}} \:\mathrm{e}^{{x}} \\ $$$$−\frac{\pi}{\mathrm{2}}+\mathrm{2tan}^{−\mathrm{1}} \:\mathrm{e}^{{x}} \:=−\mathrm{i}\:\mathrm{ln}\:\mathrm{tan}\:\frac{\pi+\mathrm{2i}{x}}{\mathrm{4}} \\ $$$$\left(−\frac{\pi}{\mathrm{2}}+\mathrm{2tan}^{−\mathrm{1}} \:\mathrm{e}^{{x}} \right)\mathrm{i}=\mathrm{ln}\:\mathrm{tan}\:\frac{\pi+\mathrm{2i}{x}}{\mathrm{4}} \\ $$$$\mathrm{e}^{\left(−\frac{\pi}{\mathrm{2}}+\mathrm{2tan}^{−\mathrm{1}} \:\mathrm{e}^{{x}} \right)\mathrm{i}} =\mathrm{tan}\:\frac{\pi+\mathrm{2i}{x}}{\mathrm{4}} \\ $$$$\mathrm{sin}\:\left(\mathrm{2tan}^{−\mathrm{1}} \:\mathrm{e}^{{x}} \right)\:−\mathrm{i}\:\mathrm{cos}\:\left(\mathrm{2tan}^{−\mathrm{1}} \:\mathrm{e}^{{x}} \right)\:=\mathrm{tan}\:\frac{\pi+\mathrm{2i}{x}}{\mathrm{4}} \\ $$$$\frac{\mathrm{2e}^{{x}} }{\mathrm{e}^{\mathrm{2}{x}} +\mathrm{1}}+\frac{\mathrm{e}^{\mathrm{2}{x}} −\mathrm{1}}{\mathrm{e}^{\mathrm{2}{x}} +\mathrm{1}}\mathrm{i}=\mathrm{tan}\:\frac{\pi+\mathrm{2i}{x}}{\mathrm{4}} \\ $$$$\frac{\mathrm{1}}{\mathrm{cosh}\:{x}}+\mathrm{i}\:\mathrm{tanh}\:{x}\:=\mathrm{tan}\:\frac{\pi+\mathrm{2i}{x}}{\mathrm{4}} \\ $$$$\:\:\:\:\:\mathrm{tan}\:\left({a}+{b}\mathrm{i}\right)\:=\frac{\mathrm{2e}^{\mathrm{2}{b}} \mathrm{sin}\:\mathrm{2}{a}\:+\left(\mathrm{e}^{\mathrm{4}{b}} −\mathrm{1}\right)\mathrm{i}}{\mathrm{e}^{\mathrm{4}{b}} +\mathrm{2e}^{\mathrm{2}{b}} \mathrm{cos}\:\mathrm{2}{a}\:+\mathrm{1}}\:\overset{{a}=\frac{\pi}{\mathrm{4}}} {=} \\ $$$$\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{cosh}\:\mathrm{2}{b}}+\mathrm{i}\:\mathrm{tanh}\:\mathrm{2}{b}\:\overset{{b}=\frac{{x}}{\mathrm{2}}} {=} \\ $$$$\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{cosh}\:{x}}+\mathrm{i}\:\mathrm{tanh}\:{x} \\ $$

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