A question about determing a residue

Question

Let $f(z)$ be analytic at $z=w$ and have a pole at $z=a$.

How does one show that the residue of $\displaystyle\frac{f(z)}{w-z}$ at $z=a$ equals the singular/principal part of $f(z)$ evaluated at $z=w$?

For example, let $ \displaystyle f(z) = \frac{\cot z}{z^{2}} = \frac{1}{z^{3}} - \frac{1}{3z} + O(z)$.

Then $ \displaystyle\text{Res} \Big[ \frac{\cot z}{z^{2}(2-z)},0 \Big] = \frac{1}{2^{3}} - \frac{1}{3(2)} = - \frac{1}{24}$.

It came up in a proof of the Mittag-Leffler partial fractions expansion theorem.

Answer 1

Good old-fashioned computation with Laurent series.

$$\begin{align} \frac{f(z)}{w-z} &= \frac{f(z)}{(w-a) - (z-a)} = \frac{1}{w-a} f(z) \frac{1}{1 - \frac{z-a}{w-a}}\\ &= \frac{1}{w-a}f(z)\sum_{\nu = 0}^\infty \left(\frac{z-a}{w-a}\right)^\nu\\ &= \frac{1}{w-a}\left(\sum_{k = -m}^\infty a_k (z-a)^k\right)\sum_{\nu = 0}^\infty \left(\frac{z-a}{w-a}\right)^\nu\\ &= \frac{1}{w-a} \sum_{s = -m}^\infty \left(\sum_{k = -m}^s\frac{a_k}{(w-a)^{s-k}}\right)(z-a)^s \end{align}$$

The residue is the coefficient of $(z-a)^{-1}$, that is

$$\frac{1}{w-a}\sum_{k = -m}^{-1} \frac{a_k}{(w-a)^{-1-k}} = \sum_{k=-m}^{-1} \frac{a_k}{(w-a)^{-k}} = \sum_{k=-m}^{-1} a_k(w-a)^k.$$

Answer 2

Expanding $(w-z)^{-1}$ about $z=a$ gives:

$$\frac{1}{w-a}\sum_{n=0}^{\infty}\left(\frac{z-a}{w-a}\right)^{n}$$

Now if

$$f(z)=\sum_{i=m}^{\infty}a_i(z-a)^i$$

Then notice that the coefficient of the term $(z-a)^{-1}$ in the expansion of $f(z)(w-z)^{-1}$ about $z=a$ will be:

$$\sum_{i+n=-1}\frac{a_i}{(w-a)^{n+1}}=\sum_{i=m}^{-1}a_i(w-a)^i$$

which is the singular part of $f$ evaluated at $w$.