Question:

Let \(z\) be a complex variable. If \(f(z) = \frac{\sin(\pi z){z^2 (z - 2)}\) and \(C\) is the circle in the complex plane with \(|z| = 3\), then:} \[ \oint_C f(z) \, dz \] is \(\_\_\_\_\).

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For contour integrals, use the residue theorem by identifying the poles, determining their residues, and summing the contributions inside the contour.
Updated On: Jan 31, 2025
  • \(\pi^2 j\)
  • \(j \pi \left(\frac{1}{2} - \pi\right)\)
  • \(j \pi \left(\frac{1}{2} + \pi\right)\)
  • \(-\pi^2 j\)
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The Correct Option is D

Solution and Explanation

Step 1: Locate the poles of \(f(z)\).
The function \(f(z)\) has two poles inside the contour \(|z| = 3\): At \(z = 0\) with order 2. At \(z = 2\) with order 1. Step 2: Use the residue theorem.
According to the residue theorem, the contour integral is: \[ \oint_C f(z) \, dz = 2\pi j \left({Residue at } z = 0 + {Residue at } z = 2\right). \] Step 3: Compute the residues.
For \(z = 0\): \[ {Residue at } z = 0 = \lim_{z \to 0} \frac{d}{dz} \left(z^2 f(z)\right) = -\pi^2. \] For \(z = 2\): \[ {Residue at } z = 2 = \frac{\sin(\pi \cdot 2)}{2^2} = \frac{\sin(2\pi)}{4} = 0. \] Step 4: Evaluate the integral.
Substituting the residues into the formula: \[ \oint_C f(z) \, dz = j \cdot (-\pi^2) = -\pi^2 j. \] Final Answer: \[\boxed{{(4) } -\pi^2 j}\]
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