# Where are the zeros of the complex sine and cosine?

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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$\cosh\,x \geq 1$ if $x$ is real... –  Ｊ. Ｍ. Aug 17 '11 at 6:49
Use de Moivre's to write sin and cos as complex exponentials. Multiply through to get a quadratic in terms of $e^{ix}$. Solve over all branches. –  anon Aug 17 '11 at 6:51
@JM: How is cosh, x≥1 related to the complex sin and cos? –  Sara Aug 17 '11 at 6:51
$\sin(x+iy)=\sin\,x\cosh\,y+i\cos\,x\sinh\,y$. You know where the zeroes of real-valued $\sin$, $\cos$, and $\sinh$ are, so... –  Ｊ. Ｍ. Aug 17 '11 at 6:54
@anon: Thanks! I solved the quadratic. –  Sara Aug 17 '11 at 6:56
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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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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 $z\in \mathbb C$.
Yes, and these infinite products converge for all complex $z$: when we say that we mean, in particular, that they do not diverge to zero. So they have value zero only if one of the factors vanishes. –  GEdgar Aug 17 '11 at 12:44