Question

The value of the contour integral \[ \oint \frac{dz}{2z - z^2} \] along the circle \( |z| = 1 \), oriented in the counterclockwise sense is

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

Hint:

N/A

Answer 1

The Correct Option is A

Step 1: Analyze the integrand and identify the poles
The integrand is: \[ f(z) = \frac{1}{2z - z^2} = \frac{1}{z(2 - z)}. \] The poles of \( f(z) \) are obtained by solving: \[ z(2 - z) = 0 \implies z = 0 \quad \text{and} \quad z = 2. \] Thus, the poles are \( z = 0 \) (inside \( |z| = 1 \)) and \( z = 2 \) (outside \( |z| = 1 \)). 

Step 2: Apply the residue theorem
Since \( z = 0 \) is the only pole inside the contour \( |z| = 1 \), we compute the residue at \( z = 0 \): \[ \text{Residue at } z = 0 = \lim_{z \to 0} z \cdot f(z) = \lim_{z \to 0} \frac{z}{z(2 - z)} = \frac{1}{2}. \] Using the residue theorem, the contour integral is given by: \[ \oint \frac{dz}{2z - z^2} = 2\pi i \cdot \text{(Residue at \( z = 0 \))}. \] Substitute the residue: \[ \oint \frac{dz}{2z - z^2} = 2\pi i \cdot \frac{1}{2} = \pi i. \] 

Step 3: Conclude the solution
Thus, the value of the contour integral is \( \pi i \).