Question

Let \(D = \{z \in \mathbb{C} : |z| < 1\}\) and \(f: D \to \mathbb{C}\) be an analytic function given by the power series \(f(z) = \sum_{n=0}^{\infty} a_n z^n\), where \(a_0 = a_1 = 1\) and \(a_n = \frac{1}{2^{2n}}\) for \(n \ge 2\).
Consider the following statements:
P: If \(z_0 \in D\), then \(f\) is one-one in some neighbourhood of \(z_0\).
Q: If \(E = \{z \in \mathbb{C} : |z| \le \frac{1}{2}\}\), then \(f(E)\) is a closed subset of \(\mathbb{C}\).
Which of the following statements is/are correct?

• A. P is TRUE
• B. Q is TRUE
• C. Q is FALSE
• D. P is FALSE

Hint:

For local injectivity of an analytic function \(f\), always check if \(f'(z_0) \neq 0\). For questions involving images of sets, remember that continuous functions map compact sets to compact sets and connected sets to connected sets. In \(\mathbb{R}^n\) or \(\mathbb{C}\), compact means closed and bounded.

Answer 1

The Correct Option is A, B

Step 1: Understanding the Concept:
This is a multiple-select question. Statement P tests the condition for an analytic function to be locally one-to-one (injective). An analytic function \(f(z)\) is locally one-to-one at a point \(z_0\) if and only if its derivative \(f'(z_0)\) is non-zero. Statement Q tests a property of continuous functions on compact sets. The continuous image of a compact set is compact. In a metric space like \(\mathbb{C}\), a compact set is closed and bounded.
Step 3: Detailed Explanation:
Analysis of Statement P:
For \(f\) to be one-one in a neighborhood of \(z_0\), it is sufficient that \(f'(z_0) \neq 0\).
Let's find the derivative of \(f(z)\).
\[ f(z) = a_0 + a_1 z + \sum_{n=2}^{\infty} a_n z^n = 1 + z + \sum_{n=2}^{\infty} \frac{1}{4^n} z^n \] The derivative is: \[ f'(z) = a_1 + \sum_{n=2}^{\infty} n a_n z^{n-1} = 1 + \sum_{n=2}^{\infty} \frac{n}{4^n} z^{n-1} \] We need to check if \(f'(z)\) can be zero for any \(z \in D\) (i.e., for \(|z| < 1\)).
Let's find an upper bound for the magnitude of the summation part for \(|z| < 1\). \[ \left| \sum_{n=2}^{\infty} \frac{n}{4^n} z^{n-1} \right| \le \sum_{n=2}^{\infty} \frac{n}{4^n} |z|^{n-1} < \sum_{n=2}^{\infty} \frac{n}{4^n} \] The series \(\sum_{n=1}^{\infty} n x^n = \dfrac{x}{(1-x)^2}\) for \(|x| < 1\). \[ \sum_{n=2}^{\infty} \frac{n}{4^n} = \left( \sum_{n=1}^{\infty} n\left(\frac{1}{4}\right)^n \right) - \frac{1}{4} = \frac{\tfrac{1}{4}}{(1-\tfrac{1}{4})^2} - \frac{1}{4} = \frac{\tfrac{1}{4}}{(\tfrac{3}{4})^2} - \frac{1}{4} = \frac{1/4}{9/16} - \frac{1}{4} = \frac{4}{9} - \frac{1}{4} = \frac{7}{36} \] So, for any \(z \in D\), we have \[ \left| \sum_{n=2}^{\infty} \frac{n}{4^n} z^{n-1} \right| < \frac{7}{36} < 1. \] Now, consider \(|f'(z)| = \left|1 + \sum_{n=2}^{\infty} \frac{n}{4^n} z^{n-1}\right|\). By the reverse triangle inequality: \[ |f'(z)| \ge 1 - \left|\sum_{n=2}^{\infty} \frac{n}{4^n} z^{n-1}\right| > 1 - \frac{7}{36} = \frac{29}{36} > 0 \] Since \(f'(z)\) is never zero in \(D\), \(f\) is locally one-to-one at every point \(z_0 \in D\). Thus, P is TRUE.
Analysis of Statement Q:
The function \(f(z)\) is analytic on the open disk \(D=\{z : |z| < 1\}\). The set \(E = \{z \in \mathbb{C} : |z| \le \tfrac{1}{2}\}\) is a closed and bounded subset of \(\mathbb{C}\). By the Heine–Borel theorem, \(E\) is compact. Since \(f\) is analytic, it is continuous. The continuous image of a compact set is compact. Therefore, \(f(E)\) is a compact subset of \(\mathbb{C}\). Every compact set in a metric space (like \(\mathbb{C}\)) is closed. Therefore, \(f(E)\) is a closed subset of \(\mathbb{C}\). Thus, Q is TRUE.
Conclusion:
Both P and Q are true statements. The correct options are (A) and (B).
Step 4: Final Answer:
P is TRUE and Q is TRUE.
Step 5: Why This is Correct:
P is true because the derivative \(f'(z)\) is shown to be non-zero everywhere in the unit disk. Q is true because it follows from the fact that the continuous image of a compact set is compact, and compact sets in \(\mathbb{C}\) are closed.