Cauchy’s Residue Theorem
We will cover following topics
Classification of Singularities
A point at which a complex function \(f(z)\) is analytic is called a regular point or ordinary point of \(f(z)\). A point \(z=a\) is a singular point of \(f(z)\) if \(f(z)\) is not analytic (or not even defined), but is analytic at some point in every deleted neighbourhood of \(a\).
Isolated Singularity
A singular point \(z=a\) is called an isolated singularity of the function \(f(z)\) if there exists a neighbourhood of \(a\) in which there is no other singularity.
A singular point which is not isolated is called a non-isolated singularity. For example,
\[f(z)=\dfrac{1}{\sin \dfrac{1}{z}} \text { has } z=\dfrac{1}{n \pi}, n \in Z\]as singular points, while 0 is a non-isolated singular point because every deleted neighbourhood of 0 contains a singularity \(\dfrac{1}{n \pi}\) for large \(n\).
Pole
An isolated singular point \(a\) of \(f(z)\) is said to be a pole of order \(m,\) if there exists a positive integer \(m\) such that \(b_{m} \neq 0\) and \(b_{m+1}=b_{m+2} \cdots=0\) in the Laurent’s series of \(f(z)\) about \(a\).
In other words, if the Laurent’s series is of the form \(f(z)=\sum_{n=0}^{\infty} a_{n}(z-a)^{n}+\dfrac{b_{1}}{z-a}+\dfrac{b_{2}}{(z-a)^{2}}+\cdots+\dfrac{b_{m}}{(z-a)^{m}}, r_{2}<|z-a|<r_{1}, \quad\) where \(b_{m} \neq 0\), then the point \(a\) is called a pole of order \(m\).
If \(m=1,\) then \(a\) is called a simple pole and if \(m=2,\) then \(a\) is called a double pole or pole of order 2. For example, \(f(z)=\dfrac{1}{z(z-1)^{2}}\) has \(z=0\) as a simple pole and \(z=1\) as a pole of order 2.
Removable Singularity
An isolated singular point \(z=a\) of \(f(z)\) is called a removable singularity of \(f(z)\) if in some neighbourhood of \(a\) the Laurent’s series expansion of \(f(z)\) has no principal part.
For example: \(f(z)=\dfrac{\sin z}{z}\),\(z \neq 0\)
\[\implies f(z)= \dfrac{1}{z}\left[z-\dfrac{z^{3}}{3 !}+\dfrac{z^{5}}{5 !}-\cdots\right]=1-\dfrac{z^{2}}{3 !}+\dfrac{z^{4}}{5 !}-\cdots \quad 0<\vert z \vert <\infty\]It has no principal part. So, \(z=0\) is a removable singularity if \(\dfrac{\sin z}{z}\) is not defined at \(z=0\).
Zero of order m
Let \(f(z)\) be analytic at \(z=z_{0} .\) If \(f\left(z_{0}\right)=0\) then \(z_{0}\) is called a zero of \(f(z) .\) If there is a positive integer \(m\) such that \(f^{\prime}\left(z_{0}\right)=0, \quad f^{\prime \prime}\left(z_{0}\right)=0, \quad f^{\prime \prime \prime}\left(z_{0}\right)=0, \cdots, \quad f^{(m-1)}\left(z_{0}\right)=0\) and \(f^{(m)}\left(z_{0}\right) \neq 0\) then \(f(z)\) is said to have a zero of order \(m\) at \(z_{0}\) Thus, \(f(z)=\left(z-z_{0}\right)^{m} g(z), g\left(z_{0}\right) \neq 0\)
Residue
Let \(z=a\) be an isolated singular point of \(f(z)\). The coefficient \(b_{1}\) of \((z-a)^{-1}\) in the Laurent’s series expansion of \(f(z)\) about \(a\) is called the residue of \(f(z)\) at \(z=a\). We denote \(b_{1}=[\operatorname{Res} f(z)]_{z=a}\) or \(R(a)\).
Methods of Finding Residue
1) If \(z=a\) is a simple pole, then \(R(a)=\lim _{z \rightarrow a}(z-a) f(z)\)
2) If \(z=a\) is a pole of order \(m,\) then \(R(a)=\dfrac{1}{(m-1) !} \lim _{z \rightarrow a}\left\{\dfrac{d^{m-1}}{d z^{m-1}}\left[(z-a)^{m} f(z)\right]\right\}\)
3) Let \(f(z)=\dfrac{g(z)}{h(z)},\) where \(g(z)\) and \(h(z)\) are analytic functions at \(z=a .\) If \(h(a)=0, h^{\prime}(a) \neq 0\) and \(g(a) \neq 0\) are finite, then \(z=a\) is a simple pole of \(f(z)\) and \(R(a)=\lim _{z \rightarrow a} \dfrac{g(z)}{h^{\prime}(z)}\).
4) Hence the residue of \(f(z)\) at infinity is given by: \(\lim _{z \rightarrow \infty}[-\mathrm{zf}(\mathrm{z})]\)
Cauchy’s Residue Theorem
Suppose \(f(z)\) is analytic in the region \(A\) except for a set of isolated singularities and Let \(C\) be a simple closed curve in \(C\) that doesn’t go through any of the singularities of \(f\) and is oriented counterclockwise.
Then, according to Cauchy’s Residue Theorem,
\[\quad \int_{C} f(z) d z=2 \pi i \sum \text { residues of } f \text { inside } C\]PYQs
Cauchy’s Residue Theorem
1) Show that an isolated singular point \(z_o\) of a function \(f(z)\) is a pole of order \(m\) if and only if \(f(z)\) can be written in the form \(f(z)=\dfrac{\phi(z)}{(z-z_0)^m}\), where \(f(z)\) is anaytic and non-zero at \(z_0\).
Moreover \(Res_{z=z_0}\) \(f(z) = \dfrac{ \phi^{(m-1)} (z_0) }{(m-1)!}\) if \(m \geq 1\).
[2019, 15M]
2) Show by applying the residue theorem that \(\int^{\infty}_0 \dfrac{dx}{(x^2+a^2)^2}=\dfrac{\pi}{4a^3}\), \(a>0\).
[2018, 15M]
3) State Cauchy’s residue theorem. Using it, evaluate the integral \(\int_{C} \dfrac{e^{z}+1}{z(z+1)(z-i)^{2}} d z\); \(C: \vert z \vert=2\).
[2015, 15M]
4) Evaluate the integral \(\int_{0}^{\pi} \dfrac{d \theta}{\left(1+\dfrac{1}{2} \cos \theta\right)^{2}}\) using residues.
[2014, 20M]
5) Using Cauchy’s residue theorem, evaluate the integral \(I=\int_{0}^{\pi} \sin ^{4} \theta d \theta\).
[2013, 15M]
6) Let \(f(z)=\dfrac{a_{0}+a_{1} \ldots \ldots+a_{n-1} z^{n-1}}{b_{0}+b_{1} z+\ldots \ldots \ldots+b_{n} z^{n}}\), \(b_{n} \neq 0\). Assume that the zeros of the denominator are simple. Show that the sum of the residues of \(f(z)\) at its poles is equal to \(\dfrac{a_{n}-1}{b_{n}}\).
[2009, 12M]
7) Find the residue of \(\dfrac{\cot z \operatorname{coth} z}{z^{3}}\) at \(z=0\).
[2008, 12M]
8) Evaluate (by using residue theorem), \(\int_{0}^{2 \pi} \dfrac{d \theta}{1+8 \cos ^{2} \theta}\).
[2007, 15M]
9) With the aid of residues, evaluate \(\int_{0}^{\pi} \dfrac{\cos 2 \theta}{1-2 a \cos \theta+\alpha^{2}} d \theta\), \(-1 < a < 1\).
[2006, 15M]