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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.

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2 Answers 2

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$$\sin(x+iy)=\sin x\cosh y+i\cos x\sinh y$$ 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.

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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 $\mathop{\rm Im} z=0$.

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  • $\begingroup$ Another great answer, thanks! $\endgroup$ Commented Dec 10, 2020 at 18:28

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