Sometimes, it is necessary to get an expression for the coefficient in the Taylor expansion of an analytic function \(f(z)\), without necessarily knowing the expression for \(f\) itself. There is a way to do this using the Cauchy theorem. We will use the Cauchy theorem in the following form.

*Proposition 1.* Let \(U \in
\mathbb{C}\) be any domain such that \(0\in U\). Then

\[ \int_{\partial U} \frac{dz}{z^n} = \begin{cases} 2\pi i, & n=1 \\ 0, & \text{otherwise}. \end{cases} \]

So, suppose we have an analytic function \(f(z)\) with a Taylor expansion \[ f(z) = \sum_{k=0}^\infty a_k z^k. \]

We claim that \[ a_k = \frac{1}{2\pi i} \int_{U} \frac{dz}{z^{n+1}} f(z), \] where \(U\) is a domain containing \(0\).

To show this, we substitute the Taylor expansion for \(f\) and work case-by-case. \[ \frac{1}{2\pi i} \int_{U} \frac{dz}{z^{n+1}} f(z) = \sum_{k=0}^\infty \frac{1}{2\pi i} \int_{U} \frac{a_k dz}{z^{n-k+1}}. \]

When \(n = k\), the summand becomes \[ \frac{1}{2\pi i} \int_{U} \frac{a_n dz}{z} = a_n, \] due to the first case in Proposition 1 above. When \(n \neq k\), the summand is zero, due to the second case in Proposition 1. So, we have shown that \[ a_k = \frac{1}{2\pi i} \int_{U} \frac{dz}{z^{n+1}} f(z). \]