Show that for every $r > 0$ there exist a $w$ $\in$ $D(0,r)$ that satisfies $\lvert h(w) \rvert > 1$ I'm learning about Complex Analysis and need some help with this problem:

Consider the function $h(z) = (1 - \frac z5 - \frac{z^3}{10})e^{iz}$.
  Show that for every $r > 0$ there exist a $w$ $\in$ $D(0,r)$ that
  satisfies $\lvert h(w) \rvert > 1$.

I'm assuming the $D(0,r)$ is the disk of radius $r$ centered at zero. I think I should use the Maximum Modulus Principle but I don't know how to apply it here. I'm also looking for solutions using other methods/theorems (if possible).
 A: What is $h(0)$? Now what does a certain big theorem tell you? It's a really big theorem. Bigger than the other theorems. Sort of a maximal theorem...
A: Since you want something more complete than David Ullrich's good hint, assume the contrary, i.e. that $|h(w)| \le 1$ for all $w$ with $|w| = r$. Then the maximum modulus principle implies that $|h(w)| \le 1$ for all $w$ with $|w| \le r$.
Next note that $h(0) = 1$, which shows that $|h|$ must have a local maximum at $z=0$, which implies that $h$ is constant on $|w| \le r$, but this is clearly a contradiction.
A: The Maximum Modulus Principle as applied to this problem says:  If $z_0 \in D(0,r)$ and $|f(z)| \le |f(z_0)|$ for all $z \in D(z_0,R)$, with $R > 0$ such that $D(z_0, R) \subseteq D(0,r)$, then $f$ is constant on $D(0,r)$.
To use it, first suppose for the sake of contradiction that there is no $w$ such that $|h(w)| > 1$ for any $r > 0$.  In other words, you have $|h(w)| \le 1$ for all $r > 0$.  So, no matter what $r > 0$ is, you have $|h(w)| \le 1$.
We also know that no matter what $r > 0$ is, we have $0 \in D(0,r)$.  Also, we know that $|h(0)| = 1$.  So for any $r > 0$ we can say the following:
$0 \in D(0,r)$ and $|h(w)| \le |h(0)|$ for all $w \in D(0,r)$ (and therefore for all $w \in D(0,R)$ for any $R > 0$ such that $D(0,R) \subseteq D(0,r)$).
So by the Maximum Modulus Principle stated above (using $0$ for $z_0$ and using $h$ instead of $f$), we see that $h$ is constant on $D(0,r)$.  But this is clearly a contradiction because the provided function $h$ is certainly not constant on any $D(0,r)$.  Therefore it must be the case that for every $r > 0$, there is a $w \in D(0,r)$ such that $|h(w)| > 1$.
A: Hint: For $x \in \mathbb {R},$ let $f(x) = 1 +x/5 -x^3/10.$ Note $f'(0)> 0,$ hence $f(x) > 1$ in some interval $(0,\delta), \delta > 0.$
