Question

The value of the integral $\displaystyle \oint_C \frac{z^3 - 6}{2z - i} \, dz$, where $C: |z| \leq 1$, is:

• A. $\dfrac{\pi}{8} - 6\pi i$
• B. $\dfrac{\pi}{8} - 5\pi i$
• C. $\dfrac{3\pi}{8} - 5\pi i$
• D. $\dfrac{5\pi}{8} - 3\pi i$

Hint:

Whenever you see an integral of the form $\oint \dfrac{f(z)}{z-a}dz$, apply Cauchy's Integral Formula directly.

Answer 1

The Correct Option is B

 

Step 1: Identify the integrand and singularities. 
The integrand is $\dfrac{z^3 - 6}{2z - i}$. The singularity is at $z = \tfrac{i}{2}$. Since $|z| \leq 1$, the pole $z = \tfrac{i}{2}$ lies inside $C$.

Step 2: Use Cauchy's Integral Formula. 
For a function $f(z)$ analytic inside $C$, \[ \oint_C \frac{f(z)}{z-a} \, dz = 2\pi i \, f(a). \] Here, $f(z) = \dfrac{z^3 - 6}{2}$ and $a = \tfrac{i}{2}$.

Step 3: Evaluate 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: Apply Cauchy's theorem. 
\[ \oint_C \frac{z^3 - 6}{2z - i} dz = 2\pi i \, f\!\left(\tfrac{i}{2}\right) = 2\pi i \cdot \frac{-i - 48}{16}. \] \[ = \frac{\pi i}{8}(-i - 48) = \frac{\pi}{8} - 5\pi i. \]

Step 5: Conclusion. 
Thus, the value of the integral is $\dfrac{\pi}{8} - 5\pi i$.