Show that $g$ is analytic and discuss the properties of $g$ Let $f$ be analytic in $\overline{B}(0; R)$ with $f(0)=0$, $f'(0) \neq 0$ and $f(z) \neq 0$ for $0<|z| \leq R$.  Put $\rho=\min\{|f(z)|:|z|=R\}>0$.  Define $g: B(0; \rho) \rightarrow \mathbb{C}$ by $$g(w) = \frac{1}{2\pi i} \int_\gamma \frac{z\,f'(z)}{f(z)-\omega} dz$$ where $\gamma$ is the circle $|z|=R$.  Show that $g$ is analytic and discuss the properties of $g$.
I was thinking about how to use the Argument Principle. If I can prove that f is injective I believe that is the end of the proof, but i have no idea how to do it. Can someone help me? 
 A: Typically, one would start by showing that $g$ is analytic by using Morera's theorem or differentiation under the integral. Both ways require to first show the continuity of $g$, which is an easy consequence of the continuity of the integrand
$$\frac{zf'(z)}{f(z)-w}$$
on $\{ z : \lvert z\rvert = R\} \times B(0;\rho)$. Since all parts of which the integrand is composed are continuous, the only obstacle to continuity of the integrand would be the vanishing of the denominator, but that doesn't happen by definition of $\rho$.
By whatever theorem about parameter-dependent integrals one prefers to invoke - it could be the dominated convergence theorem, but since the integrand is continuous and the domain of integration is compact one could also use a theorem about uniform convergence for Riemann integrals - one then has the continuity of $g$.
Next one uses the fact that the integrand depends holomorphically on $w$ to deduce that $g$ is holomorphic. When using Morera's theorem, that yields the vanishing of the inner integral
$$\int_{\partial \Delta} \frac{zf'(z)}{f(z)-w}\,dw$$
after changing the order of integration (justified by a version of Fubini's theorem, again due to the regularity of the integrand a very weak version suffices) and hence the vanishing of $\int_{\partial \Delta} g(w)\,dw$ for every triangle $\Delta \subset B(0;\rho)$. When using differentiation under the integral, the holomorphic dependence of the integrand on $w$ gives the Cauchy-Riemann equations pointwise, and hence for the integral - whether in the real or the complex form of the Cauchy-Riemann equations.
An alternative way of showing the analyticity of $g$ is to expand $\frac{1}{f(z) - w}$ into a geometric series, and since that series converges uniformly on the circle $C_R = \{ z : \lvert z\rvert = R\}$ (for any fixed $w$, but the convergence is even uniform on $C_R \times \overline{B(0;\varepsilon)}$ for every $\varepsilon \in (0,\rho)$) we can interchange summation and integration to obtain
$$g(w) = \sum_{n = 0}^{\infty} \Biggl(\frac{1}{2\pi i}\int_{\gamma} \frac{zf'(z)}{f(z)^{n+1}}\,dz\Biggr)w^n.$$
Regardless of whether we already know that $g$ is analytic, if $f$ attains the value $w$ with multiplicity $k$ in $z_0$ then the residue of $\frac{zf'(z)}{f(z) - w}$ at $z_0$ is $k\cdot z_0$, so by the residue theorem $g(w)$ is the sum of all points in $B(0;R)$ where the value $w$ is attained, where each point is summed according to the multiplicity with which $w$ is attained. Since you write

If I can prove that $f$ is injective I believe that is the end of the proof

it seems that you want to use this fact to conclude that $g$ is the inverse of $f$ (more precisely, a local inverse of $f$), which then gives you the analyticity of $g$ by yet another argument.
However, for that to work you must also know that $B(0;\rho) \subseteq f(B(0;R))$. But you don't quite need the injectivity of $f$, you only need that every $w\in B(0;\rho)$ is attained exactly once in $B(0;R)$ (counting multiplicities). And indeed, the premises do not imply that $f$ is injective on $B(0;R)$ - consider $f(z) = (z+R)^3 - R^3$ for an example.
But the premises imply that every $w\in B(0;\rho)$ is attained exactly once in $B(0;R)$. To see that, consider
$$N(w) := \frac{1}{2\pi i} \int_{\gamma} \frac{f'(z)}{f(z)-w}\,dz.$$
By the residue theorem/argument principle, $N(w)$ is the number of times $w$ is attained in $B(0;R)$ (counting multiplicities). By essentially the same argument(s) as above, we find that $N$ is continuous (or even holomorphic) on $B(0;\rho)$. Since $N$ is integer-valued, and $B(0;\rho)$ is connected, it follows that $N$ is constant on $B(0;\rho)$, i.e. $N(w) = N(0)$ for all $w\in B(0;\rho)$. But we know $N(0) = 1$, since the premises are that $f$ attains the value $0$ with multiplicity $1$ at $0$, and nowhere else in $\overline{B(0;R)}$.
Thus, since every $w\in B(0;\rho)$ is attained exactly once (counting multiplicities) in $B(0;R)$, indeed $g$ is a local inverse of $f$,
$$g = \bigl(f\lvert_{f^{-1}(B(0;\rho))}\bigr)^{-1},$$
and as the inverse of a biholomorphic function, $g$ is analytic.
