Problem 9 - Complex numbers from A to...Z

Question

$$ \text { Suppose } P \text { is a polynomial with complex coefficients and an even degree. } $$ If all the roots of \(P\) are complex nonreal numbers with modulus 1 , prove that $$ P(1) \in \mathbb{R} \text { if and only if } P(-1) \in \mathbb{R} \text { . } $$

Step-by-Step Solution

Answer

Question: If all of the roots of a polynomial with even degree have a modulus of 1 and are complex nonreal numbers, show that if P(1) is a real number, then so is P(-1), and vice versa. Answer: Since all roots of the even-degree polynomial P are complex nonreal numbers with a modulus of 1, the polynomial can be expressed as a product of quadratic polynomials with roots z and z*, where z* represents the complex conjugate of z. We simplify P(x) and calculate P(1) and P(-1). By showing that if P(1) is real, then P(-1) is also real, and vice versa, we establish the given statement.

Step 1: Find the polynomial using conjugate pairs

Since we are given that P has an even degree and all its roots are complex nonreal numbers with modulus 1, we can express P as the product of quadratic polynomials with roots z and z*, where z* represents the complex conjugate of z: $$ P(x) = a\prod_{i=1}^{n}\left((x - z_i)(x - z_i^*)\right) $$ Where a is any non-zero complex constant, and n is half the degree of P.

Step 2: Simplify the polynomial

Simplify the polynomial further using the properties of complex conjugates: $$ P(x)=a\prod_{i=1}^{n}\left(x^2 - (z_i+z_i^*)x + z_iz_i^*\right) $$ Note that since \(z_i\) and \(z_i^*\) are complex conjugates, their sum \((z_i+z_i^*)\) is a real number, as is their product \(z_iz_i^*=|z_i|^2=1^2=1\).

Step 3: Calculate P(1) and P(-1)

Calculate P(1) and P(-1) using the simplified polynomial expression: $$ P(1)=a\prod_{i=1}^{n}\left(1^2 - (z_i+z_i^*)1 + 1\right) $$ $$ P(-1)=a\prod_{i=1}^{n}\left((-1)^2 - (z_i+z_i^*)(-1) + 1\right) $$

Step 4: Show that P(1) and P(-1) are real

Now, we need to show that if P(1) is real, then P(-1) is real too, and vice versa. Let's first assume that P(1) is real: $$ P(1)=a\prod_{i=1}^{n}\left(2 - (z_i+z_i^*)\right) $$ Since a and \((z_i+z_i^*)\) are both real numbers, P(1) being real implies that the product of the factors in the expression is also real. Thus, when x=-1, we have: $$ P(-1)=a\prod_{i=1}^{n}\left(2 + (z_i+z_i^*)\right) $$ Because both a and \((z_i+z_i^*)\) are real numbers, P(-1) is real. Similarly, we can show that if P(-1) is real, then P(1) is real as well. This concludes our proof.

Key Concepts

Complex Conjugate

Understanding complex conjugates is crucial when dealing with polynomials with complex roots. If you have a complex number expressed as \( z = a + bi \), where \( a \) and \( b \) are real numbers, the complex conjugate \( z^* \) is \( a - bi \). This simply involves changing the sign of the imaginary part.

Complex conjugates have some interesting properties:

• The sum of any complex number and its conjugate is a real number: \( z + z^* = 2a \).
• The product of a complex number and its conjugate is also a real number: \( zz^* = a^2 + b^2 = |z|^2 \).

When building polynomials with complex roots, these conjugate pairs ensure that coefficients can remain real, as seen with roots in the form of conjugate pairs \( (z_i, z_i^*) \). In our exercise, we used this property to simplify the polynomial, maintaining real coefficients, which is vital for showing that both \( P(1) \) and \( P(-1) \) are real.

Modulus of Complex Numbers

The modulus (or absolute value) of a complex number \( z = a + bi \) is the distance from the origin to the point \( (a, b) \) in the complex plane. It is given by \( |z| = \sqrt{a^2 + b^2} \).

For complex numbers with modulus 1, like those mentioned in our problem, they lie on the unit circle in the complex plane. When a complex number \( z \) has modulus 1, it means:

• \( |z| = 1 \) implying \( zz^* = 1 \), where \( z^* \) is the complex conjugate.
• The real part of the complex number squared plus the imaginary part squared equals 1, as in \( a^2 + b^2 = 1 \).

These properties are helpful in ensuring that the coefficients of the polynomial remain real and in verifying that the roots contribute meaningfully to both \( P(1) \) and \( P(-1) \) being real.

Polynomial Roots

Polynomial roots are essential points where the polynomial equals zero. In polynomials, particularly those with complex coefficients, roots often appear as complex conjugate pairs. This means if \( z \) is a root, \( z^* \) is also a root.

In a polynomial \( P(x) = a(x - z_1)(x - z_1^*)...(x - z_n)(x - z_n^*) \), the presence of conjugate pairs ensures that any complex coefficient polynomial can have real coefficients when expanded. This happens because:

• Each pair of conjugate roots \( (x - z_i)(x - z_i^*) \) simplifies to \( x^2 - 2 \text{Re}(z_i)x + |z_i|^2 \), which are all real expressions.

This property is what underlies our ability to factorize the polynomial in the exercise and guarantees that calculations at specific points like \( P(1) \) and \( P(-1) \) will yield real numbers, given real coefficients.

Real Numbers

Real numbers are the set of all numbers that can be found on the number line. They include both rational and irrational numbers, which can be expressed as fractions or as non-repeating, non-terminating decimals.

In the context of polynomials, ensuring that certain values of the polynomial are real involves confirming that all imaginary elements cancel out or contribute no imaginary part. In our exercise,

• The terms \( (z_i + z_i^*) \) and products \( z_iz_i^* \) contribute only real numbers due to their properties, ensuring values like \( P(1) \) and \( P(-1) \) are real.

Maintaining that \( P(1) \) and \( P(-1) \) are real also involves understanding that these specific polynomial values will change identically under conjugate root pairs. The root symmetry and properties of complex conjugates work together to maintain real outputs in these calculations.