The value of the integral $\displaystyle \oint_C \frac{z^3 - 6}{2z - i} \, dz$, where $C: |z| \leq 1$, is:
The problem requires evaluating a contour integral. The steps are as follows:
Step 1: Identify the integrand and singularities. The integrand is $\dfrac{z^3 - 6}{2z - i}$. The singularity occurs at $z = \tfrac{i}{2}$. Given the condition $|z| \leq 1$, the pole $z = \tfrac{i}{2}$ is located within the contour $C$.
Step 2: Apply Cauchy's Integral Formula. The formula for a function $f(z)$ analytic inside $C$ is given by: \[ \oint_C \frac{f(z)}{z-a} \, dz = 2\pi i \, f(a). \] In this case, we set $f(z) = \dfrac{z^3 - 6}{2}$ and $a = \tfrac{i}{2}$.
Step 3: Evaluate $f(a)$. We compute the value of $f(z)$ at $a = \tfrac{i}{2}$: \[ f\!\left(\tfrac{i}{2}\right) = \frac{\left(\tfrac{i}{2}\right)^3 - 6}{2} = \frac{\tfrac{-i}{8} - 6}{2} = \frac{-i - 48}{16}. \]
Step 4: Calculate the integral using Cauchy's theorem. Applying the formula: \[ \oint_C \frac{z^3 - 6}{2z - i} dz = 2\pi i \, f\!\left(\tfrac{i}{2}\right) = 2\pi i \cdot \frac{-i - 48}{16}. \] Simplifying this expression yields: \[ = \frac{\pi i}{8}(-i - 48) = \frac{\pi}{8} - 5\pi i. \]
Step 5: State the result. The value of the integral is $\dfrac{\pi}{8} - 5\pi i$.