Evaluate $\int_{c} \frac{1}{(z - 1)^{3} . (z - 3)} d z$
where $c$ is the rectangular region defined by $z = 0, x = 4, y = - 1$ and $y = 1$

1

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\frac{π}{2} i

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π (3 + 2 i)

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Solution

The correct option is B $0$
$I = \int_{c} \frac{1}{(z - 1)^{3} . (z - 3)} d z$
Poles of integrad are
$z = 3$ (simple pole)
$z = 1$ (pole of order 3)
$R_{1} = R e s = F (z) (z = 3) l i m z \to 3 (z - 3) F (z) = \frac{1}{8}$

$R_{2} = Res F (z) (z = 1) = \frac{1}{(3 - 1)!} {(\frac{d^{2}}{{dz}^{2}} {(z - 1)}^{3} F (z)]}_{z = 1}$
$= - \frac{1}{8}$
Now by cauchy-Residue theorem
$I = \oint_{c} f (z d z) = 2 π i [R_{1} + R_{2}] = 2 π i [\frac{1}{8} - \frac{1}{8}] = 0$

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Evaluate ∫c1(z−1)3.(z−3)dz where c is the rectangular region defined by z=0,x=4,y=−1 and y=1

The correct option is B 0I=∫c1(z−1)3.(z−3)dz Poles of integrad are z=3 (simple pole) z=1 (pole of order 3) R1=Res= F(z)(z=3)limz→3(z−3)F(z)=18 R2 = Res F(z)(z = 1) = 1(3 − 1)!(d2dz2(z − 1)3F(z)]z = 1 =−18 Now by cauchy-Residue theorem I=∮cf(zdz)=2πi[R1+R2]=2πi[18−18]=0

Evaluate $\int_{c} \frac{1}{(z - 1)^{3} . (z - 3)} d z$
where $c$ is the rectangular region defined by $z = 0, x = 4, y = - 1$ and $y = 1$