Zeros and Singularities

The zeros and singularities of a complex analytic function are points where the given function vanishes and ceases to be analytic, respectively, within a domain of that function. An analytic complex function is differentiable at each point of its domain of the complex plane. The zero of analytic function is a point at which the function vanishes, or its value becomes zero, which is analogous to the zero of a real polynomial function.

Singularities of a complex function are points in its domain where the function does not act as an analytic function. There are several types of singularities when we are dealing with analytic functions.

Zeros of a Complex Function

A zero of an analytic function f(z) is any value of z for which f vanishes or we can say,

A zero of order k, a function f(z) which is analytic in a domain D, has a zero of order k at point

z = a in D if and only if the f⁽ⁿ⁾(a) = 0 for n = 1, 2, 3, …., k – 1 but f^(k)(a) ≠ 0. A zero of order one is known as a simple zero.

For example, consider the complex function f(z) = z sin(z²) = z³ – z⁷/3! + z¹¹/5! – … , this function has zero at z = 0 of order 3

f’(z) = 2z²cos(z²) + sin(z²) which is equal to 0 at z = 0

f’’(z) = 6zcos(z²) – 4z³sin(z²) which is equal to 0 at z = 0

f’’’(z) = 6 cos(z²) – 8z⁴cos(z²) – 24z²sin(z²) which equal to 6 at z = 0

Hence, f has a zero at z = 0 of order 3.

Singularities of a Complex Function

Singularities are few finite points of a bounded domain D of an analytic function f(z) on which the function stops being an analytic function, that is, when z equals a point of singularity in the domain then f is not differentiable.

For example, if f(z) = 1/(1 – z) then f has a singularity at z = 1.

There are various classifications of singularities. We shall discuss all four types of singularities in brief.

Isolated Singularity

A point a in the domain D of function f(z) is said to be a point of isolated singularity, if f(z) is analytic at each point in some neighbourhood |z – a| < R of a, except not being analytic at a.

For example, f(z) = (z + 1)/[z(z² + 2)] have isolated singularities at points z = 0, √2i and -√2i, except these points have no singularities within the neighbourhood of these points.

Pole

Pole is another kind of singularity defined as if there exist a positive integer m such that

For example, let f(z) = 1/(z – 5)³, then

Hence, f has a pole of 3 at z = 5.

Isolated Essential Singularity

The definition of isolated essential singularity is, if there does not exist a finite value m such that

For example, let f(z) = sin[1/(z – a)] where sin[1/(z – a)] = 1/(z – a) – 1/(z – a)³3! + 1/(z – a)⁵5! – …

Here the function has infinite terms in negative power of (z – a), so it is not possible to find a finite value of m.

Removable Singularity

A singularity z = a is said to be a removable singularity of a complex function f(z) if

For example, let f(z) = (sin z)/z = 1/z (z – z³/3! + z⁵/5! – …) = 1 – z²/3! + z⁴/5! – …

clearly , f has a singularity at z = 0, but

Hence, z = 0 is a removable singularity of the function f.

Related Articles

• Complex Numbers
• Analytic Functions
• Limits and Continuity
• Differentiabilty
• Integration
• Complex Conjugate

Solved Examples on Zeros and Singularities

Example 1:

Find the type of singularity of the function f(z) = e^z.

Solution:

Given, f(z) = e^z, clearly f has singularity at z = ∞ which will be same as that of the function f(1/z) at z = 0

f(1/z) = e^1/z = 1/z + 1/2z² + 1/3z³ + …..

There are infinite terms in negative power of z, hence we cannot find a finite value of m for which

Hence, f has an isolated essential singularity at z = ∞.

Example 2:

Find the kind of singularity of function f(z) = sin [1/(1 – z)] at z = 1.

Solution:

Given function f(z) = sin [1/(1 – z)] = 1/(1 – z) – 1/3!(1 – z)³ + 1/5!(1 – z)⁵ – ….

Since, we cannot find we cannot find a finite value of m for which

Hence, z = 1 is an isolated essential singularity.