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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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$\cosh\,x \geq 1$ 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
up vote 8 down vote accepted

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 $\text{Im}(z)=0$. – Minethlos Nov 18 '14 at 11:03

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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Looks a bit like using a sledge-hammer to crack a nut... – Did Aug 17 '11 at 9:51
@anon - btw often in the literature I've seen that referred to as "diverging" to zero. – John M Aug 17 '11 at 12:34
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
@GEdgar, John M: For some strange reason I've only heard of convergence refer to the existence of a limit, but I see the definition you refer to right on Wikipedia. Yes, I know the Wallis product converges, so it was meaningless to point out an irrelevant technicality. – anon Aug 17 '11 at 18:11
@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. – John M Aug 18 '11 at 2:20

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