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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.

Random Variable
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2 Answers2

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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

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

Daniel Fischer
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  • Could you explain a bit more how you went from line 3 to line 4? – Random Variable Jul 29 '13 at 22:02
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    Cauchy product of Laurent series $(\sum_{k = -\infty}^\infty a_kz^k)(\sum_{n = -\infty}^\infty b_nz^n) = \sum_{r = -\infty}^\infty c_r z^r$, where $c_r = \sum_{k = -\infty}^\infty a_k b_{r-k}$. Here, we have finite principal parts, $k \geqslant -m$ (where $m$ is the order of the pole of $f$) and $\nu \geqslant 0$, so only the products with $k \geqslant -m$ and $s - k \geqslant 0$, that is, $-m \leqslant k \leqslant s$, contribute to the coefficient $c_s$. – Daniel Fischer Jul 29 '13 at 22:08
  • Thanks. Similar to the Cauchy product of power series, do we just need one of the Laurent series to converge absolutely? – Random Variable Jul 29 '13 at 23:01
  • I'm not sure whether that would be enough. Normally, one considers the product of Laurent series (with potentially infinite principal part) only in the annulus in which both converge, and there both converge absolutely (locally uniformly). If, like here, both have finite principal part, it's just a product of power series with index-shift, so then it's the same. – Daniel Fischer Jul 29 '13 at 23:07
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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 $w$.

Jared
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