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.
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:
$$ \arcsin(z)=\frac{\pi}{2}+i\ln(z+i\sqrt{1-z^2}) $$