Yesenia Sherman

2022-06-24

How to simplify $\mathrm{\Re }\left[\sqrt{2}{\mathrm{tan}}^{-1}\frac{x}{\sqrt{i}}\right]$?

enfujahl

$\mathrm{\Re }\left[\sqrt{2}{\mathrm{tan}}^{-1}\frac{x}{\sqrt{i}}\right]=\mathrm{\Re }\left[\sqrt{2}{\mathrm{tan}}^{-1}x\cdot {e}^{-i\pi /4}\right]=\frac{\left(\sqrt{2}{\mathrm{tan}}^{-1}x\cdot {e}^{-i\pi /4}\right)+\left(\sqrt{2}{\mathrm{tan}}^{-1}x\cdot {e}^{i\pi /4}\right)}{2}$
$=\frac{1}{\sqrt{2}}\left({\mathrm{tan}}^{-1}x\cdot {e}^{i\pi /4}+{\mathrm{tan}}^{-1}x\cdot {e}^{-i\pi /4}\right)=\frac{1}{\sqrt{2}}{\mathrm{tan}}^{-1}\frac{x\cdot {e}^{i\pi /4}+x\cdot {e}^{-i\pi /4}}{1-x\cdot {e}^{i\pi /4}\cdot x\cdot {e}^{-i\pi /4}}$
$=\frac{1}{\sqrt{2}}{\mathrm{tan}}^{-1}\frac{2x\mathrm{cos}\frac{\pi }{4}}{1-{x}^{2}}=\frac{1}{\sqrt{2}}{\mathrm{tan}}^{-1}\frac{\sqrt{2}x}{1-{x}^{2}}$
where line $2\to 3$ is achieved by using that $\mathrm{tan}\left(x+y\right)=\frac{\mathrm{tan}x+\mathrm{tan}y}{1-\mathrm{tan}x\mathrm{tan}y}$
The differs from the expression you give by a constant, after further noting that ${\mathrm{tan}}^{-1}x+{\mathrm{tan}}^{-1}\left(\frac{1}{x}\right)=\frac{\pi }{2}$ for x>0

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