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--- math121a-f23:september_18_monday [2023/09/18 16:50]
pzhou created
+++ math121a-f23:september_18_monday [2026/02/21 14:41] (current)
@@ Line 8: / Line 8: @@
 The coefficient $a_{-1}$ is called the residue, and denoted as $Res_{z_0} f := a_{-1}$. ($:=$ read 'is defined as')
-Thm: let $f$ has a pole (of some order) at $z_0$, and let $C_{\epsilon}(z_0)$ be a small circle of radius $\epsilon$ around $z_0$ positively oriented (i.e in clockwise direction) so that $f$ is holomorphic on $C$, and $C$ encloses only one pole. Then $$\oint_{C_{\epsilon}(z_0)} f(z) dz = (2\pi i) Res_{z_0} f. $$
+==== How to find the residue ? ====
+. If the pole is order $1$, then $Res_{z_0}f(z) = \lim_{z \to z_0} f(z) (z-z_0)$.
+Example: residue of $f(z) = (z-2) / (z-1)$ at $z=1$.
+. If the pole at $z_0$ is of order $m>1$, then just Laurent expand. More precisely, do Taylor expansion of
+$$(z-z_0)^m f(z) = a_{-m} + \cdots + a_{-1} (z-z_0)^{m-1} + \cdots $$
+and then you get $a_{-1}$.
+Example: residue of $f(z) = [(z-2) / (z-1)]^2$ at $z=1$.
+==== Residue Theorem ====
+**Thm:** let $f$ has a pole (of some order) at $z_0$, and let $C$ be a small circle of radius $\epsilon$ around $z_0$ positively oriented (i.e in counter-clockwise direction) so that $f$ is holomorphic on $C$, and $C$ encloses only one pole. Then $$\oint_{C} f(z) dz = (2\pi i) Res_{z_0} f. $$
+Proof sketch: write $f$ as a Laurent expansion, then integrate term by term.

Lecture Notes