Question

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

Solution

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

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