Question

If f(z)=zⁿ+a_n-1z^n-1 +......+a₁zⁿ+a₀ ∈ R[z] is a polynomial in z with no root over R,then deg(f)is

• A. 9
• B. always≤4
• C. an odd number
• D. always≥4
• E. an even number

Answer 1

Answer: (E)

Given: \( f(z) = z^n + a_{n-1}z^{n-1} + \cdots + a_1z + a_0 \in \mathbb{R}[z] \) has no real roots.

We recall the following properties of real polynomials: 

1. Every real polynomial of odd degree has at least one real root (by the Intermediate Value Theorem) 

2. Real polynomials of even degree may have no real roots (e.g., \( z^2 + 1 \))

Since \( f \) has no real roots, it cannot be of odd degree. Therefore, the degree must be even.

Options analysis: (A) Specific degree (9) - incorrect as it's odd (B) Degree ≤4 - false (e.g., \( z^6 + 1 \) has no real roots) (C) Odd degree - contradicts given condition (D) Degree ≥4 - false (e.g., \( z^2 + 1 \) has no real roots) (E) Even degree - correct

The correct answer is (E) An even number.

Answer 2

Let \( f(z) = z^n + a_{n-1}z^{n-1} + \dots + a_1z + a_0 \) be a polynomial in \( z \) with real coefficients. If \( f(z) \) has no real roots, then the degree of \( f(z) \) must be an even number.

Reasoning:

Complex roots of polynomials with real coefficients come in conjugate pairs. That is, if \( a + bi \) is a root, then \( a - bi \) is also a root (where \( a \) and \( b \) are real numbers, and \( i \) is the imaginary unit).

Since real numbers are a subset of complex numbers (a real number is a complex number with a zero imaginary part), and the polynomial has no real roots, all of its roots must be complex. Because these complex roots come in conjugate pairs, the number of roots must be even. Since the degree of the polynomial equals the number of roots (counting multiplicity), the degree must also be even.

Therefore, the degree of \( f(z) \) is even number.