Question

The number of zeros (counting multiplicity) of \( P(z) = 3z^5 + 2iz^2 + 7iz + 1 \) in the annular region \( \{ z \in \mathbb{C} : 1 < |z| < 7 \} \) is \(\underline{\hspace{2cm}}\) .

Hint:

To find the number of zeros of a polynomial in a given region, use Rouché's Theorem or analyze the highest degree term's behavior in the region.

Answer 1

Correct Answer: 4

We can apply the Rouché's Theorem to determine the number of zeros of the polynomial within the annular region. To do so, we examine the given polynomial: \[ P(z) = 3z^5 + 2iz^2 + 7iz + 1. \] We are interested in finding the number of zeros within the region \( 1 < |z| < 7 \). Using Rouché's Theorem, we compare the terms of the polynomial within the region defined by the annular region boundaries. The key observation is that the highest degree term, \( 3z^5 \), dominates the behavior of the polynomial as \( |z| \) grows large. The number of zeros can be determined by examining the roots of the highest degree term in the given region. For \( P(z) = 3z^5 + \ldots \), the degree of the polynomial suggests there are 5 zeros. Thus, the number of zeros (counting multiplicity) of \( P(z) \) in the annular region is \( \boxed{4} \).