Holomorphic function Let $f(z)$ be a holomorphic  function  that  maps  the  unit  disk  to  the  unit  disk. Prove that $$|f^{(5)}(0)| \leq 120.$$  I use some concrete example it seems that this statement work out but i do not know how to come up with a formal proof help any one
 A: Perhaps you should use Cauchy's integral formula
$$f^{(5)}(0)=\frac{5!}{2\pi i}\int_\gamma\frac{f(z)}{z^6}dz$$
where $\gamma$ is the unit circle.
Then, use the Estimation Lemma:
$$|f^{(5)}(0)|\le \frac{120}{2\pi}|\gamma|M$$
where $|\gamma|$ is the length of $\gamma$ and $M$ is an upper bound for $f(z)/z^6$ on $\gamma$.
Can you finish?
A: Assuming $f^5$ means really $f^{(5)}$. Then the hypothesis says that $f$ is holomorphic in the open unit disk $D(0,1)$, and that $|f(z)|\leq 1$ for all $z \in D(0,1)$. Then by the general Cauchy´s Integral Formula
$$
|f^{(5)}(0)| = \left| \frac{5!}{2\pi i} \int_{|z|=1} \frac{f(z)}{z^6}dz\right| \leq \frac{|5!|}{|2\pi i|} \int_{|z|=1} \frac{|f(z)|}{|z|^6}|dz| \leq  \frac{5!}{2\pi} \int_{|z|=1}|dz| = 5! = 120
$$
since $|f(z)|/|z|^6 \leq 1$ and $\int_{|z|=1}|dz|=2\pi$.
A: Maybe overly pedantic, but we can't really integrate over $|z| = 1$ without some justification, since $f$ is only assumed to be holomorphic on $|z| < 1$. It's not that big of a deal though, since we can integrate over the circle $|z| = r$ for any $r < 1$ to get:
$$
|f^{(5)}(0)| = \bigg| \frac{5!}{2\pi i}\int_{|z|=r} \frac{f(z)}{z^6}\,dz \bigg| \le \frac{120}{2\pi} \cdot 2\pi r \cdot \frac{1}{r^6} = \frac{120}{r^5}
$$
since $|f(z)| < 1$ on $|z|=r$ by assumption. The above inequality holds for all $r$, so by letting $r \to 1^-$, it follows that $|f^{5}(0)| \le 120$.
