Question

The number of zeros of the polynomial \[ 2z^7 - 7z^5 + 2z^3 - z + 1 \] in the unit disc \( \{ z \in \mathbb{C} : |z|<1 \} \) is __________.

Hint:

Rouché's Theorem is useful for determining the number of zeros of a polynomial inside a region by comparing it with simpler functions on the boundary of the region.

Answer 1

Correct Answer: 5

We are asked to find the number of zeros of the polynomial \[ p(z) = 2z^7 - 7z^5 + 2z^3 - z + 1 \] in the unit disk \( \{ z \in \mathbb{C} : |z|<1 \} \). To determine the number of zeros inside the unit disk, we can apply Rouché's Theorem.
Step 1: Rouché's Theorem.
Rouché's Theorem states that if two holomorphic functions \( f \) and \( g \) satisfy \( |f(z) - g(z)|<|g(z)| \) on the boundary of some domain, then \( f \) and \( g \) have the same number of zeros in that domain.
Step 2: Compare the terms of the polynomial.
We now analyze the dominant terms of the polynomial in the unit disk. The highest degree term is \( 2z^7 \), which dominates on the boundary of the unit disk \( |z| = 1 \). The other terms are of lower degree and smaller in magnitude compared to \( 2z^7 \). Thus, we can approximate the polynomial by \( 2z^7 \) for \( |z| = 1 \).
Step 3: Conclusion.
Since \( 2z^7 \) has exactly 7 zeros inside the unit disk, by Rouché's Theorem, the given polynomial has 7 zeros inside the unit disk.