Prove that $\arcsin(z)=\frac{\pi}{2}+i\log(z+i\sqrt{1-z^2})$ I'm stuck here:

Prove that $w=\arcsin(z)=\frac{\pi}{2}+i\log(z+i\sqrt{1-z^2})$

By using the definition of $z=\sin(w)$, I found that 
$$\arcsin(z)=\frac{1}{i}\log(iz+\sqrt{1-z^2})$$
But where that $\frac{\pi}{2}$ comes from? How can I rearrange my expression to have $\frac{\pi}{2}+i\log(z+i\sqrt{1-z^2})$?
Thanks for your time.
 A: In order to get to the final expression you need to use the log identity:
$$
\ln(a+b)=\ln(a)+\ln(1+\frac{b}{a})
$$
So in your case:
$$
\frac{1}{i}\left[\ln(iz+\sqrt{1-z^2})\right] = \frac{1}{i}\left[\ln(iz) + \ln\left(1+\frac{\sqrt{1-z^2}}{iz}\right)\right]
$$
We know that when $z>0$, 
$$
\ln(iz)=\frac{i\pi}{2}+\ln(z)
$$
Also, by doing some algebra:
$
\ln\left(1+\frac{\sqrt{1-z^2}}{iz}\right) = \ln\left(1+\frac{i\sqrt{1-z^2}}{(-1)z}\right) = \ln\left(\frac{z}{z}-\frac{i\sqrt{1-z^2}}{z}\right)
$
$
= \ln\left(\frac{z-i\sqrt{1-z^2}}{z}\right)=\ln\left(z-i\sqrt{1-z^2}\right) - \ln(z)
$
Putting it all together:
$
\frac{1}{i}\left[\ln(z)+\frac{i\pi}{2}+\ln(z-i\sqrt{1-z^2}) -\ln(z)\right]
$
$
=\frac{\pi}{2}+\frac{i}{i^2}\ln(z-i\sqrt{1-z^2}) = 
=\frac{\pi}{2}-i\ln(z-i\sqrt{1-z^2})
$
Doing some more algebra:
$
-i\ln(z-i\sqrt{1-z^2}) = i \ln\left(\left(\frac{1}{z-i\sqrt{1-z^2}}\right)\left(\frac{z+i\sqrt{1-z^2}}{z+i\sqrt{1-z^2}}\right)\right)
$
$
= i\ln\left(\frac{z+i\sqrt{1-z^2}}{z^2-i^2(1-z^2)}\right)=i\ln(z+i\sqrt{1-z^2})
$
Hence the expression is true:
$$
\arcsin(z)=\frac{\pi}{2}+i\ln(z+i\sqrt{1-z^2})
$$
