Estimating absolute value of the derivative of a complex function Let $f:\mathbb{C}\to\mathbb C$ be holomoprhic and $|f(z)|\leq 1$ for $|z|\leq 1$. Show that $|f'(z)|\leq 4$ for $|z|\leq\frac{1}{2}$.
I can't seem to find a good start for this proof. Considering $\Re^2+\Im^2\leq1$ for $z\in\overline{C}_1(0)$ doesn't seem to lead me anywhere. Maybe a trick or an integration does?
 A: "Cauchy Derivative Estimate":
If $f$ is holomorphic on a region $U$ and maps the disc of radius $r$ about $z_0$ into a disc of radius $s$, then $|f'(z_0)| \leq \frac{s}{r}$.
In your case, $s = 1$ and $r = \frac{1}{4}$.  The theorem follows immediately.  Note that setting $r = \frac{1}{2}$ shows the stronger statement that $|f'(z)| \leq 2$ on the closed $\frac{1}{2}$-ball about $0$.  If you need a proof of this theorem based on the Schwarz Lemma, just ask, but it's only a couple of lines.
Here is a proof of the Cauchy Derivative Estimate based on the Cauchy Integral Formula.  We use the same notation.
Let $g(z) = f(z_0 + z) - f(z_0)$.  Pick an $r_1$ with $0 < r_1 < r$.  Then $g$ maps the disc of radius $r_1$ about the origin to the disc of radius $s$ about the origin.  Then by the Cauchy Integral Formula, $$f'(z_0) = g'(0) = \frac{1}{2 \pi i}\int_{C_{r_1}} \frac{g(z)}{z^2} dz$$
where $C_r$ is the circle of radius $r$ centered about the origin.  Thus $$|f'(z_0)| = \Big{|} \frac{1}{2 \pi i}\int_{C_{r_1}} \frac{g(z)}{z^2} dz \Big{|} \leq \frac{1}{2 \pi} \int_{C_{r_1}} \big{|} \frac{g(z)}{z^2} \big{|} dz \leq$$ $$\leq \frac{1}{2 \pi} \int_{C_{r_1}} \big{|} \frac{s}{r_1^2} \big{|} dz = \frac{2 \pi s r_1}{2 \pi r_1^2} = \frac{s}{r_1}$$
Now take the limit $r_n \rightarrow r$ to obtain the desired result.
A: For $|z_0|\leq1/2$ and $0<r<1/2$, the closed disk $|z-z_0|\leq r$ is contained in the unit disk $|z|<1$. By Cauchy Integral Formula:
$$f'(z_0)=\frac{1}{2\pi i}\int_{|z-z_0|=r}\frac{f(z)}{(z-z_0)^2}\mathrm{d}z.$$
So
\begin{align*}
|f'(z_0)|
  &=\left|\frac{1}{2\pi i}\int_{|z-z_0|=r}\frac{f(z)}{(z-z_0)^2}\mathrm{d}z\right|\\
  &\leq\frac{1}{2\pi}\int_{|z-z_0|=r}\frac{|f(z)|}{|z-z_0|^2}\mathrm{d}s\\
  &\leq\frac{1}{2\pi}\frac{1}{r^2}\cdot2\pi r\\
  &=\frac1r.
\end{align*}
Let $r\to1/2$, we get
$$|f'(z_0)|\leq2,\ \forall|z_0|\leq1/2 .$$
