Using complex numbers, can sine be greater than $1?$ I worked on a brain teaser that started with $\sin{(z)}=10.$ It then solved for $\cos{(z)}=\sqrt{-99}.$
I tried solving for $z,$ using $z=x+iy$ then $z=ae^{bi}$ but got nowhere. So is this even possible?

Yes, it can.
Let us find such $z$ so that $\sin(z) = 10$.
Firstly, as a consequence of the Euler's formula:
$ e^{iz} = \cos(z) + i \cdot \sin(z)$,
we have the following two identities which hold for all complex numbers $z$:
$$
\sin(z) = \frac{e^{iz} - e^{-iz}}{2i} \\
\cos(z) = \frac{e^{iz} + e^{-iz}}{2}
$$
Then:
$$
\sin(z) = 10 \\
\frac{e^{iz} - e^{-iz}}{2i} = 10 \\
e^{iz} - e^{-iz} - 20i = 0 \\
(e^{iz})^2 - 20i\cdot(e^{iz}) - 1 = 0 \\
$$
Solving the quadratic equation:
$$
\sqrt{D}=\sqrt{(-20i)^2 - 4 \cdot 1 \cdot (-1)} = \sqrt{-396} = 6i\sqrt{11} \\
(e^{iz}) = \frac{-(-20i) \pm \sqrt{D}}{2} = i \cdot (10 \pm 3\sqrt{11}) \\
$$
Complex logarithm:
$$
iz = Ln\left(i \cdot \left(10 \pm 3\sqrt{11}\right)\right) 
= \ln\left(i \cdot \left(10 \pm 3\sqrt{11}\right)\right) +
  i \cdot 2\pi k,
\qquad k \in \mathbb{Z}\\
iz = \ln(i) + \ln\left(10 \pm 3\sqrt{11}\right) + 2\pi ki \\
iz = \ln\left(e^{i\frac{\pi}{2}}\right) + 
  \ln\left(10 \pm 3\sqrt{11}\right) + 2\pi ki \\
iz = i\frac{\pi}{2} + \ln\left(10 \pm 3\sqrt{11}\right) + 2\pi ki \\
$$
The result:
$$
z = \frac{\pi}{2} + 2\pi k - i \cdot \ln\left(10 \pm 3\sqrt{11}\right),
\qquad k \in \mathbb{Z}
$$
$\cos(z) = \sqrt{-99}$ can be solved in exactly the same way, giving the result:
$$
z = \frac{\pi}{2} + 2\pi k - i \cdot \ln\left(3\sqrt{11} \pm 10\right),
\qquad k \in \mathbb{Z}
$$
The idea is taken from this video:
https://www.youtube.com/watch?v=3C_XD_cCeeI
