Extracting a coefficient of an analytic function

Using the Cauchy formula

Posted on February 28, 2023 by Egor Lappo

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 \[ a_k = \frac{1}{2\pi i} \int_{U} \frac{dz}{z^{n+1}} f(z). \]