Adela Brown

## Answered question

2022-01-03

Proving $\frac{i}{2}lsn\frac{x+i}{x-i}=\mathrm{arctan}x$

### Answer & Explanation

usaho4w

Beginner2022-01-04Added 39 answers

HINT:
Let $\mathrm{arctan}x=y⇒x=\mathrm{tan}y$
$\frac{x+i}{x-i}=\frac{\mathrm{sin}y+i\mathrm{cos}y}{\mathrm{sin}y-i\mathrm{cos}y}=\frac{i\left(\mathrm{cos}y-i\mathrm{sin}y\right)}{-i\left(\mathrm{cos}y+i\mathrm{sin}y\right)}=-{e}^{-2iy}$

John Koga

Beginner2022-01-05Added 33 answers

I dont

Vasquez

Skilled2022-01-11Added 457 answers

Identity is :
$\mathrm{arctan}z=\frac{i}{2}\mathrm{ln}\frac{i+z}{i-z}$
Proof :
$\phantom{\rule{0ex}{0ex}}\mathrm{tan}z=\frac{{e}^{iz}-{e}^{-iz}}{i\left({e}^{iz}+{e}^{-iz}\right)}\phantom{\rule{0ex}{0ex}}{e}^{i\frac{i}{2}\mathrm{ln}\frac{i+z}{i-z}}=\left(\frac{i+z}{i-z}{\right)}^{-\frac{1}{2}}\phantom{\rule{0ex}{0ex}}{e}^{-i\frac{i}{2}\mathrm{ln}\frac{i+z}{i-z}}=\left(\frac{i+z}{i-z}{\right)}^{\frac{1}{2}}\phantom{\rule{0ex}{0ex}}\mathrm{tan}\left(\frac{i}{2}\mathrm{ln}\frac{i+z}{i-z}\right)=\frac{1-\frac{i+z}{i-z}}{i\left(1+\frac{i+z}{i-z}\right)}=z$

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