Let (-1 - j), (3 - j), (3 + j) and (-1 + j) be the vertices of a rectangle C in the complex plane. Assuming that C is traversed in counter-clockwise direction, the value of the countour integral ${\displaystyle {\oint }_C{\displaystyle \frac{dz}{z^2(z-4)}}}$ is

The correct answer is -jπ/8.

To find the value of the contour integral ${\oint }_C\frac{dz}{z^2(z-4)}$, we can use Cauchy's Residue Theorem.

Step 1: Identify the contour C
The contour C is a rectangle in the complex plane with the following vertices:
$z_1=-1-j$
$z_2=3-j$
$z_3=3+j$
$z_4=-1+j$

Thus, the region enclosed by C has:
Real part x in the range $-1\le x\le 3$
Imaginary part y in the range $-1\le y\le 1$

Step 2: Identify the singularities of the integrand
The integrand is given by:
$f(z)=\frac{1}{z^2(z-4)}$

The singularities (poles) of $f(z)$ are:
1. $z=0$ (a pole of order 2)
2. $z=4$ (a simple pole)

Step 3: Determine which poles lie inside the contour C
- For the pole at $z=0$, both the real part (0) and imaginary part (0) lie strictly within the ranges $-1<0<3$ and $-1<0<1$. Therefore, $z=0$ lies inside C.
- For the pole at $z=4$, the real part (4) lies outside the range since $4>3$. Therefore, $z=4$ lies outside C.

Step 4: Compute the residue at the pole inside C
Since $z=0$ is a pole of order 2, the residue is calculated as:
${\text{Res}}_{z=0}f(z)=\underset{z\to 0}{lim}\frac{d}{dz}[z^{2}f(z)]$

Substituting $f(z)$ into the equation:
${\text{Res}}_{z=0}f(z)=\underset{z\to 0}{lim}\frac{d}{dz}\left[\frac{1}{z-4}\right]$

Differentiating $\frac{1}{z-4}$ with respect to z:
$\frac{d}{dz}{(z-4)}^{-1}=-{(z-4)}^{-2}=-\frac{1}{{(z-4)}^2}$

Now, evaluating the limit as z approaches 0:
${\text{Res}}_{z=0}f(z)=-\frac{1}{{(0-4)}^2}$

Step 5: Apply Cauchy's Residue Theorem
According to Cauchy's Residue Theorem, the value of the contour integral along a counter-clockwise path C is:
${\oint }_{C}f(z)dz=2\pi j·\sum \text{Residues inside C}$

Substituting the computed residue value:
${\oint }_C\frac{dz}{z^2(z-4)}=2\pi j(-\frac{1}{16})=-\frac{j\pi }{8}$