Question

Which of the following statements involving contour integrals (evaluated counter-clockwise) on the unit circle \( C \) in the complex plane is/are TRUE?

• A. \( \oint_C e^z \, dz = 0 \)
• B. \( \oint_C z^n \, dz = 0, { where } n { is an even integer} \)
• C. \( \oint_C \cos z \, dz \neq 0 \)
• D. \( \oint_C \sec z \, dz \neq 0 \)

Hint:

For contour integrals, always check the analyticity of the function within the contour and apply Cauchy's integral theorem for analytic functions.

Answer 1

The Correct Option is A, B

Step 1: Evaluate \( \oint_C e^z \, dz \). 
The function \( e^z \) is analytic everywhere in the complex plane, including inside and on the unit circle \( C \). According to Cauchy's integral theorem, the contour integral of any analytic function over a closed curve is zero: \[ \oint_C e^z \, dz = 0. \] Step 2: Evaluate \( \oint_C z^n \, dz \). 
For \( n \neq -1 \), the function \( z^n \) is analytic inside and on the unit circle. Hence, by Cauchy's theorem, the integral is zero: \[ \oint_C z^n \, dz = 0 \quad {(for \( n \neq -1 \))}. \] Since the question specifies \( n \) as an even integer, it satisfies the condition for \( n \neq -1 \). 
Step 3: Evaluate \( \oint_C \cos z \, dz \). 
The function \( \cos z \) is analytic everywhere, so its contour integral over the unit circle is zero: \[ \oint_C \cos z \, dz = 0. \] Step 4: Evaluate \( \oint_C \sec z \, dz \). 
The function \( \sec z \) has singularities at odd multiples of \( \pi/2 \), so it does not satisfy the conditions of Cauchy's theorem and the integral does not vanish: \[ \oint_C \sec z \, dz \neq 0. \] Thus, the correct answers are (A) and (B).