Let, $ tanh^{-1} (x+iy) \;=\; α+iβ \\ \; \\ \; \\ \therefore α+iβ=tanh^{-1}⁡(x+iy)=\dfrac{i(tanh^{-1}⁡(x+iy)}{i} \\ \; \\ \; \\ = \dfrac{(tanh^{-1}⁡(i(x+iy))}{i} \; \; \; \ldots \{ \because i tanh^{-1}⁡θ= tan^{i⁡θ} \} \\ \; \\ \; \\ \therefore α+iβ \;=\; \dfrac{tanh^{-1}⁡(ix+i^2y)}{i} \;=\; \dfrac{tanh^{-1}⁡(ix-y)}{i} \; \; \; \ldots (i) \\ \; \\ \; \\ We\; can \; prove \; that \; since \; 〖tanh〗^{-1}⁡(x+iy)=α+iβ \; ,hence \; α-iβ=〖tanh〗^{-1}⁡(x-iy) \\ \; \\ \; \\ \therefore α-iβ=tanh^{-1}⁡(x-iy)= i\dfrac{tanh^{-1}⁡(x-iy) }{i} \\ \; \\ \; \\ = {tan^(-1)⁡(i(x-iy)}/{i} \ldots(\because i 〖tanh〗^{-1}⁡θ=〖tan〗^{-1}⁡(iθ) ) \\ \; \\ \; \\ \therefore α-iβ=tan^{-1}⁡ \dfrac{〖(ix-i^2 y)〗}{i}/ \;= \;⁡ \dfrac{tan^{-1}(ix+y)}{i} \; \; \ldots (ii) \\ \; \\ \; \\ $

Adding equations (I) and (II),

$ \therefore 2α=\dfrac{1}{i} [tan^{-1}⁡(ix-y)+tan^{-1}⁡(ix+y) ]=\dfrac{1}{i} tan^{-1} \dfrac{⁡((ix-y)+(ix+y))}{(1-(ix-y)(ix+y) )} \\ \; \\ \; \\ \ldots \bigg(\because〖tan〗^{-1}⁡A+〖tan〗^{-1}⁡B=〖tan〗^{-1}⁡〖(A+B)/(1-AB)〗 \bigg) \\ \; \\ \; \\ \therefore 2α=\dfrac{1}{i} tan^{-1} \dfrac{⁡2ix}{1-(i^2 x^2-y^2)} \;=\; \dfrac{1}{i} tan^{-1} \dfrac{(i(2x)}{(1+x^2+y^2 )} \;=\; \dfrac{i}{i} tan^{-1} \dfrac{2x}{1+x^2+y^2} \;=\; tan^{-1} \dfrac{2x}{1+x^2+y^2} \\ \; \\ \; \\ \therefore α=\dfrac{1}{2} tan^{-1} \dfrac{2x}{1+x^2+y^2} \\ \; \\ \; \\ $

Now subtracting equation (II) from (I),

$ \therefore 2iβ= \dfrac{1}{i} [tan^{-1}⁡(ix-y)-tan^{-1}⁡(ix+y) ]=\dfrac{1}{i} tan^{-1} \dfrac{⁡((ix-y)-(ix+y))}{(1+(ix-y)(ix+y) )} \\ \; \\ \; \\ \ldots \bigg(\because〖tan〗^{-1}⁡A -〖tan〗^{-1}⁡B=〖tan〗^{-1}⁡〖(A-B)/(1+AB)〗 \bigg) \\ \; \\ \; \\ \therefore 2iβ= \dfrac{1}{i} tan^{-1} \dfrac{-2y}{1+(i^2 x^2-y^2)} \;=\; \dfrac{-1}{i} tan^{-1} \dfrac{(2y)}{1+- x^2-y^2} \\ \; \\ \; \\ \therefore 2β= \dfrac{-1}{i^2} tan^{-1}⁡ \dfrac{2y}{1-x^2-y^2}\;=\; tan^{-1}⁡\dfrac{2y}{1-x^2-y^2} \\ \; \\ \; \\ \therefore β= \dfrac{1}{2} tan^{-1}⁡\dfrac{2y}{1-x^2-y^2} \\ \; \\ \; \\ \; \\ \; \\ \therefore tanh^{-1}(x+iy) \;=\; \dfrac{1}{2} tan^{-1}⁡\dfrac{2x}{1+x^2+y^2} + i \bigg[ \dfrac{1}{2} tan^{-1}⁡\dfrac{2y}{1-x^2-y^2} \bigg] $