How does one find $z\in \mathbb{C}$ such that $\sin z=100?$ I am self-studying Complex Analysis and I am suppose to find $z\in \mathbb{C}$ such that $\sin z=100.$ I know that $$\sin z=\sin x \cosh y+i\cos x\sinh y$$
So I must have $\sin x \cosh y=100.$ I looked in Wolfram and I found that $y=i(x-\sin^{-1}(100))$.  I was not able to solve this question. How does one find that solution? If fact, I was not expecting to find $y$ in function of $x$.
 A: You're looking at the wrong term first. Your goal is
$$ 100 = \sin x \cosh y + i \cos x\sinh y$$
The most fruitful thing we can do here is to look at the imaginary part of each side which gives us $0=\cos x\sinh y$. A product can only be zero if at least one of its factors are, so we have either $\cos x = 0$ or $\sinh y = 0$. The latter possibility is true only for $y=0$ -- when $y=0$ the sine of $x+iy$ is indeed real, but it is also between $-1$ and $1$, so we won't find a solution there.
So we must have $\cos x=0$ which implies $\sin x = \pm 1$. Then look at the real part which now collapses into $100 = \pm \cosh y$ which you can now solve for $y$. It turns out that $\pm$ must be $+$ and $y$ must be $\pm\cosh^{-1}(100)$.
Now put your $x$ and $y$ together to find
$$ z = \frac{\pi}2 + 2\pi k \pm i\cosh^{-1}(100) \qquad k\in\mathbb Z $$
The hyperbolic arccosine happens to be expressible in terms of simpler functions, so we can also write
$$ z = \frac{\pi}2 + 2\pi k \pm i\log\left(100+\sqrt{9999}\right) \qquad k\in\mathbb Z $$
A: Rewrite the sine function using complex exponentials,
$$\sin z = \frac{e^{iz}-e^{-iz}}{2i}.$$
If you then multiply both sides by $e^{iz}$ you'll have a quadratic equation in the variable $w=e^{iz}$...
