Suppose $f(z)$ is analytic on closed disk of radius $r$ and $f(0)=0$, $f'(z) \neq 0$. Show that $f$ has an analytic inverse on $\{|z| \leq m\}$ where $m$ is the minimum of $|f(z)|$ on $\{|z| = r\}$. And
$$f^{-1}(u) = \frac{1}{2\pi i}\oint_{|z|=r}\frac{zf'(z)}{f(z)-u}dz$$
Thank you very much.
EDIT:
$f(z) \neq 0$ for $z \neq 0$, and $\{|u| < m\}$ instead of $\{|z| \leq m\}$.
Consider the closed curve $\gamma \colon [0,2\pi] \to \mathbb{C}$ given by
$$\gamma(t) = f\bigl(re^{it}\bigr).$$
Since $m = \min \{ \lvert \gamma(t)\rvert : t \in [0,2\pi]\} > 0$, the curve doesn't intersect the disk $\{ u : \lvert u\rvert < m\}$, and therefore the winding number of $\gamma$ around $u$ is constant on that disk. Now, the winding number is
$$n(\gamma,u) = \frac{1}{2\pi i} \int_\gamma \frac{dw}{w-u} = \frac{1}{2\pi i} \int_{\lvert z\rvert = r} \frac{f'(z)}{f(z)-u}\,dz,$$
and we recognise the latter as the number of times (counting multiplicity) $f$ attains the value $u$ in the disk $\{z : \lvert z\rvert < r\}$.
By assumption $n(\gamma,0) = 1$, so $n(\gamma,u) = 1$ for $\lvert u\rvert < m$. That is, each value $u$ with $\lvert u\rvert < m$ is attained exactly once on $\{ z : \lvert z\rvert < r\}$.
Hence the meromorphic function
$$g_u(z) = \frac{zf'(z)}{f(z)-u}$$
has exactly one singularity in the disk $\{ z : \lvert z\rvert < r\}$ for every $u$ with $\lvert u\rvert < m$, and
$$\frac{1}{2\pi i} \int_{\lvert z\rvert = r} \frac{zf'(z)}{f(z)-u}\,dz$$
is just the residue of $g$ in $f^{-1}(u)$.
If a holomorphic funtion $h$ attains a value $a$ in $z_0$ with multiplicity $k$, then we can write
$$h(z) = a + (z-z_0)^k\cdot \tilde{h}(z)$$
with a holomorphic function $\tilde{h}$, where $\tilde{h}(z_0) \neq 0$. Then we have
$$\frac{h'(z)}{h(z)-a} = \frac{k(z-z_0)^{k-1}\tilde{h}(z) + (z-z_0)^k\tilde{h}'(z)}{(z-z_0)^k\tilde{h}(z)} = \frac{k}{z-z_0} + \frac{\tilde{h}'(z)}{\tilde{h}(z)},$$
so the residue of $\frac{h'(z)}{h(z)-a}$ in $z_0$ is the multiplicity with which $a$ is attained at $z_0$, and
$$\frac{zh'(z)}{h(z)-a} = \frac{kz}{z-z_0} + \frac{z\tilde{h}'(z)}{\tilde{h}(z)} = \frac{kz_0}{z-z_0} + k + \frac{z\tilde{h}'(z)}{\tilde{h}(z)},$$
so the residue of $\frac{zh'(z)}{h(z)-a}$ in $z_0$ is $k\cdot z_0$.
Here we have $k = 1$ for $\lvert u\rvert < m$, so the residue of $g_u$ in $f^{-1}(u)$ is just $f^{-1}(u)$, and the integral formula
$$f^{-1}(u) = \frac{1}{2\pi i} \frac{zf'(z)}{f(z)-u}\,dz$$
is established for $\lvert u\rvert < m$. It holds for all $u$ in the connected component of $\mathbb{C} \setminus \{ \gamma(t) : t\in [0,2\pi]\}$ containing $0$, which usually is strictly larger than the disk $\{u : \lvert u\rvert < m\}$.