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Question Number 116560 by Bird last updated on 04/Oct/20

$$ifarctan(x+iy)=a+ib$$$$withaandbrealsdetermine$$$$aandb$$

Commented by MJS_new last updated on 05/Oct/20

$$\arctan (x+\mathrm{i}y)=a+\mathrm{i}b$$$$a=\frac{\pi }{2}\mathrm{sign}x+\frac{1}{2}(\arctan \frac{y-1}{x}-\arctan \frac{y+1}{x})$$$$b=\frac{1}{4}\ln \frac{x^2+{(y+1)}^2}{x^2+{(y-1)}^2}$$$$\mathrm{if}x=0$$$$a=\mathrm{real}\left(\frac{\mathrm{i}}{2}\ln \frac{1+y}{1-y}\right)$$$$b=\mathrm{imag}\left(\frac{\mathrm{i}}{2}\ln \frac{1+y}{1-y}\right)$$

Answered by Olaf last updated on 05/Oct/20

$$x+iy=\tan (a+ib)$$$$x+iy=\frac{\tan a+\tan ib}{1-\tan a\tan ib}$$$$$$$$\tan ib=\frac{\sin ib}{\cos ib}$$$$\tan ib=\frac{\frac{e^{i\left(ib\right)}-e^{-i\left(ib\right)}}{2i}}{\frac{e^{i\left(ib\right)}+e^{-i\left(ib\right)}}{2}}=-i\frac{e^{-b}-e^{ib}}{e^{-ib}+e^{ib}}$$$$\tan ib=i\frac{e^b-e^{-ib}}{e^{ib}+e^{-ib}}=i\frac{\sinh b}{\cosh b}=i\tanh b$$$$$$$$x+iy=\frac{\tan a+i\tanh b}{1-i\tan a\tanh b}$$$$x+iy=\frac{(\tan a+i\tanh b)(1+i\tan a\tanh b)}{1+{\tan }^{2}a{\tanh }^{2}b}$$$$x+iy=\frac{\tan a(1-{\tanh }^{2}b)+i\tanh b(1+{\tan }^{2}a)}{1+{\tan }^{2}a{\tanh }^{2}b}$$$$$$$$\frac{y}{x}=\frac{\tanh b(1+{\tan }^{2}a)}{\tan a(1-{\tanh }^{2}b)}$$$$\frac{y}{x}=\frac{\tanh b{\cosh }^{2}b}{\tan a{\cos }^{2}a}=\frac{\sinh b\cosh b}{\sin a\cos a}$$$$\frac{y}{x}=\frac{\tanh b{\cosh }^{2}b}{\tan a{\cos }^{2}a}=\frac{\sinh 2b}{\sin 2a}\left(1\right)$$$$$$$$x^2+y^2=\frac{{\tan }^{2}a}{{\cosh }^{4}b}+\frac{{\tanh }^{2}b}{{\cos }^{4}a}$$$$x^2+y^2=\frac{{\sin }^{2}a{\cos }^{2}a+{\sinh }^{2}b{\cosh }^{2}b}{{\cos }^{4}a{\cosh }^{4}b}$$$$x^2+y^2=\frac{{\sin }^{2}2a+{\sinh }^{2}2b}{{4\cos }^{4}a{\cosh }^{4}b}\left(2\right)$$$$\mathrm{work}\mathrm{in}\mathrm{progress}...$$$$$$

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