TITLE: How do we know that the Sine function has no Non-Real Roots?
QUESTION [19 upvotes]: In this question and answers (How was Euler able to create an infinite product for sinc by using its roots?) we use the fact that the real roots of $f(x)=\sin x$ occur when $x$ is an integer multiple of $\pi$ to obtain an infinite product for $\sin x$ in terms of of its factors.
My question is, how do we know that $f(x)=\sin x$ has no non-real roots?
Thanks in advance.

REPLY [8 votes]: Since $\sin z = \frac{1}{2i} ( e^{iz} - e^{-iz} )$, we see that $\sin z = 0$ if and only if $e^{iz} = e^{-iz}$, or $e^{2iz}=1$. Taking the modulus of both sides, we see that
$$
1 = |1| = |e^{2iz}| = e^{\Re(2iz)} = e^{-2\mathop{\rm Im} z},
$$
which implies that $-2\mathop{\rm Im} z=0$ and thus $\mathop{\rm Im} z=0$.