Problem 5 - Complex Analysis for Mathematics and Engineering

Question

Let \(P(z)=a_{n} z^{n}+a_{n-1} z^{n-1}+\cdots+a_{1} z+a_{0}\) be a polynomial of degree \(n .\) (a) Suppose that \(a_{n}, a_{n-1}, \ldots, a_{1}, a_{0}\) are all real. Show that if \(z_{1}\) is a root of \(P\), then \(\overline{z_{1}}\) is also a root. In other words, the roots must be complex conjugates, something you likely learned without proof in high school. (b) Suppose not all of \(a_{\mathrm{m}}, a_{n-1}, \ldots, a_{1}, a_{0}\) are real. Show that \(P\) has at least one root whose complex conjugate is not a root. Hint: Prove the contrapositive. (c) Find an example of a polynomial that has some roots occurring as complex conjugates, and some not.

Step-by-Step Solution

Answer

Real-coefficient polynomials have conjugate roots. Non-real coefficients disrupt this rule.

Step 1: Understand Given Polynomial and Roots

The polynomial given is \[ P(z) = a_n z^n + a_{n-1} z^{n-1} + \cdots + a_1 z + a_0 \]with coefficients \(a_n, a_{n-1}, \ldots, a_0\) being real numbers. We need to show that if \(z_1\) is a root, then its complex conjugate \(\overline{z_1}\) is also a root.

Step 2: Conjugate Symmetry for Real Coefficients

Since the coefficients are real, for any complex number \(z_1\) that satisfies \(P(z_1) = 0\), it follows that \(P(\overline{z_1}) = \overline{P(z_1)} = 0\). This is because substituting \(\overline{z}_i\) in the polynomial retains the real coefficients and conjugates the imaginary parts, maintaining the equation's balance when set to zero.

Step 3: Discuss Non-real Coefficients Scenario

If the coefficients are not all real, then the polynomial \(P(z)\) may contain complex coefficients. The given hint suggests proving the contrapositive: that a polynomial with non-real coefficients implies the existence of at least one root whose complex conjugate is not a root. Because complex coefficients can disrupt the symmetry, \(P(z)\) does not necessarily respect the conjugate rule as in the real case.

Step 4: Provide Polynomial Example

To find a polynomial with some roots as complex conjugates and some not, consider: \[ Q(z) = z^3 - (1+i)z^2 + iz - 1 \]This polynomial, for example, might have roots like 1 and -1 with a non-conjugate imaginary or complex component, dependent on complex arithmetic solving.

Key Concepts

Complex Conjugates

When dealing with complex numbers, each number has a counterpart called its complex conjugate. If you have a complex number, for example, \( z = a + bi \) where \( a \) and \( b \) are real numbers, its complex conjugate is \( \overline{z} = a - bi \). Complex conjugates have significant properties that make them especially useful in mathematics, particularly in polynomial equations and physics.

• The product of a complex number and its conjugate is always a real number: \( z \cdot \overline{z} = a^2 + b^2 \).
• If a polynomial with real coefficients has a complex root, its conjugate is also a root. This means the roots come in conjugate pairs.

In the given polynomial problem, the task is to prove that if \( z_1 \) is a root of a real-coefficient polynomial, \( \overline{z_1} \) must also be a root. This stems directly from the definition of complex conjugates aligning with real coefficients, maintaining the polynomial's equation in balance when these conjugates are substituted.

Polynomial Roots

Polynomial roots are the values of \( z \) that satisfy the polynomial equation \( P(z) = 0 \). Finding the roots is equivalent to solving the polynomial equation, which is a fundamental concept in algebra and complex analysis.

• For a polynomial \( P(z) \) with degree \( n \), there are exactly \( n \) roots, considering multiplicity.
• If the coefficients of the polynomial are real, complex roots occur in conjugate pairs. That is, if \( a + bi \) is a root, then \( a - bi \) is also a root.

The original exercise explores the nature of these roots, showing that when coefficients are real, complex roots must appear in pairs to keep the polynomial equations real-valued. This explains why polynomials with only real coefficients cannot have a single complex root without its conjugate pairing.

Real Coefficients

Real coefficients in a polynomial imply that all the values \( a_n, a_{n-1}, \ldots, a_1, a_0 \) are real numbers. This feature significantly influences the nature of the polynomial's roots and the symmetry present in its structure.

• Real coefficients ensure that any non-real roots appear in conjugate pairs. This is essential in maintaining the real-valued nature of polynomial expressions.
• If the polynomial has real coefficients and a complex number is a root, its conjugate must also be a root. This underlying principle is leveraged while solving polynomial equations, maintaining the balance when these complex values are substituted back into the equation.

When one of the coefficients is not real, this symmetry breaks down, as reflected in the original solution where a complex polynomial may have roots not corresponding their conjugates. The exercise lets us see both sides of this rule by showing examples and requiring proofs, bolstering our understanding of why symmetry only holds for polynomials with exclusively real coefficients.