Question

The value of the integral \[ \int_{C} \frac{z^{100}}{z^{101} + 1} \, dz \] where C is the circle of radius 2 centered at the origin taken in the anti-clockwise direction is

• A. $-2\pi i$
• B. $2\pi$
• C. 0
• D. $2\pi i$

Hint:

When applying the residue theorem, ensure that you identify the singularities within the contour and compute the residues correctly to find the value of the contour integral.

Answer 1

The Correct Option is D

We are given the integral: \[ \int_{C} \frac{z^{100}}{z^{101} + 1} \, dz \] where \( C \) is the circle of radius 2 centered at the origin, and the contour is taken in the anti-clockwise direction. To solve this, we first look for the singularities of the integrand. The denominator \( z^{101} + 1 = 0 \) gives us the equation \( z^{101} = -1 \), which has 101 distinct roots. These roots are the 101st roots of -1, and they are given by: \[ z_k = e^{\frac{2k\pi i}{101}} \quad \text{for} \quad k = 0, 1, 2, \dots, 100. \] These roots lie on the unit circle \( |z| = 1 \). Since the contour \( C \) is a circle of radius 2, which encloses all of the 101 roots, we can apply the residue theorem. The integrand \( \frac{z^{100}}{z^{101} + 1} \) has simple poles at the 101 roots of \( z^{101} + 1 = 0 \). The residue at each pole \( z_k \) is given by: \[ \text{Res}\left( \frac{z^{100}}{z^{101} + 1}, z_k \right) = \lim_{z \to z_k} (z - z_k) \frac{z^{100}}{z^{101} + 1}. \] Using the fact that \( z^{101} + 1 = (z - z_k) \cdot \frac{d}{dz} (z^{101} + 1) \) at each pole, we can compute the residue and sum all residues. Finally, by the residue theorem, the integral is \( 2\pi i \) times the sum of the residues inside the contour, which is \( 2\pi i \). Therefore, the value of the integral is \( 2\pi i \).
Step 2: Final Answer.
The correct answer is (D) \( 2\pi i \).