Question

The analytic function \(f\left( z \right) = \frac{{z - 1}}{{{z^2} + 1}}\) has singularities at

1. 1 & -1
2. 1 & 1
3. 1 & -i
4. i & -i

Answer 1

Correct Answer - Option 4 : i & -i

Concept:

Pole:

The value for which f(z) fails to exists i.e. the value at which the denominator of the function f(z) = 0.

When the order of a pole is 1, it is known as a simple pole.

The point where the function is not defined i.e. the value where f(z) is discontinuous is called singularities.

Concept:

Given:

Analytic function

\(\begin{array}{l} f\left( z \right) = \frac{{z - 1}}{{{z^2} + 1}}\\ f\left( z \right) = \frac{{z - 1}}{{\left( {z - i} \right)\left( {z + i} \right)}} \end{array}\)

function is not defined at z = i or z = -i

i.e. the point of singularities of function is i & -i respectively.