Question

By using Cauchy integral theorem, the value of an integral (integration being taken in counter clock wise direction) $\oint \frac{z^{3}-4}{3z-i} dz$; is:

• A. 1
• B. 0
• C. $\frac{2\pi}{81}-\frac{4\pi i}{3}$
• D. $\frac{2\pi}{81}-\frac{8\pi i}{3}$

Hint:

Always ensure the coefficient of $z$ in the denominator is 1 before applying Cauchy's Integral Formula. Factoring out the constant (like the 3 in $3z-i$) is a common first step.

Answer 1

The Correct Option is D

Concept: According to Cauchy's Integral Formula, if $f(z)$ is analytic within and on a closed contour $C$, and $a$ is a point inside $C$: \[ \oint_C \frac{f(z)}{z-a} dz = 2\pi i f(a) \]

Step 1: Rewrite the integral in standard form.
The given integral is $\oint \frac{z^3-4}{3z-i} dz$. We need the denominator in the form $(z-a)$: \[ \oint \frac{z^3-4}{3(z - i/3)} dz = \frac{1}{3} \oint \frac{z^3-4}{z - i/3} dz \] Here, $f(z) = z^3 - 4$ and $a = i/3$.

Step 2: Evaluate $f(a)$.
Substitute $z = i/3$ into $f(z)$: \[ f(i/3) = (i/3)^3 - 4 = \frac{i^3}{27} - 4 = \frac{-i}{27} - 4 \]

Step 3: Apply the formula.
\[ \text{Integral} = \frac{1}{3} [2\pi i \cdot f(i/3)] = \frac{2\pi i}{3} \left( -\frac{i}{27} - 4 \right) \] \[ \text{Integral} = \frac{2\pi (-i^2)}{81} - \frac{8\pi i}{3} \] Since $i^2 = -1$: \[ \text{Integral} = \frac{2\pi}{81} - \frac{8\pi i}{3} \]