Question

Given \[ \oint_C \frac{f(z)}{z - a} \, dz = 0, \] where \( C: |z| = K \). Consider the following conclusions:

• [I.] \( K<|a| \) and \( f(z) \) is analytic inside and on \( C \).
• [II.] \( f(z) = (z - a)^n g(z) \) for \( n \in \mathbb{N} \), and \( g(z) \) is analytic inside and on \( C \).

Which of the following is true?

• A. I alone is true
• B. II alone is true
• C. Both I and II are true
• D. Neither I nor II is true

Hint:

If \( f(z) = (z - a)^n g(z) \) with \( g(z) \) analytic, then \( \frac{f(z)}{z - a} \) is analytic too — and its integral over a closed contour is 0.

Answer 1

The Correct Option is B

Cauchy's integral theorem states that if \( f(z) \) is analytic on and within a closed contour \( C \), then \[ \oint_C \frac{f(z)}{z - a} \, dz = 2\pi i f(a) \] However, the given condition is that the integral is 0, which implies \( f(a) = 0 \) or \( f \) is not analytic at \( a \), or that \( a \) lies outside the contour.
Statement I says \( K<|a| \), meaning \( a \) lies outside the contour. This would make the function \( \frac{f(z)}{z - a} \) analytic on and inside \( C \), and by Cauchy's theorem, the integral would be 0. But this is only valid if \( f(z) \) is analytic — which isn't ensured here. So, I is insufficient.
Statement II represents \( f(z) \) having a zero at \( z = a \), of order \( n \), so \( f(z)/(z - a) \) is still analytic at \( z = a \) (since the singularity is removable). So the integrand is analytic on and inside \( C \), and hence the integral is 0.
Thus, only statement II is valid.