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

Question

Prove that $$ \left|\frac{6 z-i}{2+3 i z}\right| \leq 1 \text { if and only if }|z| \leq \frac{1}{3} . $$

Step-by-Step Solution

Answer

In this problem, we were given a complex function and asked to prove the inequality \(\left|\frac{6z-i}{2+3iz}\right| \le 1\) if and only if \(|z| \le \frac{1}{3}\). We started by finding the modulus of the function and simplifying it to \(\frac{|12z+18z^2| + |2z-3z^2|}{13}\). Then, we proved the statement in both directions. For the direct implication (\(\implies\)), we showed that if \(\left|\frac{6z-i}{2+3iz}\right| \le 1\), then the term inside the parentheses does not go below 39. Thus, if the inequality holds, then \(|z| \le \frac{1}{3}\). For the contrapositive implication (\(\Leftarrow\)), we showed that if \(|z| > \frac{1}{3}\), then \(\left|\frac{6z-i}{2+3iz}\right| > 1\). So we have proved the given inequality: $$ \left|\frac{6 z-i}{2+3 i z}\right| \leq 1 \text { if and only if }|z| \leq \frac{1}{3} $$

Step 1: Find the modulus of the function and simplify

To find the modulus of the function, we can first multiply the numerator and denominator by the complex conjugate of the denominator to get rid of the imaginary part in the denominator. $$ \left|\frac{6z-i}{2+3iz}\right| = \left|\frac{(6z-i)(2-3iz)}{(2+3iz)(2-3iz)}\right| $$ Expanding the expression and calculating the modulus: $$ \left|\frac{12z - 18i^2 z^2 - i(2-3iz)}{13}\right| = \frac{|12z+18z^2 - (2z-3z^2)|}{13} $$ Since \(|a+b| \le |a|+|b|\), we have: $$ \frac{|12z+18z^2 - (2z-3z^2)|}{13} \le \frac{|12z+18z^2| + |2z-3z^2|}{13} $$ Thus, we have simplified the modulus of the function.

Step 2: Prove the direct implication (\(\implies\))

If \(\left|\frac{6z-i}{2+3iz}\right| \le 1\), then: $$ \frac{|12z+18z^2| + |2z-3z^2|}{13} \le 1 $$ This implies that the sum of the moduli of the two complex numbers \(12z+18z^2\) and \(2z-3z^2\) is less than or equal to 13. Therefore: $$ |12z+18z^2| + |2z-3z^2| \le 13 $$ By noticing that \(|z|\) can be factored out on both terms: $$ |z|\left(|12+18z| + |2-3z|\right) \le 13 $$ Since \(|z|\) must be a non-negative number, to prove that \(|z| \le \frac{1}{3}\), it is enough to show that the term inside the parenthesis does not go below 39: Now let us analyse what happens when we increase the value of \(|z|\): Case 1: If \(|z| \ge \frac{2}{3}\), then \(|2-3z| \ge |3|\), thus inside parentheses the sum is greater or equal to \(39\), therefore we cannot decrease the value of \(|z|\) any further. Case 2: If \(|z| \le \frac{1}{3}\), then \(|12+18z|\) is also increasing, thus for smaller values of \(|z|\) we have a greater value on the inside of the parenthesis, therefore if \(|z| \le \frac{1}{3}\), \(\left|\frac{6z-i}{2+3iz}\right| \le 1\). This concludes the proof for the direct implication (\(\implies\)).

Step 3: Prove the contrapositive implication (\(\Leftarrow\))

To prove the contrapositive, we must show that if \(|z| > \frac{1}{3}\), then \(\left|\frac{6z-i}{2+3iz}\right| > 1\). We have shown in Step 2 that if \(|z| \ge \frac{2}{3}\), the sum inside the parentheses is greater than or equal to 39. Therefore, the inequality holds for any \(|z| \ge \frac{2}{3}\): $$ \left|\frac{6z-i}{2+3iz}\right| = \frac{|z|\left(|12+18z| + |2-3z|\right)}{13} \ge \frac{13}{13} = 1 $$ For the range \(|z| > \frac{1}{3}\) and \(|z| < \frac{2}{3}\), we can choose a smaller \(|z|\) to make \(|12+18z|\) and \(|2-3z|\) smaller. Therefore, if \(|z| > \frac{1}{3}\), \(\left|\frac{6z-i}{2+3iz}\right| > 1\). This concludes the contrapositive proof and the problem is solved. Hence, we have proved the given inequality: $$ \left|\frac{6 z-i}{2+3 i z}\right| \leq 1 \text { if and only if }|z| \leq \frac{1}{3} $$

Key Concepts

modulus of complex numbers

Understanding the modulus of complex numbers is essential in working with these numbers efficiently. The modulus, often represented as \(|z|\) where \(z = a + bi\), is the distance of the complex number from the origin in the complex plane. It is calculated using the formula:\[|z| = \sqrt{a^2 + b^2}\]Here, \(a\) and \(b\) are the real and imaginary parts, respectively.
The modulus tells us how big a complex number is in terms of magnitude. It's similar to finding the length of the hypotenuse of a right triangle.
In practice, the modulus helps in simplifying complex expressions, especially when dealing with inequalities or rational expressions involving complex numbers.
To determine inequalities involving complex moduli, like the problem presented, you often equate the modulus of the entire function to a real number and solve it using fundamental algebraic techniques.

• For example, multiplying by the complex conjugate to eliminate imaginary parts.
• Then applying absolute value properties make it easier to handle the real components.

inequalities involving complex numbers

Inequalities involving complex numbers are slightly more complex than real number inequalities.This is because they involve both real and imaginary components.
The primary approach involves comparing magnitudes rather than direct real-imaginary separation.
When proving inequalities involving complex expressions, such as \(\left| \frac{6z-i}{2+3iz} \right| \le 1\), we focus on the relationship between the modulus of the numerator and the denominator.
The strategy:

• Factor out the modulus of \(z\).
• Simplify and compare the results step by step using property \(|a+b| \leq |a|+|b|\). This helps in arriving at conclusions smoothly.

As demonstrated in the solution, comparing moduli allows us to derive conditions such as \(|z| \le \frac{1}{3}\), which ensures the original inequality holds.
Ultimately, mastering these inequalities paves the way for tackling advanced problems in complex analysis.

complex conjugates

In complex numbers, the complex conjugate plays a significant role in simplifying expressions involving division.The complex conjugate of a complex number \(a + bi\) is \(a - bi\).
The conjugate effectively eliminates the imaginary component in a denominator through multiplication, allowing for easier real-number comparison.

By multiplying the numerator and denominator of a complex fraction by the conjugate of the denominator, we arrive at:\[\frac{6z-i}{2+3iz} \cdot \frac{2-3iz}{2-3iz} = \frac{(6z-i)(2-3iz)}{(2+3iz)(2-3iz)}\]This results in a purely real denominator, allowing us to compute the modulus more straightforwardly.

• Utilizing complex conjugates is a standard technique in complex arithmetic, simplifying calculations considerably.

Understanding and using complex conjugates is crucial for working through many complex number problems, including proofs or solving equations involving modulus or inequalities.