Solution:

$Given: \tanh^{-1} (x+iy) $

$\tanh^{-1} x= \dfrac 1 2 \log \bigg[\dfrac {(1+z)}{(1-z)}\bigg]$

$\tanh^{-1}( x+iy)= \dfrac 1 2 \log \bigg[\dfrac {(1+x+iy)}{(1-x-iy)}\bigg]$

$= \dfrac 1 2 \log [ {(1+x+iy)}-\log{(1-x-iy)}]$

$= \dfrac 1 2 \bigg[\dfrac 1 2 \log [ {(1+x)^2+y^2]}+i .\tan^{-1}\bigg(\dfrac {y} {1+x}\bigg)\bigg]$

$= -\dfrac 1 2 \bigg[\dfrac 1 2 \log [ {(1-x)^2+y^2]}-i .\tan^{-1}\bigg(\dfrac {y} {1-x}\bigg)\bigg]$

$= \dfrac 1 2 \bigg[\dfrac 1 2 \log \bigg[ \dfrac{(1+x)^2+y^2]}{(1-x)^2+y^2]}+i \bigg[\tan^{-1}\dfrac {y} {1+x}+\tan^{-1}\dfrac {y} {1-x}\bigg]$

$= \dfrac 1 2 \bigg[\dfrac 1 2 \log \bigg[ \dfrac{(1+x)^2+y^2]}{(1-x)^2+y^2]}+i.\tan^{-1} \bigg[\dfrac {y} {1+x}+\dfrac {y} {1-x}/1-\dfrac{y}{1+x}\times\dfrac{y}{1-x} \bigg]$

$= \dfrac {1} {2} \bigg[\dfrac 1 2 \log \bigg( \dfrac{(1+x)^2+y^2}{(1-x)^2+y^2}\bigg)+i.\tan^{-1} \bigg(\dfrac {2y} {1-x^2-y^2}\bigg)\bigg]$

Separate real and imaginary parts,

Real parts: $= \dfrac 1 4 \log \bigg( \dfrac{(1+x)^2+y^2}{(1-x)^2+y^2}\bigg)$

Imaginary parts: $ \dfrac {1} {2}\tan^{-1} \bigg(\dfrac {2y} {1-x^2-y^2}\bigg)$