The value of the contour integral \frac{1}{2 \pi j} ∮(z+\frac{1}{z})^{2} d z evaluated over the unit circle |z|=1 is ____________.

GATE Electronics and Communications (EC) 2019 Solved Paper

Question : 35 of 65

Marks: +1, -0

The value of the contour integral

\;{1}/{2 π j} ∮(z+\;{1}/{z})^{2} d z

evaluated over the unit circle

|z|=1

is ____________.

Solution:

For poles:
Consider

z^{2}=0 ⇒ z=0,0

Now

f(z)=(z^{2}+1)^{2}

\;{2 π i}/{n-1) !} f^{n-1}(a)

=\;{1}/{2 π {i}}[\;{2 π {i}}/{(2-1) !} {f}^{2-1}( {a})]= {f}{'}( {a})= {f}{'}(0)

Now

f{'}(z)=2(z^{2}+1)(2 z)

f{'}(0)=2(0+1)(0)=0

∴

So answer is zero

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