Order of Pole: How to Find | Residue at Pole

To find the residue at pole it is very important to know the order of pole. In this page, we will learn when a singular point can be a pole and how to find the order of that pole.

Definition of a Pole

A point z=a is called a pole of a complex function f(z) if

1. f(z) is not analytic at z=a.
2. lim_z→a f(z) = ∞.

Order of a Pole

Let z=a be a pole of f(z). A natural number n is called the order of the pole if n is minimum such that

lim_z→a (z-a)ⁿf(z) = a finite number.

Note that if z=a is a pole of f(z) of order n, then we can write

f(z) = $\dfrac{\phi(z)}{(z-a)^n}$

where $\phi(z)$ is analytic at z=a and $\phi(a) \neq 0$.

Residue at Pole

$\boxed{\color{blue}\text{Theorem:}}$ Let z=a be a pole of f(z) of order n, that is,

f(z) = $\dfrac{\phi(z)}{(z-a)^n}$

where $\phi(z)$ is analytic at z=a and $\phi(a) \neq 0$. Then the residue of f(z) at z = a, denoted by Res_z=a f(z), is given by

Res_z=a f(z) = lim_z→a $\dfrac{\phi^{(n-1)}(z)}{(n-1)!}$.

Solved Problems

Order of Pole of sinz/z³

Q1: Find the order of the pole of $\dfrac{\sin z}{z^3}$.

Answer:

Let $f(z)=\dfrac{\sin z}{z^3}$.

Note that z=0 is a singular point of f(z). Now,

lim_z→0 f(z) = lim_z→0 $\dfrac{\sin z}{z^3}$ = lim_z→0 $\left( \dfrac{\sin z}{z} \times \dfrac{1}{z^2} \right)$ = 1 × ∞ = ∞.

Thus, z=0 is a pole of sinz/z³.

Now, to find the order of the pole z=0, we need find minimum n ∈ ℕ such that

lim_z→0 (z-0)ⁿf(z) is finite.

⇒ lim_z→0 $\dfrac{z^n \sin z}{z^3}$ is finite.

Using the fact lim_z→0 $\dfrac{\sin z}{z}$ =1, we conclude that n=2.

Therefore, the order of the pole z=0 of sinz/z³ is 2.

HomePage of Engineering Math II

Q2: Find the residue of $\dfrac{1}{z(z-1)^2}$.

Answer:

Let $f(z)=\dfrac{1}{z(z-1)^2}$.

The singular points of f(z) are z=0, 1.

As lim_z→0 f(z) = ∞ and lim_z→1 f(z) = ∞, both z=0, 1 are poles of f(z). Also, z=0 is a pole of order 1 and z=1 is a pole of order 2.

$\boxed{\text{Residue at } z=0:}$

Let us write $f(z)=\dfrac{\frac{1}{(z-1)^2}}{z}$, so φ(z) = $\dfrac{1}{(z-1)^2}$.

As z=0 is a pole of order 1, the residue at z=0 is given by

Res_z=0f(z) = $\lim\limits_{z \to 0}$ $\dfrac{\phi^{(1-1)}(z)}{(1-1)!}$

= $\lim\limits_{z \to 0}$ φ(z)

= $\lim\limits_{z \to 0}$ $\dfrac{1}{(0-1)^2}$

= 1.

$\boxed{\text{Residue at } z=1:}$

Write $f(z)=\dfrac{\frac{1}{z}}{(z-1)^2}$, so φ(z) = $\dfrac{1}{z}$.

As z=1 is a pole of order 2, the residue at z=2 is given by

Res_z=1 f(z) = lim_z→1 $\dfrac{\phi^{(2-1)}(z)}{(2-1)!}$

= $\lim\limits_{z \to 1}$ φ^‘(z)

= $\lim\limits_{z \to 1}$ $\dfrac{-1}{z^2}$

= $\dfrac{-1}{1^2}$

= -1.