How to solve $\arcsin i$?

I tried letting $\sin(z) = i$ and then using $\sin(x+iy) = i$ and expanding into $\sin(x)\cos(iy)+\cos(x)\sin(iy) = i$ by the method of comparing coefficients. How do I proceed from here?

Solution 1:

Use the complex exponential definition of $\sin z$:

$$\sin z = {e^{iz}-e^{-iz}\over 2i} = i$$

Then:

$$e^{iz}-e^{-iz} = -2$$

And solve the resulting quadratic.

Solution 2:

You can complete your work using the identities : $$ \cos (ix)=\cosh x \qquad \sin (ix)=i\sinh x $$

so your equation becomes: $$ \sin x \cosh y +i \cos x \sinh y=i $$ equivalent to: $$ \begin{cases} \sin x \cosh y=0\\ \cos x\sinh y=1 \end{cases} $$

since $\cosh y\ne 0 \quad\forall y$, this gives:

$\sin x=0 \rightarrow \cos x=\pm 1 \rightarrow \sinh y=\pm 1$

can you complete from this?