6.CSIR NET MATHS 2020-JUNE PART B

 

CSIR NET MATHEMATICS 2020-JUNE

HELD ON 26TH-NOVEMBER 

PART-B

6.Let $\gamma$ be the positively  oriented circle in the complex plane given by $\lbrace Z \in C : \vert Z-1 \vert =1 \rbrace $ then $\frac{1}{2\pi i}\int_{\gamma}\frac{dz}{z^3-1}$ equals

 

(A) 3

(B)1/3

(C)1/2 

(D)2

 

Solution

By cauchy's integral formula is 

 

$ \int_C \frac{f(z)}{z-a}dz=2\pi i f(a) $
 

Given 

 

$ \frac{1}{2\pi i}\int_{\gamma}\frac{dz}{z^3-1} =\frac{1}{2\pi i}\int_{\gamma}\frac{dz}{(z-1)(z^2+z+1)} $          [ $a^3-b^3=(a-b)(a^2+ab+b^2)$ ]

 
=$ \frac{1}{2\pi i}\int_{\gamma}\frac{\frac{1}{z^2+z+1}}{z-1}dz  $

 

 Also Given$ Z \in C : \vert Z-1 \vert =1 $

$ \vert Z-1 \vert =$ 1                 [Given ]

 
$ \vert x+iy-1 \vert =$ 1              Given Z=x+iy
 

$\vert x-1+iy \vert$  =1 

$ (x-1)^2+y^2$ =1                  This is circle equation

 

Here centre is (1,0) and radius r=1  [$x-1=0\implies x=1$ and $y=0$ ]  

 

 

 

z-1=0 $ \implies$  z=1      lies inside   C:$ \vert z-1 \vert $ =1 

Here $f(z)=\frac{1}{z^2+z+1}$ is analytic inside C

$ \int_C \frac{f(z)}{z-a} $ dz = $ 2\pi i f(a) $

$\frac{1}{2\pi i}\int_{\gamma}\frac{\frac{1}{z^2+z+1}}{z-1}$   dz = $ \frac{1}{2\pi i} 2\pi i f(1) $

$ =\frac{1}{2\pi i} 2\pi i \frac{1}{1^2+1+1}  $             [$ f(z)=\frac{1}{z^2+z+1} $]
 

$ \frac{1}{2\pi i}\int_{\gamma}\frac{dz}{z^3-1} $ = $ \frac{1}{3} $

Hence correct option (B) 1/3