TITLE: Where are the zeros of the complex sine and cosine?
QUESTION [8 upvotes]: Do sin(z) and cos(z) have any zeroes where the imaginary part of z is non-zero? How could I prove that (or show that it's reasonable)?

REPLY [12 votes]: We can use 
$$\sin z=\frac{e^{iz}-e^{-iz}}{2i}$$
Set this equal to $0$.  A little manipulation yields $(e^{iz})^2=1$.  
If the imaginary part of $z$ is non-zero, the norm of $e^{iz}$ is greater than $1$, contradicting the fact that $(e^{iz})^2=1$.  A mild variant of the same argument works for $\cos z$.

REPLY [4 votes]: There is none. It follows, for example, from the Weierstrass products
$$
\cos z=\prod_{n=1}^\infty \left(1-\frac{4z^2}{\pi^2(2n-1)^2}\right),
$$
$$
\sin z=z\prod_{n=1}^\infty \left(1-\frac{z^2}{(\pi n)^2}\right),
$$
which are valid for all $z\in \mathbb C$.