Problem 13 - Complex numbers from A to...Z

Question

$$ \text { Suppose that } z \neq 1 \text { is a complex number such that } z^{n}=1, n \geq 1 \text { . Prove } $$ $$ \begin{aligned} &\text { that }\\\ &|n z-(n+2)| \leq \frac{(n+1)(2 n+1)}{6}|z-1|^{2} . \end{aligned} $$

Step-by-Step Solution

Answer

Question: Prove that for any complex number \(z \neq 1\) such that \(z^n = 1\) for some integer \(n \geq 1\), the inequality \(|nz-(n+2)| \leq \frac{(n+1)(2n+1)}{6}|z-1|^2\) holds true. Answer: We proved the inequality \(|nz-(n+2)| \leq \frac{(n+1)(2n+1)}{6}|z-1|^2\) by analyzing the given information, simplifying the inequality expression, applying the triangle inequality, and showing that the desired inequality holds true for any complex number \(z \neq 1\) with \(z^n = 1\) for some integer \(n \geq 1\).

Step 1: Find the modulus of z

Given that \(z^n=1\), let's take the modulus of both sides of this equation. $$|z^n| =|1|$$ We know that \(|z^n| =|z|^n\) and \(|1| = 1\). Therefore, $$|z|^n = 1$$ To find \(|z|\) when \(n \neq 0\), take the nth root of both sides: $$|z| = 1$$

Step 2: Simplify and rewrite the inequality expression

We are given the inequality to prove: $$|nz-(n+2)| \leq \frac{(n+1)(2n+1)}{6}|z-1|^2$$ Now, we can express \(z\) as \(z=x+iy\), where \(x,y\) are real numbers and \(i=\sqrt{-1}\). Therefore, $$|n(x+iy)-(n+2)| \leq \frac{(n+1)(2n+1)}{6}|(x+iy)-1|^2$$

Step 3: Use the triangle inequality

Let's use the triangle inequality to simplify the modulus on the left side of the inequality: $$|n(x+iy)-(n+2)| = |nx+niy-n-2|\leq |nx-n|+|niy-2|$$ Since \(|z|=1\), we have \(x^2+y^2=1\). Now, we square the modulus on the right side of the inequality: $$\frac{(n+1)(2n+1)}{6}|(x+iy)-1|^2 = \frac{(n+1)(2n+1)}{6}|x-1+iy|^2 = \frac{(n+1)(2n+1)}{6}((x-1)^2+y^2)$$ which simplifies to $$\frac{(n+1)(2n+1)}{6}(x^2-2x+1+y^2)$$

Step 4: Show the inequality holds true

Now, we need to show that: $$|nx-n|+|niy-2| \leq \frac{(n+1)(2n+1)}{6}(x^2-2x+1+y^2)$$ Squaring both sides of the inequality and cancelling out a common term \(x^2+y^2-1\), we get: $$n^2(x^2+y^2-1)+4n(1-2x)+4 \leq \frac{(n+1)(2n+1)}{3}(x^2-2x)$$ Since we know that \(x^2+y^2=1\), we have: $$4n(1-2x)+4 \leq \frac{(n+1)(2n+1)}{3}(x^2-2x)$$ Multiplying both sides by 3 and rearranging terms, we get the inequality: $$12n(1-2x)+12\leq(2n^2+3n)(x^2-2x)$$ which simplifies to: $$\frac{(2n+1)(n+1)}{6}(2x-1)^2 \geq 0$$ Since \(\frac{(2n+1)(n+1)}{6}\) is always non-negative for \(n \geq 1\), this inequality will always hold true. Therefore, the given inequality $$|nz-(n+2)| \leq \frac{(n+1)(2n+1)}{6}|z-1|^2$$ is proven to be true.

Key Concepts

Triangle Inequality

The Triangle Inequality is a fundamental theorem in mathematics that relates to complex numbers as well as many other branches of math. In the context of complex numbers, it states that for any complex numbers \( z_1 \) and \( z_2 \), the following inequality holds:
\[ |z_1 + z_2| \leq |z_1| + |z_2| \] This means the modulus (or length) of the sum of two complex numbers is always less than or equal to the sum of their moduli.

• This property is incredibly useful, particularly when dealing with expressions involving complex numbers, helping to establish upper bounds.
• In the context of the exercise, it allows us to simplify expressions that have potentially larger, more complex forms into something more manageable.

Using the triangle inequality, we can dissect mathematical expressions and inequalities in order to analyze and prove them effectively.

Complex Modulus

The modulus of a complex number is akin to its "size" or "length" from the origin in the complex plane. If \( z = x + iy \), then the modulus is given by \( |z| = \sqrt{x^2 + y^2} \).
This concept is essential in various mathematical operations involving complex numbers. Usually denoted by vertical bars \( | \cdot | \), it provides valuable insight into the properties of the number.

• In the context of an equation like \( z^n = 1 \), where \( n \geq 1 \) and \( z eq 1 \), finding the modulus helps to simplify parts of the equation.
• From step 1 of the solution: Since \( |z^n| = |1| \), and \( |z^n| = |z|^n \), we finalize that \( |z| = 1 \). This crucial step aids in virtually reducing the complexity of our expressions further.

Understanding the modulus provides clarity when simplifying or analyzing more convoluted mathematical expressions.

Roots of Unity

Roots of Unity are specific complex numbers that are solutions to the equation \( z^n = 1 \) where \( n \) is a positive integer. These roots are significant mainly because they are evenly distributed points on the unit circle in the complex plane.

• The exercise considers a number \( z eq 1 \) such that \( z^n = 1 \). Thus, \( z \) is one of these roots of unity.
• When viewed in the complex plane, these numbers form a symmetric circle all having the modulus of 1, which simplifies the expression greatly.

Being roots of unity implies that these numbers when raised to the nth power, rotate back to 1 in the complex plane. This periodicity unlocks various symmetrical properties and patterns, instrumental for deriving simplified representations or expressions during problem entailing equal divisions of a circle.

Inequality Proofs

Inequality proofs are methods used to show that one side of an inequality is always less than or equal to another. These proofs often require a combination of algebraic manipulation, logical reasoning, and sometimes application of mathematical theorems.

• In the given problem, the goal was to establish: \( |nz - (n+2)| \leq \frac{(n+1)(2n+1)}{6}|z-1|^2 \).
• We leverage inequalities like the triangle inequality, properties of modulus, and other algebraic techniques to reach the goal.

These proofs can be extensive, involving multiple steps such as considering worst-case scenarios, bounding expressions in simplest forms, and eliminating potential maximum expansion by strategically managing expressions. It's crucial to understand that inequality proofs not only verify the relationship between different expressions but also provide deepened insights into the behavior and attribute of equality beyond numeric value.