Complex sine not bounded

Question

How do you show that the complex sine is not bounded, i.e. that there does not exist an $M > 0$ such that $|\sin(z)| < M$ for all $z \in \mathbb{C}$?

so far I've got |sin(z)|= sqrt(sin^2(x) + sinh^2(y)) then I know that the limit of sinh(y) as y approaches infinity then sinh(y) also approaches infinity, it then follows that since sinh(y) is unbounded at large values of y then the above modulus can increase (as y does) without bound.. is this enough to finish the proof? — Amelia, Nov 12 '13 at 20:06

DonAntonio · Answer 1 · 2016-07-06T18:27:51.377

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Liouville's theorem indeed helps, but also (putting $\;z=x+iy\;,\;x,y,\in\Bbb R\;$)

$$\left|\sin z\right|=\left|\frac{e^{zi}-e^{-zi}}{2i}\right|=\frac12\left|\frac{e^{2zi}-1}{e^{zi}}\right|=\frac12\left|\frac{1-e^{2xi}e^{-2y}}{e^{xi}e^{-y}}\right|\ge\frac12\frac{1-e^{-2y}}{e^{-y}}=\frac12\frac{e^{2y}-1}{e^y}$$and observe that choosing numbers with negative imaginary part big enough in absolute value (and, say, $\;x=0\;$ for simplicity), the above can be made as large as we want.

edited Jul 06 '16 at 18:27

answered Nov 12 '13 at 19:58

DonAntonio

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Thanks for the editing, @dfeuer. – DonAntonio Nov 12 '13 at 20:10
but the denominator $e^{zi}$ should become $e^{xi}e^y$, not $e^xe^{yi}$... then we obtain $\geq \frac{|e^{-2y}-1|}{2e^y}$.. if I'm not mistaken then we can fix $x$ and let $y$ approach $-\infty$. Then it will be arbitrarily large – massy255 Jun 30 '16 at 23:22
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@massy255 That's an unimportant typo for the main argument but thanks, I edited my answer (after three years...!) – DonAntonio Jul 06 '16 at 18:28
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Essentially the same thing, but less general and possibly more easily discoverable, use Euler's theorem to expand both $e^{i(-ni)}$ and $e^{i(ni)},$ then subtract the resulting equations to get $e^{-n} - e^n = 2i\sin(ni),$ and observe for this last equation that the left side is unbounded as $n$ ranges over the positive integers. – Dave L. Renfro Jul 06 '16 at 19:08
@DaveL.Renfro Nice idea, Dave. Thanks. – DonAntonio Jul 06 '16 at 19:20

score 12 · Answer 2 · answered Nov 12 '13 at 19:39

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HINT Liouville's theorem will help you a great deal.

answered Nov 12 '13 at 19:39

M.B.

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Do we mean to say that sin(z) is not constant hence it is not bounded? – humble Sep 20 '16 at 05:32

score 2 · Answer 3 · edited Sep 12 '20 at 17:13

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Assume $\sin z$ is bounded. We know that $\sin z$ is analytic on $\mathbb C$, that is $\sin z$ is an entire function. By Liouville’s theorem $\sin z$ is constant which is a contradiction. Hence sinz is unbounded.

edited Sep 12 '20 at 17:13

Mohsen Shahriari

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answered Sep 12 '20 at 17:01

Salman

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Complex sine not bounded

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