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I recently learned that $\sin(z)$ has an extension into the complex plane, namely:

$$\frac{e^{iz}-e^{-iz}}{2i}$$

Is there any complex number $z=a+bi$, with $b≠0$ such that $\sin(z)=0$ ?

I am already aware of the 'trivial' zeros when $a=πk$ with $k\in\Bbb Z$ and $b=0$.

Is there any way to prove this? Either complete solutions/proofs or hints would both be helpful.

Cameron Williams
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KR136
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  • There are no more zeros, but you can simply compute $\sin(a+ib)$ using the formula and see at what you arrive. – John B Feb 05 '16 at 01:57
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    Note that if $w$ has a real component then $|e^w|\neq |e^{-w}|$. – Thomas Andrews Feb 05 '16 at 01:59
  • Thanks these answer my question. I kept making a silly mistake with the exponentials. – KR136 Feb 05 '16 at 02:01
  • By the way, note that on the reals, the $\sin$ function is bounded ($|\sin x|\le1$), but it's unbounded on the complex numbers. There's a theorem that says that no differentiable, nonconstant function is bounded on the complex numbers. ("Differentiable" is actually a fairly strong criterion for complex functions, by the way; for example, $|z|$ and ${\rm Im}(z)$ are not differentiable at any point. This is partly because the $h$ in the expression $\lim_{h\to0}$, in the definition of the derivative, is now allowed to be a complex number approaching $0$ from any direction in the plane.) – Akiva Weinberger Feb 05 '16 at 03:09
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    Maybe of interest http://math.stackexchange.com/questions/941036/the-zeros-of-sinz-where-z-is-a-complex-number – cgiovanardi Feb 18 '17 at 11:20

1 Answers1

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You can use the lovely trigonometrical identity $$ \lvert \sin{(x+iy)} \rvert^2 = \sin^2{x}+\sinh^2{y} $$ (use the angle-addition formulae, the relationships between the trigonometric and hyperbolic functions, and the Pythagorean identity): since $\sinh^2{y}>0$ unless $y=0$, the only zeros occur when $y=0$ and $x$ is a root of $\sin$, i.e. the usual real ones.

Chappers
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