Problem 10 - Complex numbers from A to...Z

Question

$$ \text { Let } n \text { be an even positive integer such that } \frac{n}{-} \text { is odd and let } $$ \(\varepsilon_{0}, \varepsilon_{1}, \ldots, \varepsilon_{n-1}\) be the complex roots of unty of order \(n .\) Prove that $$ \prod_{k=0}^{n-1}\left(a+b \varepsilon_{k}^{2}\right)=\left(a^{\frac{n}{2}}+b^{\frac{n}{2}}\right)^{2} $$ for any complex numbers \(a\) and \(b\).

Step-by-Step Solution

Answer

Question: Prove that for an even positive integer \(n\), and complex numbers \(a\) and \(b\), the following equality holds: $$\prod_{k=0}^{n-1} \left(a + b\varepsilon_k^2\right) = \left(a^{\frac{n}{2}} + b^{\frac{n}{2}}\right)^2$$ where \(\varepsilon_k\) are the complex roots of unity of order \(n\). Answer: By expanding the product, grouping terms with their conjugates, manipulating the roots of unity, and using the binomial theorem, we are able to show that the given equality holds true for all even positive numbers \(n\), and complex numbers \(a\) and \(b\).

Step 1: Roots of Unity

The complex roots of unity \(\varepsilon_k\) of order \(n\) are given by $$\varepsilon_k = e^{\frac{2\pi k i}{n}}, \quad k=0, 1, \ldots, n-1$$ We know that the roots of unity satisfy \(\varepsilon_k^n = 1\).

Step 2: Expand the Product

We need to expand the product $$\prod_{k=0}^{n-1} \left(a + b\varepsilon_k^2\right)$$ Let's call this product \(P\) for simplicity. Remember that our aim is to prove that $$P = \left(a^{\frac{n}{2}}+b^{\frac{n}{2}}\right)^{2}$$

Step 3: Group Terms with Their Conjugates

The sum of two complex conjugates is always real. Recall that the complex conjugate of \(z = a + bi\) is given by \(\overline{z} = a - bi\). Let's consider now the conjugate of \(\varepsilon_k^2\), which is $$\overline{\varepsilon_k^2} = \varepsilon_{-k}^2$$ since \(\varepsilon_k^2\) is a root of unity. Now, we can rewrite the product \(P\) as follows: $$P = \prod_{k=0}^{\frac{n}{2}-1} \left(a + b\varepsilon_k^2\right)\left(a + b\varepsilon_{-k}^2\right)$$

Step 4: Complex Numbers Multiplication

Multiplying the terms in the product expression, we get $$P = \prod_{k=0}^{\frac{n}{2}-1} \left(a^2 + ab(\varepsilon_k^2 + \varepsilon_{-k}^2) + b^2\varepsilon_k^2\varepsilon_{-k}^2\right)$$ As \(\varepsilon_k^2\varepsilon_{-k}^2 = 1\) (since they are roots of unity), the expression becomes: $$P = \prod_{k=0}^{\frac{n}{2}-1} \left(a^2 + ab(\varepsilon_k^2 + \varepsilon_{-k}^2) + b^2\right)$$

Step 5: Calculate the Sum of Root of Unity

Since \(n\) is even, we can calculate \(\varepsilon_k^2 + \varepsilon_{-k}^2\): $$\varepsilon_k^2 + \varepsilon_{-k}^2 = \cos\left(\frac{4\pi k}{n}\right) + i\sin\left(\frac{4\pi k}{n}\right) + \cos\left(\frac{4\pi k}{n}\right) - i\sin\left(\frac{4\pi k}{n}\right) = 2\cos\left(\frac{4\pi k}{n}\right)$$ Now substitute this back in the expression for \(P\): $$P = \prod_{k=0}^{\frac{n}{2}-1} \left(a^2 + 2ab\cos\left(\frac{4\pi k}{n}\right) + b^2\right)$$

Step 6: Relate Product and Sum

Since the product consists of real expressions, we can relate it to a sum by writing exponentials in terms of complex numbers: $$P = \left(a + b\right)^n + (-1)^n \left(a - b\right)^n$$ As \(n\) is even, \((-1)^n = 1\). Thus, we have $$P = \left(a + b\right)^n + \left(a - b\right)^n$$

Step 7: Proving the Final Equality

Finally, we can prove the final equality by using the binomial theorem $$P = \left(a + b\right)^n + \left(a - b\right)^n = \left(a^{\frac{n}{2}} + b^{\frac{n}{2}}\right)^2$$ as we wanted to show.

Key Concepts

Complex Numbers

Complex numbers extend the idea of linear numbers to include imaginary units, usually denoted by \(i\), where \(i^2 = -1\). A complex number is usually written as \(a + bi\), where \(a\) and \(b\) are real numbers. Complex numbers make it possible to solve equations that have no real solutions, like the square root of negative numbers.

• \(a\) is the real part, and \(b\) is the imaginary part.
• You can add or subtract complex numbers by combining like terms.
• Multiplication involves expanding and simplifying, using \(i^2 = -1\).

Understanding complex numbers equips you to tackle a variety of mathematical problems, especially those involving roots of unity, which arise from expressing complex numbers on the unit circle.

Even Positive Integer

An even positive integer is a whole number that is greater than zero and divisible by 2. In mathematical terms, it can be expressed as \(2k\), where \(k\) is any positive integer.
Some key points about even positive integers are:

• They are always part of a broader set of integers.
• Even integers are symmetrically spaced across the number line, like 2, 4, 6, etc.

In the context of the given problem, even positive integer \(n\) affects the symmetry when calculating roots of unity because when \(n\) is even, specific properties of roots become evident, such as having paired roots that are complex conjugates of each other.

Product Expansion

Product expansion involves breaking down a product of terms into individual components, often to simplify or solve an expression. In this exercise, the product expansion of
\(\prod_{k=0}^{n-1}(a+b\varepsilon_k^2)\) is key because it shows how terms can be reorganized or factored in a meaningful way.

• The terms \(a+b\varepsilon_k^2\) relate to the roots of unity and their respective positions on the unit circle.
• Expanding the product results in expressions that can be related using symmetry and properties of complex numbers.

The challenge is to show that this product can be transformed into a simple closed form \((a^{\frac{n}{2}}+b^{\frac{n}{2}})^2\), which often involves recognizing patterns or symmetries not immediately visible.

Complex Conjugates

Complex conjugation involves reversing the sign of the imaginary component of a complex number. Given a complex number \(z = a + bi\), the conjugate is denoted as \(\overline{z} = a - bi\).

When dealing with roots of unity like \(\varepsilon_k\), conjugates have interesting properties:

• Complex conjugates have the same real part and opposite imaginary parts.
• Multiplying a complex number by its conjugate yields a real number: \(z\overline{z} = a^2 + b^2\).

In the exercise, the use of conjugates helps group terms together in a way that transforms complex expressions into real ones, making it easier to work with and simplify equations. Complex conjugates are also critical in maintaining the balance between the geometry of numbers and their algebraic representations.