Test: Laurent Series - Engineering Mathematics MCQ

Concept:

The given region is a ring-shaped region bounded by two concentric circles with center at origin.

The expansion will be Laurent's series.

Calculation:

Given function is 
By partial fractions,

Now using standard expansions,

∴ the coefficient of z² is – 1/8

Concept:

Laurentz Series is obtained by the arrangement and manipulation of standard series or expansions, i.e.

(1 - x)^-1 = 1 + x + x² + x³ + …… |x| < 1

(1 + x)^-1 = 1 – x + x² – x³ + ….. |x| < 1

(1 - x)^-2 = 1 + 2x + 3x² + ….. |x| < 1

(1 + x)^-2 1 – 2x + 3x² – 4x² + …. |x| < 1

Observe that in all the expansions; |x| should be less than 1.

 ∴ We need to manipulate the variable to satisfy the above condition.

Calculation:

Given:

= -1 [1 - (z - 1)]^-1 - [z - 1]^-1
= -[z - 1]^-1 - [1 + (z - 1) + (z - 1)² + (z - 1)³ +…]
Co-efficient of 1^z−1 is -1

Concept:

Laurent series of f(z) is given by:

Calculation:
Given:

The given region is 1 < |z| < 2

Co-efficient of 1/z² is -1

Concept:
Laurent Series:
If f(z) is analytic at every point inside and on the boundary of a ring-shaped region 'R' bounded by two concentric circle C₁ and C₂ having centre at 'a' & respective radii r₁ and r₂ (r₁ > r₂).


Calculation:

Given:

 and 1 < |z| < 2

Here region of convergence is 1 < |z| and |z| < 2

e^x – 1 has a zero at ‘0’ of multiplicity one and hence f(z) has pole at 0 of order 1. So, the Laurent series f(z) is given by

Since (e^z – 1) f(z) = 1

By comparing both the sides,
a-1 = 1

Calculation:
Let z +1 = u ⇒ z = u – 1

Given 1 < z + 1 < 3
⇒ 1 < u < 3

Now it is known that 
Applying this result to the second term ( n = 6) of the expression of I we get,

Again, applying this result to the term of the expression of I we get,