How to solve $\arcsin i$?

Question

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?

Answer 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.

Answer 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?

Answer 3

You can also get an answer via the integral definition of the complex arcsine (multi-)function:

$$\arcsin(i)=\int_0^i{dz\over\sqrt{1-z^2}}$$

One path for the countour integral is $z=iy$ with $0\le y\le 1$, which gives

$$\arcsin(i)=\int_0^1{i\,dy\over\sqrt{1+y^2}}=i\int_0^{\pi/4}{\sec^2\theta\over\sec\theta}d\theta=i\int_0^{\pi/4}\sec\theta\,d\theta=\cdots$$