Does the inequality $|\sin(x)|\leq |x|$ extend to the complex numbers? 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
$$|\sin(z)|\leq |z|\quad  \forall z \in \mathbb{C}?$$
 A: 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 entire bounded function is constant)
A: 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.")
A: 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. 
A: 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 $C>0$.
