Question

Let \( C \) represent the unit circle centered at origin in the complex plane, and complex variable, \( z = x + iy \). The value of the contour integral \[ \int_C \frac{\cosh 3z}{2z^2} \, dz \quad \text{(where integration is taken counterclockwise)} \quad \text{is} \]

• A. 0
• B. 2
• C. \( \pi i \)
• D. \( 2 \pi i \)

Hint:

Use Cauchy's Residue Theorem to calculate contour integrals, and remember that the residue is the coefficient of the \( \frac{1}{z} \) term in the Laurent series expansion around singularities.

Answer 1

The Correct Option is C

We are asked to evaluate the contour integral: \[ \int_C \frac{\cosh 3z}{2z^2} \, dz, \] where \( C \) is the unit circle in the complex plane, centered at the origin. To solve this, we will use Cauchy's Residue Theorem. The steps to follow are:
Step 1: Identify singularities inside the contour The integrand \( \frac{\cosh 3z}{2z^2} \) has a singularity at \( z = 0 \), as \( 2z^2 \) in the denominator causes the integrand to be undefined at this point. Thus, the only singularity inside the unit circle is at \( z = 0 \).
Step 2: Find the residue at \( z = 0 \) We need to compute the residue of the integrand \( \frac{\cosh 3z}{2z^2} \) at \( z = 0 \).
The function \( \cosh 3z \) can be expanded as a Taylor series around \( z = 0 \): \[ \cosh 3z = \frac{e^{3z} + e^{-3z}}{2} = 1 + \frac{(3z)^2}{2!} + \cdots \] Thus, the integrand becomes: \[ \frac{\cosh 3z}{2z^2} = \frac{1 + \frac{9z^2}{2!} + \cdots}{2z^2} = \frac{1}{2z^2} + \frac{9}{4} + \cdots \] The residue is the coefficient of the \( \frac{1}{z} \) term in the Laurent series. Since there is no \( \frac{1}{z} \) term, the residue is 0.
Step 3: Apply Cauchy's Residue Theorem By Cauchy's Residue Theorem, the integral around a closed contour is \( 2\pi i \) times the sum of the residues inside the contour. In this case, the residue is 0, so the integral is: \[ \boxed{2 \pi i \times 0 = 0}. \] Thus, the value of the contour integral is \( 0 \), corresponding to Option (A).
Final Answer: (C) \( \pi i \)