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) = \frac{1}{\sin \frac{1}{z}} has z = \frac{1}{n π}, n \in Z

as singular points, while 0 is a non-isolated singular point because every deleted neighbourhood of 0 contains a singularity $\frac{1}{n π}$ 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 $b_{m + 1} = b_{m + 2} \dots = 0$ in the Laurent’s series of $f (z)$ about $a$ .

In other words, if the Laurent’s series is of the form $f (z) = \sum_{n = 0}^{\infty} a_{n} (z - a)^{n} + \frac{b_{1}}{z - a} + \frac{b_{2}}{(z - a)^{2}} + \dots + \frac{b_{m}}{(z - a)^{m}}, r_{2} < | z - a | < r_{1},$ 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, $f (z) = \frac{1}{z (z - 1)^{2}}$ has $z = 0$ as a simple pole and $z = 1$ as a pole of order 2.

Essential Singularity

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: $f (z) = \frac{\sin z}{z}$ , $z \neq 0$

⟹ f (z) = \frac{1}{z} [z - \frac{z^{3}}{3!} + \frac{z^{5}}{5!} - \dots] = 1 - \frac{z^{2}}{3!} + \frac{z^{4}}{5!} - \dots 0 < | z | < \infty

It has no principal part. So, $z = 0$ is a removable singularity if $\frac{\sin z}{z}$ is not defined at $z = 0$ .

Zero of order m

Let $f (z)$ be analytic at $z = z_{0} .$ If $f (z_{0}) = 0$ then $z_{0}$ is called a zero of $f (z) .$ If there is a positive integer $m$ such that $f^{'} (z_{0}) = 0, f^{''} (z_{0}) = 0, f^{'''} (z_{0}) = 0, \dots, f^{(m - 1)} (z_{0}) = 0$ and $f^{(m)} (z_{0}) \neq 0$ then $f (z)$ is said to have a zero of order $m$ at $z_{0}$ Thus, $f (z) = {(z - z_{0})}^{m} g (z), g (z_{0}) \neq 0$

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 $b_{1} = [Res f (z)]_{z = a}$ or $R (a)$ .

Methods of Finding Residue

1) If $z = a$ is a simple pole, then $R (a) = lim_{z \to a} (z - a) f (z)$

2) If $z = a$ is a pole of order $m,$ then $R (a) = \frac{1}{(m - 1)!} lim_{z \to a} {\frac{d^{m - 1}}{d z^{m - 1}} [(z - a)^{m} f (z)]}$

3) Let $f (z) = \frac{g (z)}{h (z)},$ where $g (z)$ and $h (z)$ are analytic functions at $z = a .$ If $h (a) = 0, h^{'} (a) \neq 0$ and $g (a) \neq 0$ are finite, then $z = a$ is a simple pole of $f (z)$ and $R (a) = lim_{z \to a} \frac{g (z)}{h^{'} (z)}$ .

4) Hence the residue of $f (z)$ at infinity is given by: $lim_{z \to \infty} [- z f (z)]$

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,

\int_{C} f (z) d z = 2 π i \sum residues of f 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 $f (z) = \frac{ϕ (z)}{(z - z_{0})^{m}}$ , where $f (z)$ is anaytic and non-zero at $z_{0}$ .

Moreover $R e s_{z = z_{0}}$ $f (z) = \frac{ϕ^{(m - 1)} (z_{0})}{(m - 1)!}$ if $m \geq 1$ .

[2019, 15M]

2) Show by applying the residue theorem that $\int_{0}^{\infty} \frac{d x}{(x^{2} + a^{2})^{2}} = \frac{π}{4 a^{3}}$ , $a > 0$ .

[2018, 15M]

3) State Cauchy’s residue theorem. Using it, evaluate the integral $\int_{C} \frac{e^{z} + 1}{z (z + 1) (z - i)^{2}} d z$ ; $C : | z | = 2$ .

[2015, 15M]

4) Evaluate the integral $\int_{0}^{π} \frac{d θ}{{(1 + \frac{1}{2} \cos θ)}^{2}}$ using residues.

[2014, 20M]

5) Using Cauchy’s residue theorem, evaluate the integral $I = \int_{0}^{π} \sin^{4} θ d θ$ .

[2013, 15M]

6) Let $f (z) = \frac{a_{0} + a_{1} \dots \dots + a_{n - 1} z^{n - 1}}{b_{0} + b_{1} z + \dots \dots \dots + b_{n} z^{n}}$ , $b_{n} \neq 0$ . 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 $\frac{a_{n} - 1}{b_{n}}$ .

[2009, 12M]

7) Find the residue of $\frac{\cot z \coth z}{z^{3}}$ at $z = 0$ .

[2008, 12M]

8) Evaluate (by using residue theorem), $\int_{0}^{2 π} \frac{d θ}{1 + 8 \cos^{2} θ}$ .

[2007, 15M]

9) With the aid of residues, evaluate $\int_{0}^{π} \frac{\cos 2 θ}{1 - 2 a \cos θ + α^{2}} d θ$ , $- 1 < a < 1$ .

[2006, 15M]

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