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Question

C is a closed path in the zplane given by |z|=3. The value of the integral C(z2z+4jz+2j)dz is

A
4π(1+j2)
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B
4π(3+j2)
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C
4π(3+j2)
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D
4π(1j2)
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Solution

The correct option is C 4π(3+j2)
Method I:

f(z)=(z2z+4jz+2j) so pole is z=2j
c:|z|=3



So pole lies inside c
R=Res f(z)(z=2j)=limz2j[(z+2j)f(z)]
=limz2j(z2z+4j)=4+6j
By Residue theorem,
cf(z)dz=2πj (Sum of residues)
=2πj(4+6j)=12π8πj
=4π(3+2j)

Method II:
I=Cz2z+4jz+2jdz,|Z|=3
z=2j is singular point which lies inside |z|=3.
By cauchy integral formula
Cz2z+4jz+2jdz=2πj(2j)
=2πj(4j2+2j+4j)
=2πj(6j4)
=2π(+6j24j)
=4π(3+2j)

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