Q:

Hyperbolic Trigonometry with Complex Numbers

I was trying to show the following complex hyperbolic trigeometric relation
$$\mathrm{arctanh}(x+iy)=\frac{1}{2}\mathrm{arctanh}\left(\frac{2x}{1+x^2+y^2}\right)+\frac{i}{2}\arctan\left(\frac{2y}{1-x^2-y^2}\right),$$
where $x,y$ are real numbers. The formula helps to separate the real and imaginary parts of the hyperbolic trigeometric function of a complex argument $z=x+iy$. I know it is correct because wolfram alpha gives practically numerical zero. I thought of making use of the relations such as $$\mathrm{tanh}(iy)=i \tan(y),$$ and $$\mathrm{tanh}(x+y)=\frac{\mathrm{tanh}(x)+\mathrm{tanh}(y)}{1+\mathrm{tanh}(x)\mathrm{tanh}(y)}.$$ In fact I tried to multiply 2 on both sides of the expression and then evaluate the $\mathrm{tanh}$ of them, and I end up in the following expression 
$$\frac{2 x + i 2 y}{1 + x^2 - y^2 + (2 i) (x y)} - \frac{ 2x/(1 + x^2 + y^2) + i 2 y/(1 - x^2 - y^2)}{1 + ( 2x/(1 + x^2 + y^2))  ( i2y/(1 - x^2 - y^2))}$$
which gives zero. Is there a straightforward way to show the validity of the above formula?
Edit: As pointed out by user5713492, there was a algebraic mistake in the original question 

A:

The following answer is based on Kf-Sansoo's hint. It is noted that $$\mathrm{tanh}^{-1}(z)=\frac{1}{2} \ln \frac{1+z}{1-z},$$ which can be obtained by reverting $$z \equiv \mathrm{tanh}(w)=\frac{e^{2w}-1}{e^{2w}+1}.$$
Then, one can calculate $u+\bar{u}$ to obtain the real part of $u=\mathrm{tanh}^{-1}(z)$. Explicit calculations give $$2\mathbf{Re}(u)=u+\bar{u}=\frac{1}{2} \ln \frac{1+z}{1-z}+\frac{1}{2} \ln \frac{1+\bar{z}}{1-\bar{z}}=\frac{1}{2} \ln \frac{1+\frac{2x}{1+x^2+y^2}}{1-\frac{2x}{1+x^2+y^2}}=\mathrm{tanh}^{-1}(\frac{2x}{1+x^2+y^2}).$$ Similarly, one has $$2\mathbf{Im}(u)=u-\bar{u}=\frac{1}{2} \ln \frac{1+z}{1-z}-\frac{1}{2} \ln \frac{1+\bar{z}}{1-\bar{z}}=\frac{1}{2} \ln \frac{1+\frac{i2y}{1-x^2-y^2}}{1-\frac{i2y}{1-x^2-y^2}}=i\mathrm{tan}^{-1}(\frac{2y}{1-x^2-y^2}).$$ This yields the desired result.

