Given z = x +iy, i = √-1 C is a circle of radius 2 with the centre at the origin. If the contour C is traversed anticlockwise, then the value of the integral $\frac{1}{2\pi }{\int }_c\frac{1}{(z-i)(z+4i)}dZ$ is ________ (round off to one decimal place.)

The correct answer is 0.2.

Step-by-step Explanation:

We are given the integral:
$$I=\frac{1}{2\pi }{\int }_C\frac{1}{(z-i)(z+4i)}dz$$
where the contour $C$ is a circle of radius 2 with its center at the origin (represented by the equation $|z|=2$).

Step 1: Identify the poles of the integrand
The complex function to be integrated is:
$$f(z)=\frac{1}{(z-i)(z+4i)}$$
The poles of $f(z)$ occur where the denominator is equal to zero:
$(z-i)(z+4i)=0$
This gives two simple poles at $z=i$ and $z=-4i$.

Step 2: Analyze the poles with respect to the contour $C$ using the provided diagram
The circle $C$ is centered at $(0,0)$ with a radius of 2.
1. For the pole $z=i$: The distance from the origin is $|i|=1$. Since $1<2$, this pole lies inside the circle. In the image, this is explicitly marked with a cross at $i$ and labeled as "Inside the closed contour C".
2. For the pole $z=-4i$: The distance from the origin is $|-4i|=4$. Since $4>2$, this pole lies outside the circle. In the image, this is explicitly marked with a cross at $-4i$ and labeled as "Outside the closed contour C".

Step 3: Calculate the residue at the pole inside the contour
Since only the pole $z=i$ lies inside the contour, we calculate the residue at this point:
$${\text{Res}}_{z=i}f(z)=\underset{z\to i}{lim}(z-i)f(z)$$
$${\text{Res}}_{z=i}f(z)=\underset{z\to i}{lim}\frac{1}{z+4i}=\frac{1}{i+4i}=\frac{1}{5i}$$

Step 4: Apply Cauchy's Residue Theorem to find the integral value
Cauchy's Residue Theorem states that:
$${\int }_{C}f(z)dz=2\pi i\times (\text{Sum of residues inside}C)$$
Substituting the calculated residue:
$${\int }_C\frac{1}{(z-i)(z+4i)}dz=2\pi i\times \frac{1}{5i}=\frac{2\pi }{5}$$

Now, substitute this result back into the expression for $I$:
$$I=\frac{1}{2\pi }\times \left({\int }_C\frac{1}{(z-i)(z+4i)}dz\right)$$
$$I=\frac{1}{2\pi }\times \frac{2\pi }{5}=\frac{1}{5}=0.2$$