Question:

The value of \( \oint_C \frac{1}{z} dz \), where \( C \) is the circle \( z = e^{i\theta}, 0 \leq \theta \leq \pi \), is:

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\[ \oint_C \frac{1}{z} dz = 2\pi i \] if \( C \) encloses the origin. A semicircle contour gives half this value.
Updated On: Feb 6, 2025
  • \( \pi i \)
  • \( -\pi i \)
  • \( 2\pi i \)
  • \( 0 \)
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The Correct Option is A

Solution and Explanation


Step 1:
Integral of \( \frac{1}{z} \) over a contour. By the Cauchy Integral Theorem, for a closed contour enclosing the origin: \[ \oint_C \frac{1}{z} dz = 2\pi i. \]
Step 2:
Consider the given semicircular contour. - Given contour \( C \) covers half of the full circle. - So, the integral is half of \( 2\pi i \), which gives: \[ \pi i. \]
Step 3:
Selecting the correct option. Since \( \pi i \) is correct, the answer is (A).
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