# If $f$ has pole of order $m$, then $\text{res}\left(f,z_0\right)=\lim_{z\to z_0}\frac{1}{(m-1)!}\left\{(z-z_0)^mf(z)\right\}^{(m-1)}$

Statement: Let $$f(z):=\sum_{k=-\infty}^\infty a_kz^k$$ have a pole of order $m$ at $z_0$. Then $$\text{res}\left(f,z_0\right)=\lim_{z\to z_0}\frac{1}{(m-1)!}\left\{(z-z_0)^mf(z)\right\}^{(m-1)}$$

Since $f$ has a pole of order $m$ $$(*)\;\;\;f(z)=\sum_{k=0}^\infty a_kz_k+\sum_{k=1}^m a_{-k}z^{-k}$$ By the Residue theorem we've got $$\text{res}\left(f,z_0\right)\text{ind}_\gamma\text{ }z_0=\frac{1}{2\pi i}\int_\gamma f(z)\;dz$$ where $\gamma$ is a nullhomologous path such that $z_0\notin\gamma^*$. Now comes the point where I'm a little bit confused: Wouldn't Cauchy's Integral theorem yield $\int_\gamma f(z)\;dz=0$? [No, it would not - as stated in the comment of Squirtle and the answer of Daniel Fischer].

However, how to proceed? I'm unsure whether $(*)$ does help here or not. Another fact that follows from the pole of order $m$ at $z_0$ is the existence of a holomorphic function $g$ with $$f(z)=\frac{1}{(z-z_0)^m}g(z)$$ But I still got problems to obtain the statement.

Notes:

• $\text{ind}_\gamma\text{ }z_0$ denotes the Winding number
• $\gamma^*$ denotes the trace
• Its only zero if there are no poles, or more precisely: $\int_\gamma f(z)dz=0$ whenever there is no pole. – Squirtle Jul 11 '14 at 18:18

## 1 Answer

where $\gamma$ is a nullhomologous path such that $z_0 \notin \gamma^\ast$

The point is: "nullhomologous with respect to which domain"?

Cauchy's integral theorem states that if $h\colon \Omega \to \mathbb{C}$ is holomorphic, then

$$\int_\kappa h(z)\,dz = 0$$

for every closed path of integration (more generally, every cycle) $\kappa$ that is nullhomologous in $\Omega$.

Here, you have $f$ meromorphic on $U$, with a pole in $z_0\in U$ (and, for simplicity, no other pole), and the residue theorem says that

$$\frac{1}{2\pi i} \int_\gamma f(z)\,dz = \operatorname{ind}_\gamma z_0\cdot \operatorname{res}(f,z_0)$$

for every closed path of integration (cycle) $\gamma$ that is nullhomologous in $U$ and does not pass through $z_0$, while Cauchy's integral theorem says

$$\frac{1}{2\pi i} \int_\gamma f(z)\,dz = 0$$

for every closed path of integration (cycle) $\gamma$ that is nullhomologous in $U\setminus \{z_0\}$.

If $\gamma$ is nullhomologous in $U\setminus \{z_0\}$, then it is also nullhomologous in $U$, but additionally $\operatorname{ind}_\gamma z_0 = 0$. So in that case, the residue theorem says exactly the same thing as the integral theorem.

If $\operatorname{ind}_\gamma z_0 \neq 0$, then $\gamma$ can be nullhomologous in $U$, but it is not nullhomologous in $U\setminus \{z_0\}$, so then Cauchy's integral theorem says nothing about the value of the integral.

Now, concerning the formulae

$$\operatorname{res}(f,z_0) = \frac{1}{(m-1)!} \lim_{z\to z_0} \left\{ (z-z_0)^m f(z)\right\}^{(m-1)}$$

and $(\ast)$:

The residue of $f$ in $z_0$ is the coefficient $a_{-1}$ in the Laurent expansion

$$f(z) = \sum_{n=-\infty}^\infty a_n (z-z_0)^n.\tag{0}$$

From the definition of the residue as an integral, we obtain that by expanding the integrand into its Laurent series, and interchanging summation and integration (which is legitimate since the Laurent series converges uniformly on all small enough circles centred at $z_0$). All terms except the $a_{-1}(z-z_0)^{-1}$ terms have a primitive on $\mathbb{C}\setminus\{z_0\}$, and hence vanishing integral.

If $f$ has a pole of order $m$ in $z_0$, then all coefficients $a_n$ for $n < -m$ vanish, and $(0)$ becomes, writing $(\ast)$ as a single series,

$$f(z) = \sum_{n=-m}^\infty a_n (z-z_0)^n.\tag{1}$$

From that, we obtain the Taylor expansion of $g(z) = (z-z_0)^m\cdot f(z)$ about $z_0$ by a simple multiplication:

$$(z-z_0)^mf(z) = \sum_{n=-m}^\infty a_n(z-z_0)^{n+m} = \sum_{k=0}^\infty a_{k-m}(z-z_0)^k = \sum_{k=0}^\infty b_k (z-z_0)^k.\tag{2}$$

By the shifting of the indices, we have

$$\operatorname{res}(f,z_0) = a_{-1} = a_{(m-1)-m} = b_{m-1},$$

so the residue of $f$ in $z_0$ is the $(m-1)^{\text{st}}$ coefficient of the Taylor expansion of $g$, which is

$$b_{m-1} = \frac{g^{(m-1)}(z_0)}{(m-1)!}.$$

Since $g$ is holomorphic, all its derivatives are continuous, and so

$$g^{(m-1)}(z_0) = \lim_{z\to z_0} g^{(m-1)}(z) = \lim_{z\to z_0} \left(\frac{d}{dz}\right)^{m-1}\left((z-z_0)^mf(z)\right)$$

and we have the limit formula.

• Thanks for your detailed explanation. It helped a lot to better understand the relationship between the two theorems. However, the main question was not about the applicability of Cauchy's integral theorem, but the proof of the given statement (maybe I didn't make that clear enough). I've updated my question to make things more clear. – 0xbadf00d Jul 12 '14 at 13:18
• Okay, give me a few minutes to read ;) – Daniel Fischer Jul 12 '14 at 13:21
• I've added an explanation of the formula. If it's not clear, yell. – Daniel Fischer Jul 12 '14 at 13:43
• That holds because $\int_{\partial U} (z-z_0)^k\,dz = 0$ for all $k\in \mathbb{Z}\setminus \{-1\}$ (these functions have a primitive, $\frac{1}{k+1}(z-z_0)^{k+1}$, so the integral over any closed path is $0$). If you expand $f$ into its Laurent series and interchange the order of summation and integration, you get $$\frac{1}{2\pi i}\int_{\partial U} f(z)\,dz = \sum_{k=-\infty}^\infty \frac{a_k}{2\pi i}\int_{\partial U} (z-z_0)^k\,dz = \frac{a_{-1}}{2\pi i}\int_{\partial U} \frac{dz}{z-z_0} = a_{-1}.$$ – Daniel Fischer Jul 12 '14 at 17:37
• $(z-z_0)^{-1}$ has local primitives, and these are indeed branches of the logarithm plus some constant (on a connected set). But the point is that it has no primitive on $\mathbb{C}\setminus\{z_0\}$, its integral over a curve winding around $z_0$ is nonzero ($2\pi i \cdot \operatorname{ind}_\gamma z_0$, as you know), thus that piece is what's left over (the residue) when integrating. – Daniel Fischer Jul 12 '14 at 21:51