Question:

If $I = \frac{1}{2\pi i} \oint \frac{z e^{1/z}}{z} \, dz$ in the unit circle $|z| = 1$, then:

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When integrating around a unit circle, look for the residue (coefficient of $1/z$) in the Laurent series to evaluate using Cauchy’s theorem.
Updated On: Jun 24, 2025
  • $I = 2$
  • $I = 1/2$
  • $I = 1$
  • $I = 1/3$
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The Correct Option is C

Solution and Explanation

Step 1: Simplify the integrand 
\[ \frac{z e^{1/z}}{z} = e^{1/z} \] Step 2: Use Laurent series of $e^{1/z}$ around $z = 0$ 
\[ e^{1/z} = \sum_{n=0}^{\infty} \frac{1}{n! z^n} \] So, \[ e^{1/z} = 1 + \frac{1}{z} + \frac{1}{2! z^2} + \dots \] Step 3: Use Cauchy's integral formula 
\[ I = \frac{1}{2\pi i} \oint e^{1/z} dz = \text{coefficient of } \frac{1}{z} \text{ in } e^{1/z} = 1 \]

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