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?
Asked
Active
Viewed 636 times
5
-
1From $e^{it}=\cos t + i \sin t$ you have $\sin z = \frac{e^{iz}-e^{-iz}}{2i}$. Does that help finding a bound? – Hagen von Eitzen Jan 14 '13 at 22:05
-
I didn't look this through, but I think taking $z=\rho e^{i\theta}$ and then multiplying the resulting equation by $2ie^{i\theta}$ might work. – Git Gud Jan 14 '13 at 22:06
-
@Freddie: Should the function be $\mathbf{2}z-\sin z$ rather than $z-\sin z$? – JohnD Jan 15 '13 at 04:39
-
No, I've stated the problem correctly. – Freddie Jan 15 '13 at 13:47
1 Answers
6
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 $1/3!$.
mrf
- 43,639