The sine function can take a complex argument. e.g. sin(x + iy)

But does it get used that way in any field? Either practical (e.g. electrical engineering) or in other fields of math? Naturally, I am not interested in trivial examples where the real or imaginary part of the argument is always zero.

I've asked elsewhere and the best anyone has come up with is that it can be a 2D solution of the Laplace equation. Anything more substantial? Any other interesting properties? Anyone stumbled upon it in deep in some analysis somewhere? Or is it really never used?

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    $\begingroup$ I've always been under the impression that engineers love the complex sine function because of its conformal properties. Namely it maps the strip $ \lbrace z = x + iy \in \mathbf{C} : |x| < \pi/2, y >0\rbrace $ to the upper half plane, making it easier to do line integrals. $\endgroup$ – stankewicz Sep 23 '10 at 12:43
  • $\begingroup$ By the time you're willing to plug in and receive complex values, why work with sine instead of complex exponentials? $\endgroup$ – Allen Knutson Sep 24 '10 at 17:45

It turns up in a functional equation for the Gamma function, $\Gamma(s)\Gamma(1-s)={\pi\over\sin\pi s}$. From there one goes on to the functional equation for the Riemann zeta-function, $$\zeta(s)=2(2\pi)^{s-1}\Gamma(1-s)\sin(\pi s/2)\zeta(1-s)$$ The million dollar question is where are the (complex) zeros of $\zeta(s)$?

  • $\begingroup$ Thanks for this answer. I've attempted to incorporate it in the Wikipedia article, but it's still a bit beyond me, so please help with the wording if it's off. en.wikipedia.org/wiki/Sine#Usage_of_complex_sin $\endgroup$ – Pengo Oct 2 '10 at 7:30
  • $\begingroup$ What you've written at Wikipedia looks OK to me. But my answer was only ever intended to start the ball rolling - I'm sure the sine function with complex arguments has many uses beyond Gamma and zeta functions, and I'm surprised there was so little activity on this question. $\endgroup$ – Gerry Myerson Oct 2 '10 at 12:26

Euler discovered an infinite product expansion for the sine function,

$$ \sin z=z\prod_{k\geq 1}\left(1-\frac{z^2}{k^2\pi^2}\right) $$ by analogy with the factorization of a polynomial with known zeroes (i.e. roots) into linear terms. In order for this to be true, it is crucial to know all zeros, real as well as complex. As a consequence of the product formula, Euler evaluated zeta function at even integers in terms of Bernoulli numbers.

Euler gave an interesting proof of the product formula based on the idea with zeros, where it is shown that the products of $n$ factors approximate $\sin z$ as $n\to\infty.$ I've read some recent papers making this proof entirely rigorous (e.g. cited in Varadarajan's Bull of AMS article). The first step is to write

$$ \sin z=\frac{1}{2i}(e^{iz}-e^{-iz})= \frac{1}{2i}\lim_{n\to\infty}\left(\left(1+\frac{iz}{n}\right)^n-\left(1-\frac{iz}{n}\right)^n\right). $$

A standard proof of the product formula in complex analysis textbooks (due to Weierstrass?) also crucially relies on the fact that $\sin z$ is a complex analytic function, rather than merely real analytic.

  • $\begingroup$ I think it's actually easier (in this specific case) to prove the product formula with real numbers directly, rather than with Weierstrauss products (although of course Weierstrauss products are very useful for other things as well). First, prove by induction that $\sin (2n+1)x = p_n(\sin x)$ for some polynomial $p_n$ of degree $\leq 2n+1$. Now note that $\sin[ k \pi/(2n+1)]$ for $k=0,\pm 1, \pm 2, \ldots, \pm n$ are $2n+1$ distinct zeros of $p_n$. $\endgroup$ – Zen Harper Oct 9 '10 at 13:41
  • $\begingroup$ Thus we can factorise $p_n(\sin x)$ very nicely; we need the limit of $\frac{\sin \theta}{\theta}$ as $\theta \rightarrow 0$. Finally, fix $z$, let $x_n=z/(2n+1)$ and let $n \rightarrow \infty$. Some care is needed in proving convergence of the infinite product, but it does all work out, with no complex numbers used at any point. The only point where we use real numbers is the monotonicity of $\sin x$ for $|x|<\pi/2$, so actually this proof works equally well with real or complex numbers, I think. (I think I saw this years ago in Bromwich's Infinite Series - a wonderfully ancient book!) $\endgroup$ – Zen Harper Oct 9 '10 at 13:48

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