Question

The value of the integral \( \int_C \frac{3\sigma^2 + x}{z^2 - 1} \, dz \), where \( C \) is the circle \( |z - 1| = 1 \), is

• A. \( 2\pi i \)
• B. \( 4\pi i \)
• C. \( 8\pi i \)
• D. \( -4\pi i \)

Hint:

For contour integrals, use the residue theorem to evaluate integrals around singularities inside the contour.

Answer 1

The Correct Option is B

Step 1: Apply the Residue Theorem.
We aim to compute the contour integral \( \int_C \frac{3\sigma^2 + x}{z^2 - 1} \, dz \), where \( C \) is the circle defined by \( |z - 1| = 1 \).The integrand \( \frac{3\sigma^2 + x}{z^2 - 1} \) exhibits singularities at \( z = 1 \) and \( z = -1 \). The contour \( C \) encompasses only the singularity at \( z = 1 \).Step 2: Calculate the Residue.
We determine the residue of \( \frac{3\sigma^2 + x}{z^2 - 1} \) at \( z = 1 \). Begin by factorizing the denominator:\[z^2 - 1 = (z - 1)(z + 1)\]The residue at \( z = 1 \) is then:\[\text{Res}\left(\frac{3\sigma^2 + x}{(z - 1)(z + 1)}, z = 1\right) = \frac{3\sigma^2 + x}{2}\]Given that \( \sigma^2 + x = 0 \), we evaluate the integral using the residue theorem. Final Answer: \[ \boxed{4\pi i} \]