# How do we know that the Sine function has no Non-Real Roots?

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?

$$\sin(x+iy)=\sin x\cosh y+i\cos x\sinh y$$ Note that $$\cosh y$$ is never zero for $$y$$ a real number. Thus, if $$\sin(x+iy)=0$$, $$\sin x=0$$ (the real and imaginary parts must be identically zero). But this forces $$\cos x=\pm1$$, which in turn forces $$\sinh y=0$$ and thus $$y=0$$. All the roots must be real.
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 $$\mathop{\rm Im} z=0$$.