Problem 14 - Complex numbers from A to...Z

Question

$$ \text { Let } x, y, z \text { be complex numbers. } $$ a) Prove that $$ |x|+|y|+|z| \leq|x+y-z|+|x-y+z|+|-x+y+z| $$ b) If \(x, y, z\) are distinct and the numbers \(x+y-z, x-y+z,-x+y+z\) have equal absolute values, prove that $$ 2(|x|+|y|+|z|) \leq|x+y-z|+|x-y+z|+|-x+y+z| $$

Step-by-Step Solution

Answer

Question: Prove the following inequalities involving absolute values of complex numbers: a) $$|x|+|y|+|z| \leq |x+y-z|+|x-y+z|+|-x+y+z|$$ b) If \(x, y, z\) are distinct and the numbers \(x+y-z, x-y+z,-x+y+z\) have equal absolute values, prove that $$2(|x|+|y|+|z|) \leq |x+y-z|+|x-y+z|+|-x+y+z|$$ Solution: a) We used the triangle inequality for complex numbers to prove that $$|x|+|y|+|z| \leq |x+y-z|+|x-y+z|+|-x+y+z|$$ b) Given that \(|x+y-z|=|x-y+z|=|-x+y+z|\), we used the triangle inequality and algebraic manipulations to prove that $$2(|x|+|y|+|z|) \leq |x+y-z|+|x-y+z|+|-x+y+z|$$

Step 1: Triangle Inequality for Complex Numbers

Recall the triangle inequality for complex numbers states that: $$ |a+b| \leq |a| + |b|, $$ where \(a, b\) are complex numbers.

Step 2: Apply the Triangle Inequality

Applying the triangle inequality to \(x+y-z\) and then to \(x-y+z\), we get: $$ |x+y-z| \geq |x|+|y|-|z|, $$ and $$ |x-y+z| \geq |x|+|z|-|y|. $$ Similarly, applying the triangle inequality to \(-x+y+z\) and then to \(-(x+y-z)\), we have: $$ |-x+y+z| \geq |y|+|z|-|x|. $$

Step 3: Sum the Inequalities

Now, we add these three inequalities: $$ (x+y-z) + (x-y+z) + (-x+y+z) \geq (|x|+|y|-|z|) + (|x|+|z|-|y|) + (|y|+|z|-|x|). $$

Step 4: Simplify the Inequalities

Simplifying the left-hand side and the right-hand side, we have: $$ 3|x| + 3|y| + 3|z| \leq |x+y-z|+|x-y+z|+|-x+y+z|. $$ Dividing both sides by 3, we obtain the desired inequality: $$ |x|+|y|+|z| \leq |x+y-z|+|x-y+z|+|-x+y+z|. $$ b) If \(x, y, z\) are distinct and the numbers \(x+y-z, x-y+z,-x+y+z\) have equal absolute values, prove that $$ 2(|x|+|y|+|z|) \leq |x+y-z|+|x-y+z|+|-x+y+z|. $$ Given that \(|x+y-z|=|x-y+z|=|-x+y+z|\).

Step 5: Add the Equal Absolute Values

Let's add these equal absolute values: $$ |x+y-z| + |x-y+z| + |-x+y+z| = 2(x+y-z). $$ This equality implies that: $$ |x-y+z| + |-x+y+z| = (x+y-z). $$

Step 6: Use Triangle Inequality

Now, using the triangle inequality for the left-hand side: $$ |x-y+z| + |-x+y+z| \geq |x-y+2z|. $$ Thus, we have \(2(x+y-z) \geq |x-y+2z|\).

Step 7: Substitute and Divide

Recall from part (a) that: $$ |x|+|y|+|z| \leq |x+y-z|+|x-y+z|+|-x+y+z|. $$ Substitute \(2(x+y-z)\) for \(|x-y+z| + |-x+y+z|\) and divide by 2: $$ 2(|x|+|y|+|z|) \leq |x+y-z|+ 2(x+y-z). $$

Step 8: Simplify the Inequality

Now simplify the inequality, we get: $$ 2(|x|+|y|+|z|) \leq |x+y-z|+|x-y+z|+|-x+y+z|. $$ So, the required inequality has been proven.

Key Concepts

Triangle Inequality

Understanding the triangle inequality in the context of complex numbers is critical for solving many mathematical problems. In essence, the triangle inequality states that for any complex numbers, say 'a' and 'b', the absolute value of their sum is less than or equal to the sum of their absolute values:

\[\begin{equation}|a + b| \<= |\a + |b|.\end{equation}\]
This principle can be visualized on a complex plane, where the sum of the lengths of any two sides of a triangle is always greater than or equal to the length of the third side. In the problem provided, applying the triangle inequality allows us to establish a relationship between the complex numbers 'x', 'y', 'z', and their sums and differences. Through strategic summation and simplification of inequalities, we conclude that the sum of the absolute values of individual complex numbers is less than or equal to the sum of the absolute values of their collective combinations.

Absolute Values

In mathematical terms, the absolute value of a number is its distance from zero on the number line, regardless of direction. For complex numbers, the absolute value, often referred to as the modulus, represents the distance from the origin in the complex plane.

The concept of absolute value is pivotal when handling inequalities involving complex numbers. It helps us measure magnitudes without considering directionality. For complex numbers, the absolute value is calculated using the formula:

\[\begin{equation}|a + bi| = \sqrt{a^2 + b^2}.\end{equation}\]
Here, 'a' is the real part and 'b' is the imaginary part of the complex number. In the context of the exercise, equating the absolute values of different expressions involving 'x', 'y', and 'z' sets the stage for proving inequalities, serving as a foundation for establishing the given relationships.

Complex Number Properties

Complex numbers, which have the form 'a + bi' where 'a' is the real part and 'b' is the imaginary part, behave somewhat differently from real numbers, yet they obey certain algebraic rules. When solving complex number inequalities, several properties come into play:

• Closure: The sum, difference, and product of any two complex numbers are also complex numbers.
• Conjugates: The conjugate of a complex number 'a + bi' is 'a - bi', and it has the same absolute value as the original number.
• Modulus: The absolute value of a complex number corresponds to its modulus, which can be used in geometric interpretations on the complex plane.

By using these properties wisely, students can maneuver through complex number calculations with confidence, which is what the step-by-step solution to our original exercise demonstrates.