What is $y$ for $\sin(x+iy) = 2$

Question

Decide $y$ For a certain complex number $z = x+ iy$ so that $$\sin(z) = 2?$$

What I think to have realised is that I need to write it in the form $$e^{x+iy}$$ but can't seem to find a way how.

Answer 1

Hint:

You may use

$$\sin(x+iy)=\sin x\cos(iy)+\cos x\sin(iy)=\sin x\cosh y+i\cos x\sinh y.$$

Then $\cos x=0,\sin x=\pm1$ (reject $-$), $y=\pm\text{arcosh}(2)$.

Answer 2

You have to solve $$ \frac{e^{iz}-e^{-iz}}{2i}=2 $$ where $z=x+iy$. This easily becomes $$ e^{2iz}-4ie^{iz}-1=0 $$ which is a quadratic in $e^{iz}$: $t^2-4it-1=0$ has roots $i(2\pm\sqrt{3})$,

We have $e^{iz}=e^{ix-y}$, so $|e^{iz}|=e^{-y}$ and we have $$ e^{-y}=2+\sqrt{3}\qquad\text{or}\qquad e^{-y}=2-\sqrt{3} $$