The value of the following complex integral, with C representing the unit circle centered at origin in the counterclockwise sense, is:${\int }_C\frac{z^2+1}{z^2-2z}dz$

The correct option is -πi.

Step-by-step Explanation:

We are required to evaluate the following complex contour integral:

${\int }_C\frac{z^2+1}{z^2-2z}dz$

where C represents the unit circle centered at the origin, traversed in the counterclockwise direction. The equation of this contour C is given by:

$\left|z\right|=1$

Step 1: Identify the Singularities of the Integrand
Let us denote the integrand as f(z):

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

The singularities (poles) of f(z) occur where the denominator is equal to zero:
z(z - 2) = 0, which gives z = 0 and z = 2.

Step 2: Determine Which Singularities Lie Inside the Contour C
The contour C is the unit circle centered at the origin with a radius of 1.
- For the pole z = 0: |0| = 0 < 1, which lies inside the contour C.
- For the pole z = 2: |2| = 2 > 1, which lies outside the contour C.

Therefore, we only need to calculate the residue of the integrand at the simple pole z = 0.

Step 3: Calculate the Residue at z = 0
Since z = 0 is a simple pole (pole of order 1), the residue is calculated as follows:

${\text{Res}}_{z=0}f\left(z\right)=\underset{z\to 0}{lim}(z-0)f\left(z\right)$

Substituting the expression for f(z):

${\text{Res}}_{z=0}f\left(z\right)=\underset{z\to 0}{lim}z·\frac{z^2+1}{z(z-2)}$

Simplifying the limit by canceling out the common factor of z in the numerator and denominator:

${\text{Res}}_{z=0}f\left(z\right)=\underset{z\to 0}{lim}\frac{z^2+1}{z-2}$

Evaluating this limit by direct substitution of z = 0:

${\text{Res}}_{z=0}f\left(z\right)=\frac{0^2+1}{0-2}=-\frac{1}{2}$

Step 4: Apply Cauchy's Residue Theorem
According to Cauchy's Residue Theorem, the integral of a function f(z) over a closed contour C is given by:

${\int }_{C}f\left(z\right)dz=2\pi i·\sum \text{Residues inside}C$

Substituting our calculated residue value into the equation:

${\int }_{C}f\left(z\right)dz=2\pi i·(-\frac{1}{2})=-\pi i$

Thus, the value of the complex integral is -πi.