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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?

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4 Answers 4

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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 $$\lim_{z\to z_0}\frac{d^{n-1}}{dz^{n-1}}(z-z_0)^nf(z)=\lim_{z\to z_0}(z-z_0)G(z)+(n-1)!a_{-1}=(n-1)!a_{-1}= (n-1)!Res_{z=z_0}(f(z)).$$

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

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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...

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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, $$\begin{eqnarray} \frac{1}{2\pi i} \oint_\gamma f(z) dz &=& \mathrm{Res}_{z=z_0} f(z)\\ &=& \frac{1}{(n-1)!} \frac{d^{n-1}}{dz^{n-1}} [(z-z_0)^n f(z)]|_{z=z_0}. \end{eqnarray}$$

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