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


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:

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


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