First derivative of holomorphic function I want to prove that $ |f'(z)| \le \frac{1}{1-|z|}$ where $f:B(0,1) \rightarrow B(0,1)$ is a holomorphic function. My idea was to use Cauchy's integral formula. The fact that $||f||\le 1$ might be helpful too. But I don't see how to get this $1-|z|$?
 A: It is rather easy to prove a stronger result. First, we prove the special case $z = 0$ using the Cauchy integral formula:
$$\lvert f'(0)\rvert = \Biggl\lvert\frac{1}{2\pi i} \int\limits_{\partial D_r} \frac{f(\zeta)}{\zeta^2}\, d\zeta\Biggr\rvert \leqslant \frac{1}{2\pi} \int\limits_{\partial D_r} \frac{1}{\lvert\zeta\rvert^2}\,\lvert\zeta\rvert\, d\varphi = \frac{1}{r}$$
for all $0 < r < 1$. Taking the limit $r \to 1$ proves the case $z = 0$.
Next, to handle the general case, consider the Möbius transformation
$$T \colon w \mapsto \frac{w + z}{1 + \overline{z}w}.$$
$T \in \operatorname{Aut}(\mathbb{D})$ is easy to verify, and obviously $T(0) = z$. Now consider $g = f \circ T \colon \mathbb{D} \to \mathbb{D}.$
By the special case we proved above, $\lvert g'(0)\rvert \leqslant 1$.
On the other hand,
$$g'(0) = f'(T(0)) \cdot T'(0) = f'(z)\cdot T'(0) \iff f'(z) = \frac{g'(0)}{T'(0)}.$$
Now,
$$T'(w) = \frac{d}{dw} \frac{w+z}{1+\overline{z}w} = \frac{(1+\overline{z}w) - \overline{z}(w+z)}{(1+\overline{z}w)^2} = \frac{1 - \lvert z\rvert^2}{(1+\overline{z}w)^2}$$
and hence $T'(0) = 1 - \lvert z\rvert^2$, and
$$\lvert f'(z)\rvert = \frac{\lvert g'(0)\rvert}{1 - \lvert z\rvert^2} \leqslant \frac{1}{1 - \lvert z\rvert^2} = \frac{1}{1 + \lvert z\rvert} \cdot \frac{1}{1 - \lvert z \rvert}.$$
A stronger result, and less easy to come by is the Schwarz-Pick lemma:
For every holomorphic $f \colon \mathbb{D} \to \mathbb{D}$, we have the inequality
$$\biggl\lvert\frac{f(z) - f(w)}{1 - \overline{f(w)}f(z)} \biggr\rvert \leqslant \biggl\lvert\frac{z-w}{1-\overline{w}z} \biggr\rvert$$
and the infinitesimal version (divide by $z-w$ and let $w \to z$)
$$\frac{\lvert f'(z)\rvert}{1 - \lvert f(z)\rvert^2} \leqslant \frac{1}{1 - \lvert z\rvert^2}, \quad z \in \mathbb{D}.$$
A: We can use the Cauchy Integral Formula:
$$
f'(z)=\frac1{2\pi i}\oint\frac{f(w)}{(w-z)^2}\,\mathrm{d}w\tag{1}
$$
Set $w=\cos(t)+i\sin(t)$ and $|z|=r$. Then we can use the fact that $|f(z)|\le1$ and the Weierstrass substitution to get
$$
\begin{align}
|f'(z)|
&\le\frac1{2\pi}\int_0^{2\pi}\frac{\,\mathrm{d}t}{(r-\cos(t))^2+\sin^2(t)}\\
&=\frac1{2\pi}\int_0^{2\pi}\frac{\,\mathrm{d}t}{r^2+1-2r\cos(t)}\\
&=\frac1{2\pi}\int_{-\infty}^\infty\frac{2\,\mathrm{d}u}{(r^2+1)(1+u^2)-2r(1-u^2)}\\
&=\frac1{2\pi}\int_{-\infty}^\infty\frac{2\,\mathrm{d}u}{(r-1)^2+u^2(r+1)^2}\\
&=\frac1{1-r^2}\\
&\le\frac1{1-r}\tag{2}
\end{align}
$$
