Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose that $z_0$ is a pole of $f$ of order $n$. Then:

$$Res_{z=z_0} (f(z)) = \frac{1}{(n-1)!} \lim _{z \to z_0} \frac{d^{n-1}}{dz^{n-1}} ((z-z_0)^{n} f(z))$$

How would I go about proving this?

share|cite|improve this question
up vote 1 down vote accepted

If $f(z)$ has a pole of order $n$ at $z_0$, then for some neighborhood $U$ of $z_0$, we have $$f(z)=g(z)+\frac{a_{-1}}{z-z_0}+\frac{a_{-2}}{(z-z_0)^2}+\cdots+\frac{a_{-n}}{(z-z_0)^n}\mbox{ for }z\in U-\{z_0\},$$ where $g$ is a holomorphic function in $U$. This implies that $$(z-z_0)^nf(z)=(z-z_0)^ng(z)+(z-z_0)^{n-1}a_{-1}+(z-z_0)^{n-2}a_{-2}+\cdots+a_{-n} \mbox{ for }z\in U-\{z_0\}.$$ Differentiate it $(n-1)$-times, we have $$\frac{d^{n-1}}{dz^{n-1}}(z-z_0)^nf(z)$$ $$=\frac{d^{n-1}}{dz^{n-1}}(z-z_0)^ng(z)+\frac{d^{n-1}}{dz^{n-1}}(z-z_0)^{n-1}a_{-1}+\frac{d^{n-1}}{dz^{n-1}}(z-z_0)^{n-2}a_{-2}+\cdots+\frac{d^{n-1}}{dz^{n-1}}a_{-n}$$ $$=\frac{d^{n-1}}{dz^{n-1}}(z-z_0)^ng(z)+\frac{d^{n-1}}{dz^{n-1}}(z-z_0)^{n-1}a_{-1}$$ $$=(z-z_0)G(z)+(n-1)!a_{-1}$$ for some holomorphic function $G$ in $U$. (In fact, $G$ can be written down explicitly in terms of $g$ and its derivative by using binormal theorem). Taking limit, we obtain $$\lim_{z\to z_0}\frac{d^{n-1}}{dz^{n-1}}(z-z_0)^nf(z)=\lim_{z\to z_0}(z-z_0)G(z)+(n-1)!a_{-1}=(n-1)!a_{-1}= (n-1)!Res_{z=z_0}(f(z)).$$

share|cite|improve this answer

Do you know that the residue is the coefficient of $(z-z_0)^{-1}$?

share|cite|improve this answer
Yes, I know what a residue is. I've tried doing case studies like when n = 2 (double poles) and I found that the formula worked but I have to show why this is true for any any n. – Low Scores Apr 11 '12 at 6:58
So if you take the expansion in powers of $z-z_0$ and do what the formula says to do, don't you get the coefficient of $(z-z_0)^{-1}$? – Gerry Myerson Apr 11 '12 at 7:56

The $(z-z_0)^{-1}$ coefficient of $f(z)$ is the $(z-z_0)^{n-1}$ coefficient of $(z-z_0)^nf(z)$, the latter of which is analytic in a neighborhood of $z_0$. Take a Taylor expansion then and note the $(n-1)$th coefficient...

share|cite|improve this answer

Consider the Cauchy integral formula. It implies for $g(z)$ analytic in neighborhood of $z_0$ $$\frac{1}{2\pi i} \oint_\gamma \frac{g(z)}{(z-z_0)^n} dz = \frac{1}{(n-1)!} \frac{d^{n-1}}{dz^{n-1}} g(z)|_{z=z_0}.$$ Let $g(z) = (z-z_0)^n f(z)$. Since the singularity of $f(z)$ at $z_0$ is isolated and of order $n$, $g(z)$ is analytic in a neighborhood of $z_0$. Therefore, $$\begin{eqnarray} \frac{1}{2\pi i} \oint_\gamma f(z) dz &=& \mathrm{Res}_{z=z_0} f(z)\\ &=& \frac{1}{(n-1)!} \frac{d^{n-1}}{dz^{n-1}} [(z-z_0)^n f(z)]|_{z=z_0}. \end{eqnarray}$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.