Question

Let \( C \) be the ellipse \( \{ z \in \mathbb{C} : |z - 2| + |z + 2| = 8 \} \) traversed counter-clockwise. The value of the contour integral \[ \int_C \frac{z^2 \, dz}{z^2 - 2z + 2} \] is equal to:

• A. \( 0 \)
• B. \( 2\pi i \)
• C. \( 4\pi i \)
• D. \( -\pi i \)

Hint:

For contour integrals, use the residue theorem and carefully evaluate residues for poles enclosed by the contour.

Answer 1

The Correct Option is C

The integrand is \( \frac{z^2}{z^2 - 2z + 2} \). To evaluate the integral, we first find the singularities of the function by solving \( z^2 - 2z + 2 = 0 \): \[ z = 1 \pm i. \] These are the singularities of the function inside the ellipse \( |z - 2| + |z + 2| = 8 \), as the ellipse encloses both \( z = 1 + i \) and \( z = 1 - i \). Residue Calculation: The function \( \frac{z^2}{z^2 - 2z + 2} \) can be rewritten as: \[ \frac{z^2}{(z - (1 + i))(z - (1 - i))}. \] For each singularity, we calculate the residues: 1. At \( z = 1 + i \), the residue is: \[ {Residue} = \lim_{z \to 1+i} \frac{z^2}{z - (1 - i)} = \frac{(1 + i)^2}{2i} = \frac{1 + 2i - 1}{2i} = i. \] 2. At \( z = 1 - i \), the residue is: \[ {Residue} = \lim_{z \to 1-i} \frac{z^2}{z - (1 + i)} = \frac{(1 - i)^2}{-2i} = \frac{1 - 2i - 1}{-2i} = i. \] Contour Integral: By the residue theorem: \[ \int_C \frac{z^2 \, dz}{z^2 - 2z + 2} = 2 \pi i ({Sum of residues}) = 2 \pi i (i + i) = 4 \pi i. \] Final Answer: \( 4 \pi i \).