Let z be a complex variable. For a counter-clockwise integration around a unit circle C , centred at origin,  the value of A is ________________-

The correct answer is 2/5.

Step-by-Step Explanation:

Based on the provided image, the integral equation we need to evaluate is:
${\oint }_C\frac{1}{5z-4}dz=A\pi i$
where $C$ is the unit circle centered at the origin, oriented counter-clockwise.

1. Identify the contour and the singularity:
The contour $C$ is the unit circle, which is mathematically represented as:
$\left|z\right|=1$
The integrand is:
$f\left(z\right)=\frac{1}{5z-4}$
To find the singular points (poles) of $f\left(z\right)$, we set the denominator to zero:
$5z-4=0$
Solving for $z$ gives:
$z=\frac{4}{5}=0.8$
Since the absolute value of this pole satisfies $\left|0.8\right|<1$, the singularity lies inside the unit circle contour $C$.

2. Calculate the residue at the pole:
We can rewrite the function $f\left(z\right)$ in standard form:
$f\left(z\right)=\frac{1}{5(z-\frac{4}{5})}=\frac{\frac{1}{5}}{z-\frac{4}{5}}$
Thus, the residue of $f\left(z\right)$ at the simple pole $z=\frac{4}{5}$ is:
$\text{Res}(f,\frac{4}{5})=\underset{z\to \frac{4}{5}}{lim}(z-\frac{4}{5})f\left(z\right)=\frac{1}{5}$

3. Apply Cauchy's Residue Theorem:
According to Cauchy's Residue Theorem, the counter-clockwise line integral along $C$ is:
${\oint }_{C}f\left(z\right)dz=2\pi i\times \text{Res}(f,\frac{4}{5})$
Substituting the calculated residue:
${\oint }_C\frac{1}{5z-4}dz=2\pi i\left(\frac{1}{5}\right)=\frac{2}{5}\pi i$

4. Determine the value of A:
Comparing the result with the given equation:
$\frac{2}{5}\pi i=A\pi i$
We get:
$A=\frac{2}{5}$