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

To solve this problem, we utilize the theory of complex analysis, particularly residues and properties of analytic functions. Given the contour integral:

\[ \oint_C \frac{f(z)}{z-a} \, dz = 0, \]

where \( C: |z| = K \), let's analyze the 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 \( C \).

If \( K < |a| \):

• Since \( |z|=K \) does not encircle the point \( a \), the point \( a \) is outside the contour \( C \).
• The function \( \frac{1}{z-a} \) has no singularity inside \( C \), so \( f(z) \) being analytic implies the integral is zero. Thus, this is consistent.

If \( f(z) = (z-a)^n g(z) \):

• If \( n \geq 1 \), \( f(z) \) has a zero at \( z = a \). Since \( g(z) \) is analytic and \( a \) could either lie inside or outside the contour \( C \), evaluate the integral using residue theorem.
• For \( n \geq 1 \), \( f(z)/(z-a) = (z-a)^{n-1} g(z) \) is still analytic at \( z = a \). The residue of an analytic function is zero, thus:
• \[ \oint_C \frac{f(z)}{z-a} \, dz = 0 \] holds by the property of analyticity and zeros.

Therefore, under both conditions I and II, the integral result is zero. However, since we are evaluating which conclusion guarantees the result for general scenarios, II alone is true because it directly provides a condition where the function \( f(z)/(z-a) \) is analytic for any \( z = a \). Condition I only covers specific cases where \( a \) is outside the contour.