9.3 Residues and the Residue Theorem
Definition 9.22. The residue of an analytic function \(f(z)\) at an isolated singularity \(z_0\) denoted by \(\,\operatorname {Res}\big [f(z), z_0\big ]\,\)
is given by
\[\operatorname {Res}\big [ f(z), z_0\big ] = \frac {1}{2\pi i}\int _{\Gamma } f(\zeta )\,d\zeta \]
where \(\Gamma \) is any simple closed curve described in a positive sense containing any \(z_0\) in its interior.
Now, from Laurent’s theorem, it follows that \(\operatorname {Res}[f(z), z_0] = a_{-1}\).
Residue at a removable singularity \(\,\operatorname {Res}[f(z), z_0] = 0\)
Residue at a Pole
If \(z_0\) is a simple pole, then \(\,\boxed {\operatorname {Res}[f(z), z_0] = \lim \limits _{z\rightarrow z_0} (z - z_0)\,f(z)}\)
Proof. If \(z_0\) is a simple pole for \(f(z)\). Then \begin {align*} f(z) & = \frac {a_{-1}}{(z - z_0)} + \sum ^{\infty }_{n = 0} a_n \, (z - z_0)^n\\\\ \implies \quad (z - z_0) f(z) & = a_{-1} + (z - z_0) \sum ^{\infty }_{n= 0} a_n (z- z_0)^n\\\\ \lim _{z \rightarrow z_0} (z-z_0)f(z) & = \operatorname {Res}[f(z), z_0] + 0\\\\ \end {align*} □
Cauchy’s Residue Theorem
Let \(f\) be analytic inside and on a positively oriented contour \(\Gamma \) except possibly for a finite number of
poles \(a_1, \ldots , a_n\) inside \(\Gamma \). Then
\[\int _{\Gamma } f(z)dz = 2\pi i \sum ^n_{k = 1} \operatorname {Res}[ f(z), a_k]\]
Proof. Let \(f_k(z)\) be the principal part of the Laurent series expansion of \(f\) about \(a_k\) (for \(k = 1, 2,\cdots n)\)
\(g = f - \displaystyle {\sum ^n_{k = 1} f_k}\) has a removable singularities at \(a_k,\, k = 1, \ldots , n\). Remove them and redefine \(g\).
Now \(g\) is analytic on and inside \(\Gamma \). By Cauchy’s theorem \begin {align*} \int _{\Gamma } g(z) dz & = 0\\\\ \implies \quad \int _{\Gamma } f(z) dz - \sum ^n_{k = 1}\int _{\Gamma } f_k (z) dz & = 0\\\\ \implies \quad \int _{\Gamma } f(z) dz & = \sum ^n_{k = 1} \int _{\Gamma } f_k (z) dz\\\\ \int _{\Gamma }f(z) dz & = 2\pi i \sum ^n_{k = 1} \operatorname {Res}\big [f(z); a_k\big ]\\\\ \end {align*}
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