Question

Choose the correct statement(s) from the following options, regarding Cauchy's theorem on complex integration \( \oint_C f(z)\,dz \), where \( C \) is a simple closed path in a simply connected domain \( D \).

• A. Cauchy's theorem cannot be directly applied to conclude that \( \oint_C f(z)\,dz = 0 \) when \( f(z) = \frac{1}{z^2} \), and \( C \) is the unit circle
• B. If \( f(z) \) is analytic in \( D \), then it can be concluded that \( \oint_C f(z)\,dz = 0 \) for any simple closed path \( C \) in \( D \)
• C. The function \( f(z) \) must be analytic in \( D \) to conclude \( \oint_C f(z)\,dz = 0 \) for any simple closed path \( C \) in \( D \)
• D. \( \oint_C f(z)\,dz \neq 0 \) when \( f(z) = \frac{1}{z^2} \), since the function is not analytic at \( z = 0 \)

Hint:

Cauchy's theorem only applies when the function is analytic throughout the domain enclosed by the contour. For singularities inside the path, other results like Cauchy's residue theorem may apply.

Answer 1

The Correct Option is A, B

Step 1: Recall Cauchy’s theorem.
Cauchy’s theorem states that if a function \( f(z) \) is analytic everywhere inside and on a simple closed contour \( C \) in a simply connected domain \( D \), then:
\[ \oint_C f(z)\,dz = 0 \]
Step 2: Analyze Option (A).
Given \( f(z) = \frac{1}{z^2} \). This function is not analytic at \( z = 0 \), and if the unit circle \( C \) encloses the origin, then \( f(z) \) is not analytic in the domain enclosed by \( C \).
Thus, Cauchy’s theorem cannot be applied here.
Hence, option (A) is correct.
Step 3: Analyze Option (B).
This is a direct application of the theorem — if \( f(z) \) is analytic throughout \( D \), then for any closed contour \( C \) in \( D \), the integral is zero.
Hence, option (B) is correct.
Step 4: Analyze Option (C).
Although the function must be analytic in \( D \), this option is slightly misleading because it's restating the precondition of the theorem, not the result. The wording may be interpreted as a truism, but it does not constitute a full conclusion by itself. Still, it's not strictly false — so it's more interpretive.
Step 5: Analyze Option (D).
This is incorrect. Even though \( f(z) = \frac{1}{z^2} \) is not analytic at \( z = 0 \), the integral may still be zero or non-zero depending on the specific contour and nature of the singularity.
In fact, \( \oint_{|z|=1} \frac{1}{z^2}\,dz = 0 \), as \( \frac{1}{z^2} \) has a second-order pole at the origin and the integral of a second-order pole over a closed loop enclosing it is zero.
Hence, option (D) is incorrect.