8.CSIR NET MATHS 2020-JUNE PART B

 

CSIR NET MATHEMATICS 2020-JUNE

HELD ON 26TH-NOVEMBER 

PART-B

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

 

(A) $i\pi e$ 

 

(B) $-i\pi e$

 

(C) $\pi e$

 

(D) $- \pi e$

 

Solution:  By cauchy's integral formula is 

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

 

Given

 

 $ \int_{\gamma}\frac{ze^{1/z}}{z^2-1}dz$ =   $ \int_{\gamma}\frac{ze^{1/z}}{(z^2-1^2}dz$            [$ a^2-b^2=(a-b)(a+b) $ ]

            = $ \int_{\gamma}\frac{ze^{1/z}}{(z-1)(z+1)}dz  $ 

 

Also Given$ Z \in C : \vert Z-1 \vert =1/2 $

 

 $ \vert Z-1 \vert = 1/2 $           [Given]

 
 $ \vert x+iy-1 \vert = 1/2  $              [given Z=x+iy ]

  $ \vert x-1+iy \vert = 1/2 $ 

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

 

Here centre is (1,0) and radius r=1/2    [$x-1=0\implies x=1$ and $y=0$ ]  

 

 z-1=0 $ \implies  $ z=1 lies inside $  C: \vert z-1 \vert =1/2 $

z+1=0  $ \implies $  z=-1 lies outside $  C: \vert z-1 \vert =1/2 $ 

 $ \int_{\gamma}\frac{ze^{1/z}}{(z-1)(z+1)}dz =\int_{\gamma}\frac{\frac{ze^{1/z}}{z+1}}{z-1}dz  $ 

 

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

Hence by Cauchy's integral formula

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

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

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

                        = $   2 \pi i \frac{e}{2} $ 

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

 

 Hence correct option (A)  i $ \pi $  e