Laurent's Series MCQ Quiz - Objective Question with Answer for Laurent's Series - Download Free PDF

Calculation:

$\frac{1}{\left(z-1\right)\left(z-2\right)}=\frac{1}{z-2}-\frac{1}{z-1}$

⇒ $\frac{1}{1-z}+\frac{1}{\left(z-1\right)-1}=\frac{1}{1-z}-(1+\left(1-z\right){)}^{-1}$

⇒ $\frac{1}{1-z}-[1+\left(1-z\right)+\frac{\left(-1\right)\left(-2\right)}{2!}(1-z{)}^2+\frac{\left(-1\right)\left(-2\right)\left(-3\right)}{3!}(1-z{)}^3\dots ]$

⇒ $\frac{1}{1-z}-[z+(1-z{)}^2-\left(1-z{)}^3+\dots \right]$

Calculation:

$\frac{1}{\left(z-1\right)\left(z-2\right)}=\frac{1}{z-2}-\frac{1}{z-1}$

⇒ $\frac{1}{1-z}+\frac{1}{\left(z-1\right)-1}=\frac{1}{1-z}-(1+\left(1-z\right){)}^{-1}$

⇒ $\frac{1}{1-z}-[1+\left(1-z\right)+\frac{\left(-1\right)\left(-2\right)}{2!}(1-z{)}^2+\frac{\left(-1\right)\left(-2\right)\left(-3\right)}{3!}(1-z{)}^3\dots ]$

⇒ $\frac{1}{1-z}-[z+(1-z{)}^2-\left(1-z{)}^3+\dots \right]$

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 $f\left(z\right)=\frac{1}{\left(z-1\right)\left(z-2\right)}$

By partial fractions,

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

Now using standard expansions,

$f\left(z\right)=-\frac{1}{2}\left(1+\frac{z}{2}+\frac{z^2}{4}+\frac{z^3}{8}+\dots \right)-\frac{1}{z}\left(1+z^{-1}+z^{-2}+z^{-3}+\dots \right)$

⇒ $f\left(z\right)=\dots -z^{-4}-z^{-3}-z^{-2}-z^{-1}-\frac{1}{2}-\frac{1}{4}z-\frac{1}{8}{z}^2-\frac{1}{16}{z}^3-\dots$

∴ the coefficient of z² is – 1/8

Explanation:

$\mathrm{I}={\int }_0^{\pi }{\sin }^2\theta {\cos }^4\theta \mathrm{d}\theta ={\int }_0^{\pi }\left(1-{\cos }^2\theta \right){\cos }^4\theta \mathrm{d}\theta ={\int }_0^{\pi }{\cos }^4\theta \mathrm{d}\theta -{\int }_0^{\pi }{\cos }^6\theta \mathrm{d}\theta$

Now it is known that $\int {\cos }^{\mathrm{n}}\theta \mathrm{d}\theta =\frac{1}{\mathrm{n}}\mathrm{s}\mathrm{i}\mathrm{n}\theta {\cos }^{\mathrm{n}-1}\theta +\frac{\mathrm{n}-1}{\mathrm{n}}\int {\cos }^{\mathrm{n}-2}\theta \mathrm{d}\theta$

Applying this result to the second term ${\int }_0^{\pi }{\cos }^6\theta \mathrm{d}\theta \left(\mathrm{n}=6\right)$of the expression of I we get,

$\mathrm{I}={\int }_0^{\pi }{\cos }^4\theta \mathrm{d}\theta -{\left[\frac{1}{6}\mathrm{s}\mathrm{i}\mathrm{n}\theta {\cos }^5\theta \right]}_0^{\pi }-\frac{6-1}{6}{\int }_0^{\pi }{\cos }^{6-2}\theta \mathrm{d}\theta =\frac{1}{6}{\int }_0^{\pi }{\cos }^4\theta \mathrm{d}\theta$

Again, applying this result to the term ${\int }_0^{\pi }{\cos }^4\theta \mathrm{d}\theta \left(\mathrm{n}=4\right)$of the expression of I we get,

$\mathrm{I}=\frac{1}{6}{\int }_0^{\pi }{\cos }^4\theta \mathrm{d}\theta =\frac{1}{6}\left\{{\left[\frac{1}{4}\mathrm{s}\mathrm{i}\mathrm{n}\theta {\cos }^{4-1}\theta \right]}_0^{\pi }-\frac{4-1}{4}{\int }_0^{\pi }{\cos }^{4-2}\theta \mathrm{d}\theta \right\}=\frac{1}{6}\frac{3}{4}{\int }_0^{\pi }{\cos }^2\theta \mathrm{d}\theta$

$\mathrm{I}=\frac{1}{16}{\int }_0^{\pi }2{\cos }^2\theta \mathrm{d}\theta =\frac{1}{16}{\int }_0^{\pi }\left(1+\mathrm{c}\mathrm{o}\mathrm{s}2\theta \right)\mathrm{d}\theta =\frac{1}{16}{\left[1-\frac{\mathrm{s}\mathrm{i}\mathrm{n}2\theta }{2}\right]}_0^{\pi }=\frac{1}{16}\times \pi -0=\frac{\pi }{16}$

$e^z-1=\sum _{n=0}^{\mathrm{\infty }}\frac{z^n}{n!}-1$ 

$=\sum _{n=1}^{\mathrm{\infty }}\frac{z^n}{n!}$ 

$=z\sum _{n=0}^{\mathrm{\infty }}\frac{z^n}{\left(n+1\right)!}$ 

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

$f\left(z\right)=\sum _{n=-1}^{\mathrm{\infty }}{a}_n{z}^n$

$=\frac{a-1}{z}+a_0+a_{1}z+a_2{z}^2+a_3{z}^3+\dots$ 

Since (e^z – 1) f(z) = 1

$\Rightarrow \left(\frac{a-1}{z}+a_0+a_{1}z+a_2{z}^2+a_3{z}^3\right)z\left(1+\frac{z}{2}+\frac{z^2}{6}+\frac{z^3}{4}+\frac{z^4}{120}+\dots \right)=1$ 

$\Rightarrow \left(a_{-1}+a_{0}z+a_1{z}^2+a_2{z}^3+a_3{z}^4+\dots \right)\left(1+\frac{z}{2}+\frac{z^2}{6}+\frac{z^3}{24}+\frac{z^4}{120}+\dots \right)=1$ 

By comparing both the sides,

a_-1 = 1

$a_0+\frac{a_{-1}}{2}=0\Rightarrow a_0=\frac{-1}{2}$ 

Similarly, $a_1=\frac{1}{12},a_2=0,a_3=\frac{-1}{720}$ 

Concept:

Laurent series of f(z) is given by:

$f\left(z\right)=\sum _{n=-1}^{\mathrm{\infty }}{a}_n{z}^n$

$=a^{-1}{z}^{-1}+a_0+a_{1}z+a_2{z}^2+a_3{z}^3+\dots$

Calculation:

Given:

$f\left(z\right)=\frac{1}{\left(z-1\right)\left(z-2\right)}=\frac{1}{z-2}-\frac{1}{z-1}$

The given region is 1 < |z| < 2

$1<\left|z\right|\Rightarrow \frac{1}{\left|z\right|}<1$

$\left|z\right|<2\Rightarrow \frac{\left|z\right|}{2}<1$

$\Rightarrow f\left(z\right)=\frac{1}{2\left(\frac{z}{2}-1\right)}-\frac{1}{z\left(1-\frac{1}{z}\right)}$

$=\frac{-1}{2\left(1-\frac{z}{2}\right)}-\frac{1}{z\left(1-\frac{1}{z}\right)}$

$=\frac{-1}{2}{\left(1-\frac{z}{2}\right)}^{-1}-\frac{1}{z}{\left(1-\frac{1}{z}\right)}^{-1}$

$=\frac{-1}{2}\left[1+\frac{z}{2}+{\left(\frac{z}{2}\right)}^2+\dots \right]-\frac{1}{z}\left[1+\frac{1}{z}+{\left(\frac{1}{z}\right)}^2+\dots \right]$

$=\left[\frac{-1}{2}-\frac{z}{4}-\frac{z^2}{8}+\dots \right]-\left[\frac{1}{z}+\frac{1}{z^2}+\frac{1}{z^3}+\dots \right]$

Co-efficient of $\frac{1}{z^2}$is -1

Explanation:

$\mathrm{I}={\int }_0^{\pi }{\sin }^2\theta {\cos }^4\theta \mathrm{d}\theta ={\int }_0^{\pi }\left(1-{\cos }^2\theta \right){\cos }^4\theta \mathrm{d}\theta ={\int }_0^{\pi }{\cos }^4\theta \mathrm{d}\theta -{\int }_0^{\pi }{\cos }^6\theta \mathrm{d}\theta$

Now it is known that $\int {\cos }^{\mathrm{n}}\theta \mathrm{d}\theta =\frac{1}{\mathrm{n}}\mathrm{s}\mathrm{i}\mathrm{n}\theta {\cos }^{\mathrm{n}-1}\theta +\frac{\mathrm{n}-1}{\mathrm{n}}\int {\cos }^{\mathrm{n}-2}\theta \mathrm{d}\theta$

Applying this result to the second term ${\int }_0^{\pi }{\cos }^6\theta \mathrm{d}\theta \left(\mathrm{n}=6\right)$of the expression of I we get,

$\mathrm{I}={\int }_0^{\pi }{\cos }^4\theta \mathrm{d}\theta -{\left[\frac{1}{6}\mathrm{s}\mathrm{i}\mathrm{n}\theta {\cos }^5\theta \right]}_0^{\pi }-\frac{6-1}{6}{\int }_0^{\pi }{\cos }^{6-2}\theta \mathrm{d}\theta =\frac{1}{6}{\int }_0^{\pi }{\cos }^4\theta \mathrm{d}\theta$

Again, applying this result to the term ${\int }_0^{\pi }{\cos }^4\theta \mathrm{d}\theta \left(\mathrm{n}=4\right)$of the expression of I we get,

$\mathrm{I}=\frac{1}{6}{\int }_0^{\pi }{\cos }^4\theta \mathrm{d}\theta =\frac{1}{6}\left\{{\left[\frac{1}{4}\mathrm{s}\mathrm{i}\mathrm{n}\theta {\cos }^{4-1}\theta \right]}_0^{\pi }-\frac{4-1}{4}{\int }_0^{\pi }{\cos }^{4-2}\theta \mathrm{d}\theta \right\}=\frac{1}{6}\frac{3}{4}{\int }_0^{\pi }{\cos }^2\theta \mathrm{d}\theta$

$\mathrm{I}=\frac{1}{16}{\int }_0^{\pi }2{\cos }^2\theta \mathrm{d}\theta =\frac{1}{16}{\int }_0^{\pi }\left(1+\mathrm{c}\mathrm{o}\mathrm{s}2\theta \right)\mathrm{d}\theta =\frac{1}{16}{\left[1-\frac{\mathrm{s}\mathrm{i}\mathrm{n}2\theta }{2}\right]}_0^{\pi }=\frac{1}{16}\times \pi -0=\frac{\pi }{16}$

Concept:

Laurent series of f(z) is given by:

$f\left(z\right)=\sum _{n=-1}^{\mathrm{\infty }}{a}_n{z}^n$

$=a^{-1}{z}^{-1}+a_0+a_{1}z+a_2{z}^2+a_3{z}^3+\dots$

Calculation:

Given:

$f\left(z\right)=\frac{1}{\left(z-1\right)\left(z-2\right)}=\frac{1}{z-2}-\frac{1}{z-1}$

The given region is 1 < |z| < 2

$1<\left|z\right|\Rightarrow \frac{1}{\left|z\right|}<1$

$\left|z\right|<2\Rightarrow \frac{\left|z\right|}{2}<1$

$\Rightarrow f\left(z\right)=\frac{1}{2\left(\frac{z}{2}-1\right)}-\frac{1}{z\left(1-\frac{1}{z}\right)}$

$=\frac{-1}{2\left(1-\frac{z}{2}\right)}-\frac{1}{z\left(1-\frac{1}{z}\right)}$

$=\frac{-1}{2}{\left(1-\frac{z}{2}\right)}^{-1}-\frac{1}{z}{\left(1-\frac{1}{z}\right)}^{-1}$

$=\frac{-1}{2}\left[1+\frac{z}{2}+{\left(\frac{z}{2}\right)}^2+\dots \right]-\frac{1}{z}\left[1+\frac{1}{z}+{\left(\frac{1}{z}\right)}^2+\dots \right]$

$=\left[\frac{-1}{2}-\frac{z}{4}-\frac{z^2}{8}+\dots \right]-\left[\frac{1}{z}+\frac{1}{z^2}+\frac{1}{z^3}+\dots \right]$

Co-efficient of $\frac{1}{z^2}$is -1

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:

$f\left(z\right)=\frac{1}{\left(z-1\right)\left(z-2\right)}$

$=\frac{1}{z-2}-\frac{1}{z-1}$

$=\frac{1}{\left(z-1-1\right)}-\frac{1}{\left(z-1\right)}$

$=\frac{-1}{\left[1-\left(z-1\right)\right]}-\frac{1}{\left(z-1\right)}$

= -1 [1 - (z - 1)]^-1 - [z - 1]^-1

= -[z - 1]^-1 - [1 + (z - 1) + (z - 1)² + (z - 1)³ +…]

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).

$f(z)=a_0+a_1(z-a)+a_2(z-a{)}^2+.....+a_n(z-a{)}^n+a_{-1}(z-a{)}^{-1}+a_{-2}(z-a{)}^{-2}+....+a_{-n}(z-a{)}^{-n}$

$\therefore \sum _{n=0}^{\mathrm{\infty }}{a}_n(z-a{)}^n+\sum _{n=1}^{\mathrm{\infty }}{a}_{-n}(z-a{)}^{-n}$

$\therefore \sum _{-\mathrm{\infty }}^{\mathrm{\infty }}{a}_n(z-a{)}^{n}where{a}_n=\frac{1}{2\pi i}\oint \frac{f(z)}{(z-a{)}^{n+1}}dz$

Calculation:

Given:

$f(z)=\frac{1}{\left(z-1\right)\left(z-2\right)}$ and 1 < |z| < 2

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

$\therefore \frac{1}{|z|}<1and\frac{|z|}{2}<1$

$f(z)=\frac{1}{\left(z-1\right)\left(z-2\right)}\Rightarrow \frac{1}{(z-2)}-\frac{1}{(z-1)}$

$\frac{1}{(z-2)}=\frac{1}{-2(1-\frac{z}{2})}\Rightarrow -\frac{1}{2}{(1-\frac{z}{2})}^{-1}$

$\frac{1}{(z-1)}=\frac{1}{z(1-\frac{1}{z})}\Rightarrow \frac{1}{z}{(1-\frac{1}{z})}^{-1}$

$\therefore \frac{1}{(z-2)}-\frac{1}{(z-1)}=-\frac{1}{2}{(1-\frac{z}{2})}^{-1}-\frac{1}{z}{(1-\frac{1}{z})}^{-1}$

$\Rightarrow -\frac{1}{2}\left(1+\frac{z}{2}+\frac{z^2}{4}+\frac{z^3}{8}+\dots \right)-\frac{1}{z}\left(1+\frac{1}{z}+\frac{1}{z^2}+\frac{1}{z^3}+\dots \right)$

$\Rightarrow \dots -z^{-3}-z^{-2}-z^{-1}-\frac{1}{2}-\frac{1}{4}z-\frac{1}{8}{z}^2-\frac{1}{16}{z}^3\dots$

$\begin{array}{l}f\left(z\right)=\frac{z}{\left(z^2+1\right)\left(z^2+4\right)}\\ =\frac{z}{3\left(z^2+1\right)}-\frac{z}{3\left(z^2+4\right)},\left|z\right|<1\Rightarrow {\left|z\right|}^2<1\\ f\left(z\right)=\frac{z}{3}\left(1-z^2+z^4-\dots \right)-\frac{z}{12}\left(1-\frac{z^2}{4}+\frac{z^4}{16}-\dots \right)\\ f\left(z\right)=\frac{1}{4}z-\frac{5}{16}{z}^3+\frac{21}{64}{z}^5\dots \end{array}$

Calculation:

Let z +1 = u ⇒ z = u – 1

$\mathrm{f}\left(\mathrm{z}\right)=\frac{7\mathrm{z}-2}{\left(\mathrm{z}+1\right)\mathrm{z}\left(\mathrm{z}-2\right)}$

$=\frac{7\left(\mathrm{u}-1\right)-2}{\left(\mathrm{u}-1+1\right)\left(\mathrm{u}-1\right)\left(\mathrm{u}-1-2\right)}=\frac{7\mathrm{u}-9}{\mathrm{u}\left(\mathrm{u}-1\right)\left(\mathrm{u}-3\right)}$

$=-\frac{3}{\mathrm{u}}+\frac{1}{\mathrm{u}\left(1-\frac{1}{\mathrm{u}}\right)}-\frac{2}{3\left(1-\frac{\mathrm{u}}{3}\right)}$

$=\frac{-3}{\mathrm{u}}+\frac{1}{\mathrm{u}}\left(1-\frac{1}{\mathrm{u}}\right)-\frac{2}{3}{\left(1-\frac{\mathrm{u}}{3}\right)}^{-1}$

Given 1 < z + 1 < 3

⇒ 1 < u < 3

$\Rightarrow \frac{1}{u}<1,\frac{u}{3}<1$

$\mathrm{f}\left(\mathrm{z}\right)=\frac{-3}{\mathrm{u}}+\frac{1}{\mathrm{u}}\left(1+\frac{1}{\mathrm{u}}+\frac{1}{{\mathrm{u}}^2}+\frac{1}{{\mathrm{u}}^3}+\dots \mathrm{\infty }\right)-\frac{2}{3}\left(1+\frac{\mathrm{u}}{3}+\frac{{\mathrm{u}}^2}{3^2}+\frac{4^3}{3^3}+\dots \mathrm{\infty }\right)$

$=[\frac{-2}{u}+\frac{1}{u^2}+\frac{1}{u^3}+\dots \mathrm{\infty }]-\frac{2}{3}\left(1+\frac{u}{3}+\frac{u^2}{3^2}+\frac{4^3}{3^3}+\dots \mathrm{\infty }\right)$

$\mathrm{f}\left(\mathrm{z}\right)=[\frac{-2}{\mathrm{z}+1}+\frac{1}{{\left(\mathrm{Z}+1\right)}^2}+\frac{1}{{\left(\mathrm{Z}+1\right)}^3}+\dots \mathrm{\infty }]-\frac{2}{3}\left(1+\frac{\mathrm{z}+1}{3}+\frac{{\left(\mathrm{Z}+1\right)}^2}{3^2}+\frac{{\left(\mathrm{z}+1\right)}^2}{3^3}+\dots \mathrm{\infty }\right)$