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 option is -jπ/8.

To find the value of the contour integral, we use Cauchy's Residue Theorem. Let the integrand be:

$f\left(z\right)=\frac{1}{z^2(z-4)}$

First, we find the singularities (poles) of the function $f\left(z\right)$ by setting the denominator to zero:

$z^2(z-4)=0$

This gives two poles:
1. $z=0$ which is a pole of order 2 (double pole).
2. $z=4$ which is a simple pole (order 1).

Next, we determine which of these poles lie inside the contour C. The contour C is a rectangle with the following vertices in the complex plane:
- Vertex 1: -1 - j
- Vertex 2: 3 - j
- Vertex 3: 3 + j
- Vertex 4: -1 + j

The region enclosed by the rectangle C contains complex numbers z = x + jy where x is between -1 and 3, and y is between -1 and 1.
- For the pole $z=0$, the real part is 0 and the imaginary part is 0. Since 0 is between -1 and 3, and 0 is between -1 and 1, the pole $z=0$ lies inside C.
- For the pole $z=4$, the real part is 4, which is outside the range [-1, 3]. Therefore, the pole $z=4$ lies outside C.

By Cauchy's Residue Theorem, the value of the contour integral is given by:

${\oint }_{C}f\left(z\right)dz=2\pi j·\text{Res}(f,0)$

Since $z=0$ is a double pole, we compute the residue using the formula:

$\text{Res}(f,0)=\underset{z\to 0}{lim}\frac{d}{dz}\left[z^{2}f\left(z\right)\right]$

Substituting the expression for $f\left(z\right)$:

$\text{Res}(f,0)=\underset{z\to 0}{lim}\frac{d}{dz}\left[\frac{1}{z-4}\right]$

Differentiating $\frac{1}{z-4}$ with respect to z gives:

$\frac{d}{dz}{(z-4)}^{-1}=-\frac{1}{{(z-4)}^2}$

Evaluating this derivative as z approaches 0:

$\text{Res}(f,0)=-\frac{1}{{(0-4)}^2}=-\frac{1}{16}$

Now, substituting the residue back into Cauchy's Residue Theorem formula:

${\oint }_C\frac{dz}{z^2(z-4)}=2\pi j·\left(-\frac{1}{16}\right)=-\frac{j\pi }{8}$