Question

Seperate ${\tan }^{-1}(x+iy)$ in to real and imaginary parts.

Answer 1

Let $\alpha +i\beta ={\tan }^{-1}(x+iy)$
Then $\alpha -i\beta ={\tan }^{-1}(x-iy)$
Adding, we get
$2\alpha ={\tan }^{-1}(x+iy)+{\tan }^{-1}(x-iy)$
$\Rightarrow {\tan }^{-1}\left(\frac{x+iy+x-iy}{1-(x+iy)(x-iy)}\right)$
$\therefore$ Real part $\alpha =\frac{1}{2}{\tan }^{-1}\left(\frac{x+iy+x-iy}{1-(x+iy)(x-iy)}\right)$
Subtracting, $2i\beta ={\tan }^{-1}(x+iy)-{\tan }^{-1}(x-iy)$
$\Rightarrow 2i\beta ={\tan }^{-1}\left(\frac{(x+iy)-(x-iy)}{1+(x+iy)(x-iy)}\right)$
$\Rightarrow 2i\beta ={\tan }^{-1}\left(\frac{2iy}{1+x^2+y^2}\right)$
$\Rightarrow 2i\beta =i\tan h^{-1}\left(\frac{2y}{1+x^2+y^2}\right)$
$\therefore$ Imaginary part $\beta =\frac{1}{2}\tan h^{-1}\left(\frac{2y}{1+x^2+y^2}\right)$