CSIR NET MATHEMATICS| Complex Analysis | How to use Cauchy’s Integral Formula? (June 2018)

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Nov 11, 2026
CSIR NET MATHEMATICS| Complex Analysis | How to use Cauchy’s Integral Formula? (June 2018)
Analytic Function Singular Point Closed Contour Circle in the Complex plane
CSIR NET DEC 2018 Answer Key
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Video Summary
AI GeneratedThis video provides a step-by-step solution to a complex analysis problem from the CSIR NET June 2018 exam, focusing on calculating a contour integral using the Cauchy Integral Formula.
In this video, we are going to discuss
the problem of complex analysis from the CSI
Net June 2018.
Example, the value of integral over the closed
contour 1 minus z equal to 1 e power z divided
by z square minus 1 dz is the first option 0,
second option pi i e, third option pi i
e minus pi i e power minus 1
and the fourth option e plus e power minus 1.
We are given the
Integration over the closed contour modulus 1 minus z is equal to 1
e power z divided by z square minus 1 dz.
Factorize the denominator z square minus 1
is equal to z minus 1 z plus 1.
Now, integrand e power z divided by z minus 1 z plus 1.
So, we have two singularities.
One is z is equal to 1 and second one is z is equal to minus 1.
So, we are getting the closed contour modulus 1 minus z
z is equal to 1.
So which represent this circle with the center 1, 0 and the radius 1.
Now this circle contain positive 1 but not contain the negative 1.
Therefore from the two singularities z is equal to 1
and z is equal to minus 1. Only z is equal to 1 is inside
the closed contour and minus 1 outside the closed contour.
Therefore z is equal to 1 can be considered
as a singularity for this function,
so this integrand can be written in the form of e power z divided
by z plus 1, this whole divided by z minus 1,
so numerator will be the analytic function for this closed contour
and z minus 1 will become a singularity.
Now using the Cauchy integral formula, if f is analytic then
Integration over the closed contour fz divided
by z minus 1 dz is equal to 2 pi i f of a.
Here f of z is equal to the numerator
e power z divided by z plus 1 which is analytic on the
closed contour modulus 1 minus z equal to 1 and a is 1.
So the integral value is 2 pi i
into f of a. So a is equal to 1.
2 pi i f of 1, so 2 pi i, if you put z is equal to 1,
e power 1 divided by 1 plus 1, so it is e by 2, and 2 to cancel,
the final answer is pi i e, which is the second option,
so this example is the answer, second option, pi i e.
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