I am trying to solve the following problem from Lang's Complex Analysis book:
Let $f$ be analytic on an open set $U$, let $a \in U$, and $f'(a) \ne 0$. Show that
$$\frac {2 \pi i} {f'(a)} = \int_\gamma \frac {dz} {f(z) - f(a)}$$
where $\gamma$ is a small circle centered at $a$.
What I tried was the following:
Use the inverse mapping theorem to shrink $U$ to a small neighborhood $a$ where $f$ is invertible, and let $\gamma$ lie in this neighborhood. Define the family of functions $g_w : U \to \mathbb C$, where
$$g_w(z) = \begin{cases} \frac {z - w} {f(z) - f(w)} & \text{if } z \ne w \\ \frac 1 {f'(w)} & \text{if } z = w \\ \end{cases} $$
Every $g_w$ is analytic at every $z \ne w$. In particular, $g_a$ is analytic at every $z \ne a$. Now take a convergent sequence $w_n$ to $a$. Then the sequence $g_{w_n}$ is also (uniformly) convergent to $g_a$. Every $g_{w_n}$ is analytic at $a$, hence so is the limit $g_a$.
Hence by Cauchy's integral formula, we have the desired result:
$$\frac {2 \pi i} {f'(a)} = \int_\gamma \frac {g_a(z)} {z - a} dz = \int_\gamma \frac {dz} {f(z) - f(a)}$$
I am not convinced about the highlighted step. I have previously proved that, if $f_n \to f$ is a uniformly convergent family of analytic functions in $U$, the limit $f$ is also analytic in $U$. But, when I took a uniformly convergent sequence of functions $g_{w_n} \to g_a$, there was no common set in which all $g_{w_n}$'s are analytic. How could I fix this proof?
I finally found the right way to solve it.
Let $f(z) = \sum_k c_k (z - a)^k$ be the power series expansion of $f$ around $a$. By hypothesis, $f(z) - f(a)$ has order 1, so there exists a power series $g(z)$ of order 0 such that $(z-a) g(z) = f(z) - f(a)$. Of course, it is $g(z) = \sum_k c_{k+1} (z-a)^k$. This series has a multiplicative inverse $h(z)$. Moreover, $h(z)$ is convergent, because $g(z)$ is convergent, because $f(z)$ is convergent. Hence $h(z)$ is the power series expansion of an analytic function in a neighborhood of $a$. It is not hard to see that this function is
$$ h(z) = \begin{cases} \frac {z - a} {f(z) - f(a)} & \text{if } z \ne a \\ \frac 1 {f'(a)} & \text{if } z = a \\ \end{cases} $$
Hence, using Cauchy's integral formula, we have the desired result:
$$ \frac 1 {2 \pi i} \int_\gamma \frac {dz} {f(z) - f(a)} = \frac 1 {2 \pi i} \int_\gamma \frac {h(z) dz} {z - a} = W(\gamma, a) h(a) = \frac 1 {f'(a)} $$