Using the Laurent-expansion of $f(z)$ around $0$
$$f(z) = \sum_{n=0}^{+\infty} a_n z^n + \sum_{n=1}^{+\infty} b_n z^{-n} \tag{1}$$
Theorem
The coefficients of the Laurent series are given by the formulae:
$$a_m = \frac{1}{2\pi i} \int_\Gamma \frac{f(z)}{z^{m+1}} \text{d}z \qquad b_m = \frac{1}{2\pi i} \int_\Gamma z^{m-1} f(z) \text{d}z$$
where $\Gamma$ is any circle $|z-z_0| = \rho,\; 0< r < \rho < R$.
I understand why $f(z)$ is uniformly convergent (because both series are uniformly convergent) over $\Gamma$. But I'm experiencing some problems in the proof of these coefficients.
Proof
Since the series (1) is uniformly convergent on $\Gamma$ the same will be true of the series obtained by muliplying (1) with:
$$\text{either}\qquad z^{m-1}\qquad \text{and}\qquad \frac{1}{z^{m+1}}$$
Then the proof continues by swapping sum and integral (for which the uniform convergence is required) ...
Why?
Could someone provide an argument why that same will be true is valid?
The magnitude of $z^{m-1}$ and $1/z^{m+1}$ will be constants on a circle centered around 0.
Suppose $f_n(z)$ converges uniformly on $\Gamma$ to $f(z)$ and that $g(z)$ is some function such that $|g(z)| = \rho$ on $\Gamma$.
$|g(z)f_n(z) - g(z)f(z)| = |g(z)||f_n(z) - f(z)| = \rho |f_n(z) - f(z)|$
So if the RHS is bounded by $\epsilon$ then the LHS is bounded by $\epsilon / \rho$ which shows $g f_n$ is uniformly convergent to $g f$.
Your situation is that $g(z) = z^{\pm m \pm 1}$ which is indeed a constant on $\Gamma$ which is a circle around $0$.