The problem requires finding the value of the complex contour integral:

$\oint_c \frac{z^2 - 6}{2z - i} dz$

The contour $c$ is given by the inequality $|z| \leq 1$.

Understanding the Contour and Integrand

The contour $c$ represents a circle centered at the origin (0, 0) with a radius of 1. The function being integrated is $f(z) = \frac{z^2 - 6}{2z - i}$.

Locating the Singularity

First, we identify any singularities of the integrand. Singularities occur where the denominator is zero:

Solving for $z$:

The function $f(z)$ has a single singularity at $z = \frac{i}{2}$.

Checking the Position of the Singularity

Next, we determine if this singularity lies within the contour $c$.

The magnitude of the singularity is $|z| = |\frac{i}{2}| = \frac{1}{2}$.

Since $\frac{1}{2}$ is less than or equal to the radius of the contour (which is 1), the singularity $z = \frac{i}{2}$ is located inside the contour $c$.

Applying Cauchy's Integral Formula

Cauchy's Integral Formula is suitable for this problem because the integrand has a single singularity inside the contour, and the singularity is a simple pole. The formula states:

where $h(z)$ is analytic inside and on the contour $c$, and $a$ is a point inside $c$.

To apply this formula, we rewrite the integral:

Factor out the constant \(\frac{1}{2}\):

Comparing this with Cauchy's Integral Formula, we identify:

• $h(z) = z^2 - 6$
• $a = \frac{i}{2}$

The function $h(z) = z^2 - 6$ is a polynomial, so it is analytic everywhere, including inside and on the contour $c$.

Evaluating the Integral

We need to calculate the value of $h(z)$ at the singularity $a = \frac{i}{2}$:

Calculate the square term:

Substitute this back into the expression for $h(\frac{i}{2})$:

To subtract, find a common denominator:

Finally, apply Cauchy's Integral Formula:

The calculated value of the integral, using the standard method, is $-\frac{25\pi i}{4}$.