Prove that the only root of the equation $z-\sin(z)$ in the unit disk is zero. Prove that the only root of the equation $z-\sin(z)$  in the unit disk is $z=0$.
My first thought is Rouche's Theorem, but I don't know any bounds on $|\sin(z)|$. Suggestions? 
 A: Not the most elegant solution, but here's something.
First of all, $f(z) = z - \sin z$ has a triple zero at $z=0$. Define
$$g(z) = \frac{z-\sin z}{z^3} = \frac{1}{3!} + \underbrace{\left(- \frac{z^2}{5!} + \frac{z^4}{7!} - \cdots \right)}_{h(z)}. $$
If $|z| < 1$, then by the triangle inequality,
\begin{align}
|h(z)| &\le \frac{1}{5!} + \frac{1}{7!} + \frac{1}{9!} + \cdots \\
&= \frac{1}{5!} \left( 1 + \frac{1}{6\cdot 7} + \frac{1}{6\cdot 7 \cdot 8 \cdot 9} + \cdots \right) \\
&\le \frac{1}{5!}\left(1 + \frac{1}{42} + \frac{1}{42^2} + \cdots + \right) \\
&= \frac{1}{5!}\cdot \frac{42}{41} < \frac{1}{3!}
\end{align}
Hence $g$ has no zero on the unit disc, which shows that the only zero for $f$ is the triple zero at the origin. (You can get by with a cruder estimate of $h$, of course.)
Added In fact, the same method shows that $z=0$ is the only zero on a much larger disc than the unit disc. If $|z| < 6$, for example, the final estimate for $h$ will be $|h(z)| < 7/120$ which is still less than $1/3!$.
