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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)?

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    @JM: Thanks, another solution!2011-08-17

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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 $\cos z$.

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    The norm is actually $\exp(-\text{Im}(z))$ which can be less than $1$, but it is equal to $1$ only if $\text{Im}(z)=0$.2014-11-18
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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 $z\in \mathbb C$.

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    @anon - actually (& I've only recently learned this myself) the reason for this is that most of the theorems regarding convergence of infinite products have to rule out the "converge to 0" case as exceptional. It's easy enough to see why if you take the log.2019-05-12