Real part of Arcsine of a Complex variable How do we show: $\operatorname{Re}({\arcsin{z}}) = \frac{1}{2}(\sqrt{x^2+y^2+2x+1} -\sqrt{x^2+y^2-2x+1})$ ?
None of the things I tried including using the formula of inverse sine seems to work... Please Help

Solving the equation $(\cosh^2 v)^2-(x^2+y^2+1)\cosh^2 v +x^2=0\,\,$  we have
$$
\cosh ^2v=\frac{1}{2}\left(x^2+y^2+1+\sqrt{(x^2+y^2+1)^2-4x^2}\right)$$
since $\cosh^2 v=1+\sinh^2 v\ge 1$.
Noting $$
\left(\frac{1}{2}\left(\sqrt{(x+1)^2+y^2}+\sqrt{(x-1)^2+y^2}\right)\right)^2=\frac{1}{2}\left(x^2+y^2+1+\sqrt{(x^2+y^2+1)^2-4x^2}\right),$$
we have$$
\cosh v=\frac{1}{2}\left(\sqrt{(x+1)^2+y^2}+\sqrt{(x-1)^2+y^2}\right).$$
Therefore $$\sin u=\frac{x}{\cosh v}=\frac{1}{2}\left(\sqrt{(x+1)^2+y^2}-\sqrt{(x-1)^2+y^2}\right),
$$which ensures $$u=\operatorname{Re}(\arcsin (x+iy))=\arcsin \left(\frac{1}{2}\left(\sqrt{(x+1)^2+y^2}-\sqrt{(x-1)^2+y^2}\right)\right).$$
