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As it is well known: $$|\sin(x)|\leq |x| \forall x \in \mathbb{R}.$$

Now, if we have a complex number $z$; can I preserve the same inequality

$$|\sin(z)|\leq |z|\quad \forall z \in \mathbb{C}?$$

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Think about the purely imaginary case. –  Noah Snyder Jun 22 '12 at 16:13
Very related... –  J. M. Jun 27 '12 at 16:16

4 Answers 4

No, for example $2i\sin( in)=e^{i(i n)}-e^{-i(i n)}=e^{-n}-e^n$ hence $$|\sin(in)|\geq \frac{e^n-e^{-n}}{2}\geq \frac{e^n-1}{2}$$ for each integer $n$.

(a sledgehammer argument would be Liouville's theorem, which says a an entier bounded function is constant)

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The answer is no. There are many counterexamples, e.g., $z=3i$.

Are you aware of the complex form of the sine function? See about a fourth of the way down on this page: http://mathworld.wolfram.com/Sine.html. It may help you understand why this occurs.

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Suppose that $f:\mathbb C\to\mathbb C$ is an analytic function such that $|f(z)|\leq |z|$ for all $z\in\mathbb C$. Define $g:\mathbb C\to\mathbb C$ by $g(0)=f'(0)$, $g(z)=\frac{f(z)}{z}$ if $z\neq 0$. Then $g$ is an analytic function satisfying $|g(z)|\leq 1$ for all $z\in \mathbb C$. By Liouville's Theorem, $g$ is constant. Hence $f(z)=cz$ for some constant $c\in\mathbb C$ with $|c|\leq 1$. In particular, $f\neq \sin$.

(This is probably roughly what Davide G. had in mind when he referred to "a sledgehammer argument.")

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The usual trigonometric formulae for $\cos$ and $\sin$ are also valid in the complex domain; furthermore $\cos(iy)=\cosh y$ and $\sin(i y)=i\sinh y$. It follows that $$\sin(x+iy)=\sin x\cos(i y)+\cos x\sin(iy)=\sin x\cosh y+i\cos x\sinh y$$ and therefore $$|\sin(x+iy)|^2=\sin^2 x\cosh^2 y +\cos^2 x\sinh^2 y=\sin^2 x+\sinh^2 y\ .$$ As $|\sinh y|$ increases exponentially with $y\to \infty$ there is no estimate of the form $$|\sin z|^2\leq C |z|^2\qquad(z\in{\mathbb C})$$ with a fixed $C>0$.

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