Zeros and Singularities: Types, Examples, Residue, Theorem

In Complex Analysis, zeroes are points where the function vanishes while singularities are points where the function loses its analytic property (differentiability). Here we study zeros and singularities along with their types, examples, residues and related theorems.

Zero of a Function

Definition: Let f(z): D → ℂ be a function. A point z=a is called a zero of f(z) if f(a) =0.

For example, consider f(z)=z². As f(0)=0² =0, so z=0 is a zero of f(z).

Order of zero

Let z=a be a zero of f(z). If f(a)=f'(a)=f”(a)=…+f^(m-1)(a)=0 but f^(m)(a)≠0, then z=a is called a zero of order m.

Example 1: f(z)=z².

z=0 is a zero of order 2.

Example 2: f(z)=sinz.

z=nπ is a zero of order 1 (simple zero).

Singular Point

Definition: A point z=a is called a singular point of f(z) if f(z) is not analytic at z=a.

Example: f(z) =1/z.

z=0 is a singular point.

Types of Singularities

There are two types of singularities:

• Isolated singularity
• non-isolated singularity

Isolated Singularity

A singular point z=a is called an isolated singularity of f(z) if f(z) is analytic in some deleted neighbourhood of a.

For example, f(z)=1/z.

z=0 is an isolated singularity. Because, it is analytic at every point of a neighbourhood of 0 other than 0.

Question: Find the isolated singularities of f(z) = $\dfrac{z-2}{z^2} \sin \left( \dfrac{1}{z-1} \right)$.

Answer: z=0, 1.

Types of Isolated Singularity

Isolated singularity can be of three types:

1. Removable singularity
2. Pole
3. Essential singularity.

Removable Singularity

An isolated singularity z=a of f(z) is called a removable singularity if f(a) is defined so that f(z) becomes analytic at z=a.

In other words, lim_z→a f(z) exists finitely.

Example, f(z) = $\dfrac{\sin z}{z}$, z≠0.

As lim_z→0 $\dfrac{\sin z}{z}$ = 1, so define f(0)=1.

Therefore, z=0 is a removable singularity of sinz/z.

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Pole

Recall, the Laurent series expansion of f(z) about z=a is given by

f(z) = $\displaystyle \sum_0^\infty A_n(z-a)^n +\displaystyle_{n=1}^\infty A_n(z-a)^{-n}$.

The part $\displaystyle \sum_{n=1}^\infty A_n(z-a)^{-n}$ is called the principal part in the Laurent series expansion of f(z).

Definition of Pole: If the principal part of the Laurent series expansion of f(z) about z=a contain a finite number of terms (say m terms), then z=a is called a pole of order m.

Example, f(z) = $\dfrac{\sin z}{z^3}$, z≠0.

z=0 is a singular point. Is this a pole?

We know sin z = $z-\dfrac{z^3}{3!}+\dfrac{z^5}{5!}-\cdots$.

So f(z) = $\dfrac{1}{z^3} \left( z-\dfrac{z^3}{3!}+\dfrac{z^5}{5!}-\cdots \right)$

= $\dfrac{1}{z^2} +\dfrac{0}{z} -\dfrac{1}{3!}+\dfrac{z^2}{5!}-\cdots$

So the principle part is $\dfrac{1}{z^2} +\dfrac{0}{z}$. Therefore, by definition, z=0 is a pole of f(z) of order 2 with residue 0.

Theorem: A point z=a is called a pole of f(z) if and only if limz→a f(z) = ∞. Moreover, limz→a (z-a)nf(z) = a finite number, for some n. Minimum such n is the order of the pole z=a.

Example, f(z) = $\dfrac{1}{e^z-1}$, z≠0.

z=0 is a singular point.

As lim_z→0 $\dfrac{1}{e^z-1}$ = ∞, the point z=0 is a pole.

Now, lim_z→0 $(z-0)^1\dfrac{1}{e^z-1}$ = 1 (0/0 form). Therefore, z=0 is a pole of order 1 (simple pole).

Related Article: Order of Pole, Residue at Pole

Essential Singularity

A singular point z=a of f(z) is called an essential singular point if it is neither a removable singularity nor a pole.

Theorem: A point z=a is an essential singularity of f(z) if and only if lim_z→a f(z) does not exist finitely or infinitely.

Example, f(z) = sin(1/z)

As we can show lim_z→0 sin(1/z) does not exist, so z=0 is a point of essential singularity of sin(1/z).

Summary:

The types of the singular point z=a of f(z) can be determined as follows:

Types of Singularity	Condition
Removable singularity	limz→a f(z) exists finitely.
Pole	limz→a f(z) = ∞
Essential Singularity	limz→a f(z) does not exist finitely or infinitely.

Residue

The Laurent series expansion of f(z) about z=a is given by

f(z) = $\displaystyle \sum_0^\infty A_n(z-a)^n +\displaystyle_{n=1}^\infty A_n(z-a)^{-n}$.

The quantity A_-1, the co-efficient of 1/(z-a), is called the residue of f(z) at z=a, and it is denoted by Res_z=af(z).

So Res_z=af(z) = A_-1.

Example, sin(1/z) = $\dfrac{1}{z}-\dfrac{1}{3!}\dfrac{1}{z^3}+\dfrac{1}{5!}\dfrac{1}{z^5}-\cdots$

So A_-1 = 1 (the co-efficient of 1/z).

Therefore, Res_z=0 sin(1/z) = 1.

How to Find Residue

Theorem: Let f(z) has a pole of order m at z0, so we can write f(z) = $\dfrac{\phi(z)}{(z-z_0)^m}$ for some analytic function $\phi(z)$ with $\phi(z_0) \neq 0$. Then we have: Res$_{z=z_0}$ f(z) = $\lim\limits_{z \to z_0} \dfrac{\phi^{(m-1)(z)}}{(m-1)!}$.

Let us now learn how to apply the above theorem in order to compute residue at a pole.

$\boxed{\color{blue}\textbf{Question}:}$ Find the residue of f(z) = $\dfrac{z}{(z+1)(z-1)^2}$ at z=1.

$\boxed{\color{red}\textbf{Answer:}}$ Note that

f(z) = $\dfrac{\frac{z}{z+1}}{(z-1)^2}$

Order of the pole z=1 is 2.

That is, m=2.

Also, $\phi(z)=\dfrac{z}{z+1}$.

Therefore, residue = $\lim\limits_{z \to 1} \dfrac{\phi^{(m-1)(z)}}{(m-1)!}$

= $\lim\limits_{z \to 1} \dfrac{\phi'(z)}{1!}$

= $\lim\limits_{z \to 1} \phi'(z)$

= $\lim\limits_{z \to 1} \dfrac{d}{dz} \left( \dfrac{z}{z+1} \right)$

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

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

= $\dfrac{1}{(1+1)^2}$

= $\dfrac{1}{4}$.

Residue at Simple Pole

Theorem: Let f(z) = $\dfrac{g(z)}{h(z)}$ where h(z) has a simple zero at z0 and g(z) is analytic at z=z0 with $g(z_0) \neq 0$. Then we have: Res$_{z=z_0}$ f(z) = $\dfrac{g(z_0)}{h'(z_0)}$.

$\boxed{\color{blue}\textbf{Question 1}:}$ Find the residue of f(z) = $\dfrac{1}{\sin z}$ at z=0.

$\boxed{\color{red}\textbf{Answer:}}$ Here, g(z)=1 and h(z)=sin z.

h(0) = sin 0 = 0

h'(0) = cos 0 ≠ 0.

So z=0 is a simple zero of h(z)=sin z.

Also, g(z) = 1 is analytic at z=0 with g(0)≠ 0.

Therefore, z=0 is a simple pole of f(z).

So by the above theorem, Res_z=0 $\dfrac{1}{\sin z}$ = $\dfrac{1}{\cos 0}$ = $\dfrac{1}{1}=1.$

$\boxed{\color{blue}\textbf{Question 2}:}$ Find the residue of f(z) = $\cot z$ at z=0.

$\boxed{\color{red}\textbf{Answer:}}$ Here, f(z) = cot z = $\dfrac{\cos z}{\sin z}$.

So g(z)=cos z and h(z)=sin z.

Now, h(0) = sin 0 = 0

h'(0) = cos 0 ≠ 0.

So z=0 is a simple zero of h(z)=sin z.

Also, g(z) = cos z is analytic at z=0 with g(0)≠ 0.

Therefore, z=0 is a simple pole of f(z).

So by the above theorem, Res_z=0 $\cot z$ = Res_z=0 $\dfrac{\cos 0}{\cos 0}=1.$

Cauchy’s Residue Theorem

If f(z) is an analytic function inside and on a simple closed curve C except at a finite number of poles z₁, z₂, …,z_n inside C, then

$\displaystyle \int_C f(z~dz)$ $=2\pi i (R_1+R_2+\cdots+R_n)$

where R_i = Res$_{z=z_i}$ f(z).

Example 1

sin(1/z) = $\dfrac{1}{z}-\dfrac{1}{3!}\dfrac{1}{z^3}+\dfrac{1}{5!}\dfrac{1}{z^5}-\cdots$

So Res_z=0 sin(1/z) = 1 (the co-efficient of 1/z).

If C is any closed contour containing 0, then

$\displaystyle \int_C \sin(\frac{1}{z})~dz$ $=2\pi i \times 1$ $=2\pi i$.

Example 2

e^z = $1+\dfrac{z}{1!}+\dfrac{z^2}{2!}+\dfrac{z^3}{3!}+\cdots$

So e^1/z = $1+\dfrac{1}{z}+\dfrac{1}{2!}\dfrac{1}{z^2}+\dfrac{1}{3!}\dfrac{1}{z^3}+\cdots$

So Res_z=0 e^1/z = 1 (the co-efficient of 1/z).

Thus, $\displaystyle \int_{|z|=1} e^{\frac{1}{z}}~dz$ $=2\pi i \times 1$ $=2\pi i$.

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