Question

Let \(D = \{z \in \mathbb{C} : |z|<1\}\) and \(f: D \to \mathbb{C}\) be defined by \[ f(z) = z - 25z^3 + \frac{z^5}{5!} - \frac{z^7}{7!} + \frac{z^9}{9!} - \frac{z^{11}}{11!} \] Consider the following statements:
P: \(f\) has three zeros (counting multiplicity) in D.
Q: \(f\) has one zero in \(U = \{z \in \mathbb{C} : \frac{1}{2}<|z|<1\}\).
Then

• A. P is TRUE but Q is FALSE
• B. P is FALSE but Q is TRUE
• C. both P and Q are TRUE
• D. both P and Q are FALSE

Hint:

When using Rouché's Theorem on a polynomial or power series, a good first attempt is to choose the term with the largest coefficient as \(h(z)\) and all other terms as \(g(z)\). Then, evaluate the magnitudes on the boundary circle to see if the required inequality \(|g(z)|<|h(z)|\) holds.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This problem requires finding the number of zeros of a complex function within specific regions (a disk and an annulus). The primary tool for this is Rouché's Theorem.

Step 2: Key Formula or Approach:
Rouché's Theorem: If \(h(z)\) and \(g(z)\) are analytic inside and on a simple closed contour C, and if \(|g(z)| < |h(z)|\) for all \(z\) on C, then \(h(z)\) and the sum \(h(z) + g(z)\) have the same number of zeros (counting multiplicities) inside C. To find zeros in an annulus, we find the zeros in the larger disk and subtract the zeros in the smaller disk.

Step 3: Detailed Calculation:
Analysis of Statement P: Zeros in \(D = \{z : |z| < 1\}\)
We apply Rouché's theorem on the circle \(C = \{z : |z| = 1\}\).
Let \(h(z) = -25z^3\) and \(g(z) = z + \dfrac{z^5}{5!} - \dfrac{z^7}{7!} + \dfrac{z^9}{9!} - \dfrac{z^{11}}{11!}\).
On \(|z| = 1\): \[ |h(z)| = |-25z^3| = 25|z|^3 = 25 \] For \(g(z)\): \[ |g(z)| \le 1 + \frac{1}{120} + \frac{1}{5040} + \frac{1}{362880} + \cdots \approx 1.0085 \] Clearly, \(|g(z)| < |h(z)|\).
By Rouché's Theorem, \(f(z)\) has the same number of zeros in \(|z| < 1\) as \(-25z^3\), which has a zero of multiplicity 3 at \(z=0\). Thus, P is TRUE.

Analysis of Statement Q: Zeros in \(U = \{z : \tfrac{1}{2} < |z| < 1\}\)
Number of zeros in U = (Zeros in \(|z| < 1\)) − (Zeros in \(|z| \le 1/2\)).
Already we have 3 zeros in \(|z| < 1\). Now, on \(|z| = 1/2\): \[ |h(z)| = 25 \left(\tfrac{1}{2}\right)^3 = 3.125 \] \[ |g(z)| \le 0.5 + \tfrac{(1/2)^5}{120} + \tfrac{(1/2)^7}{5040} + \cdots \approx 0.50026 \] So again, \(|g(z)| < |h(z)|\). Thus, \(f(z)\) has 3 zeros inside \(|z| < 1/2\). Therefore, zeros in the annulus = \(3 - 3 = 0\). So, Q is FALSE.

Step 4: Final Answer:
P is TRUE but Q is FALSE.

Step 5: Why This is Correct:
Rouché's Theorem shows the dominant term \(-25z^3\) dictates the number of zeros in both disks. Hence, all three zeros lie inside \(|z| < 1/2\), leaving none in the annulus.