--- title: "Residue theorem" sort_title: "Residue theorem" date: 2021-11-13 categories: - Mathematics - Complex analysis layout: "concept" --- A function $$f(z)$$ is **meromorphic** if it is [holomorphic](/know/concept/holomorphic-function/) except in a finite number of **simple poles**, which are points $$z_p$$ where $$f(z_p)$$ diverges, but where the product $$(z - z_p) f(z)$$ is nonzero and still holomorphic close to $$z_p$$. In other words, $$f(z)$$ can be approximated close to $$z_p$$: $$\begin{aligned} f(z) \approx \frac{R_p}{z - z_p} \end{aligned}$$ Where the **residue** $$R_p$$ of a simple pole $$z_p$$ is defined as follows, and represents the rate at which $$f(z)$$ diverges close to $$z_p$$: $$\begin{aligned} \boxed{ R_p \equiv \lim_{z \to z_p} (z - z_p) f(z) } \end{aligned}$$ **Cauchy's residue theorem** for meromorphic functions is a generalization of Cauchy's integral theorem for holomorphic functions, and states that the integral on a contour $$C$$ purely depends on the simple poles $$z_p$$ enclosed by $$C$$: $$\begin{aligned} \boxed{ \oint_C f(z) \dd{z} = i 2 \pi \sum_{z_p} R_p } \end{aligned}$$ {% include proof/start.html id="proof-theorem" -%} From the definition of a meromorphic function, we know that we can decompose $$f(z)$$ like so, where $$h(z)$$ is holomorphic and $$z_p$$ are all its poles: $$\begin{aligned} f(z) = h(z) + \sum_{z_p} \frac{R_p}{z - z_p} \end{aligned}$$ We integrate this over a contour $$C$$ which contains all poles, and apply both Cauchy's integral theorem and Cauchy's integral formula to get: $$\begin{aligned} \oint_C f(z) \dd{z} &= \oint_C h(z) \dd{z} + \sum_{p} R_p \oint_C \frac{1}{z - z_p} \dd{z} = \sum_{p} R_p \: 2 \pi i \end{aligned}$$ {% include proof/end.html id="proof-theorem" %} This theorem might not seem very useful, but in fact, by cleverly choosing the contour $$C$$, it lets us evaluate many integrals along the real axis, most notably [Fourier transforms](/know/concept/fourier-transform/). It can also be used to derive the [Kramers-Kronig relations](/know/concept/kramers-kronig-relations).