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 $