# Prove that the coefficients in a Laurent expansion are unique?

My uni's complex analysis lecture notes simply state that the coefficients in a Laurent expansion are unique, and don't give a proof. It occurred to me to prove in the manner shown in this answer: Coefficients for Laurent series expansion are unique

However, it's not immediately clear to me at all why the Laurent coefficients of the 0 function are identically 0, and the answer given in that link doesn't give an explanation.

In short: why, if $$\sum_{n=-\infty}^{n=\infty} a_nz^n=0$$ for all $$z$$ in some annulus, do we have $$a_n=0$$ for all n?

For Taylor series there's the usual trick of differentiating $$k$$ times and evaluating at 0 to show that the $$k$$-th coefficient is 0, but of course that doesn't work here

## 1 Answer

The following proof is essentially taken from Wikipedia: Laurent series:

Let $$f(z) = \sum_{n=-\infty}^\infty a_n z^n$$ be a Laurent series in the annulus $$r < |z| < R$$ and assume that $$f(z) = 0$$ for all $$z$$.

Let $$k$$ be an integer and choose any $$\rho \in (r, R)$$. The series converges uniformly on $$|z| = \rho$$ so that we can change the order of integration and summation in the following calculation: $$0 = \frac{1}{2\pi i} \int_{|z| = \rho }\sum_{n=-\infty}^\infty a_n z^{n-k-1} \, dz = \sum_{n=-\infty}^\infty a_n \frac{1}{2\pi i} \int_{|z| = \rho } z^{n-k-1} \, dz = \sum_{n=-\infty}^\infty a_n \delta_{nk} = a_k \, ,$$ i.e. all $$a_k$$ are zero.