GATE Electrical Engineering (EE) 2018 Solved Paper

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Question : 54 of 65

Marks: +1, -0

If

C

is a circle

|z|=4

and

f(z)=\;{z^{2}}/{(z^{2}-3 z+2)^{2}}

, then

∮↙{C} f(z) d z

is

$1$
$0$
$-1$
$-2$

Solution:

∫ \;{z^{2}}/{(z^{2}-3 z+2)} d z

∫ \;{z^{2}}/{(z-1)(z-2)^{2}} d z

{\Res.}↙{z=1} \;\;f(z){Lt}↙{z→1} \;{1}/{1 !} \;{d}/{d z}((z-1)^{2} ⋅ \;{z^{2}}/{(z-1)^{2}(z-2)^{2}})

=L t_{z → 1}(\;{2 z(z-2)^{2}-2 z^{2}(z-2)}/{(z-2)^{4}})

=L_{z → 1}(\;{2 z(z-2)-2 z^{2}}/{(z-2)^{3}})

=\;{-4}/{-1}=4

{\Res.}↙{z=1} \;\;f(z){Lt}↙{z→2} {1}/{1 !} \;{d}/{d z}((z-2)^{2} ⋅ \;{z^{2}}/{(z-1)^{2}(z-2)^{2}})

=L t_{z → 2}(\;{(z-1)^{2} ⋅ 2 z-z^{2}(z-1)}/{(z-1)^{4}})

=L_{z → 2}(\;{2 z(z-1)-2 z^{2}}/{(z-1)^{3}})

=\;{4-8}/{1}=-4

By residue theorem,

I=2 π i(4-4)=0

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