Question

In the complex z-domain, the value of the integral \[ \oint_{C} \frac{z^{3} - 9}{3z - i} \, dz \] is

• A. \(\frac{2\pi}{81} - 6i\pi\)
• B. \(\frac{2\pi}{81} + 6i\pi\)
• C. \(-\frac{2\pi}{81} + 6i\pi\)
• D. \(-\frac{2\pi}{81} - 6i\pi\)

Hint:

For simple poles, the residue of \( \frac{f(z)}{g(z)} \) at \(g(z_0)=0\) is \( \frac{f(z_0)}{g'(z_0)} \).

Answer 1

The Correct Option is A

Step 1: Identify the Pole

The integrand is \( \frac{z^3 - 9}{3z - i} \). The denominator becomes zero when:

\( 3z - i = 0 \Rightarrow z = \frac{i}{3} \). This pole lies inside the contour \( C \).

Step 2: Rewrite the Integrand to Compute the Residue

The integrand is \( \frac{z^3 - 9}{3z - i} \). The residue at \( z_0 = \frac{i}{3} \) is:

\( \text{Res} = \frac{z^3_0 - 9}{3} \)

Step 3: Evaluate the Numerator at the Pole

At \( z_0 = \frac{i}{3} \), we have:

\( z_0^3 = \left( \frac{i}{3} \right)^3 = \frac{i^3}{27} = \frac{-i}{27} \)

Thus, the numerator is \( -\frac{i}{27} - 9 \).

Step 4: Compute the Residue

Now we can compute the residue as:

\( \text{Res} = -\frac{i}{27} - 9 = -\frac{i}{27} - \frac{243}{27} = -\frac{i + 243}{27} \)

Step 5: Apply the Residue Theorem

Using the residue theorem, we have:

\( \int_C f(z) \, dz = 2\pi i \cdot \text{Res} = 2\pi i \left( -\frac{i + 243}{27} \right) \)

Since \( i^2 = -1 \), we can simplify further:

\( = -6\pi i + \frac{2\pi}{81} \)

Quick Tip

For simple poles, the residue of \( f(z) \) is computed using the formula as shown, leading to a result matching option (A).