Problem 70 - A Complete Resource Book in Mathematics for JEE Main

Question

f \(\left|z-\frac{4}{z}\right|=2\), then the maximum value of \(|z|\) is equal to (A) \(\sqrt{3}+1\) (B) \(\sqrt{5}+1\) (C) 2 (D) \(2+\sqrt{2}\)

Step-by-Step Solution

Answer

The maximum value of \(|z|\) is \(1+\sqrt{5}\) (option B).

Step 1: Rewrite the given condition

The given condition is \( |z - \frac{4}{z}| = 2 \). Let's denote \( z = re^{i\theta} \) where \( r = |z| \) and \( e^{i\theta} \) is the unit complex number on the complex plane. The condition then becomes \( | re^{i\theta} - \frac{4}{re^{i\theta}} | = 2 \).

Step 2: Apply the Triangle Inequality

According to the Triangle Inequality, we have:\[ | re^{i\theta} - \frac{4}{re^{i\theta}} | \geq ||z| - \frac{4}{|z|} |. \]Therefore, the given equality \( | re^{i\theta} - \frac{4}{re^{i\theta}} | = 2 \) implies:\[ ||z| - \frac{4}{|z|} | \leq 2. \]

Step 3: Simplify the Inequality

Substitute \( |z| = r \) into the inequality:\[ | r - \frac{4}{r} | \leq 2. \]This can be split into two separate inequalities:\( r - \frac{4}{r} \leq 2 \) and \( \frac{4}{r} - r \leq 2 \).

Step 4: Solve the first inequality

Solve \( r - \frac{4}{r} \leq 2 \):\[ r - \frac{4}{r} = 2 \] leads to the quadratic equation \( r^2 - 2r - 4 = 0 \), which can be solved using the quadratic formula: \[ r = \frac{2 \pm \sqrt{(2)^2 + 4\times4}}{2} = \frac{2 \pm \sqrt{18}}{2} = 1 \pm \sqrt{5}. \] Thus, the solution that satisfies \( r \geq 0 \) is \( r = 1 + \sqrt{5} \).

Step 5: Solve the second inequality

Solve \( \frac{4}{r} - r \leq 2 \):Start by rearranging to \( \frac{4}{r} \leq r + 2 \), which simplifies to \( 4 \leq r^2 + 2r \).Rearrange further: \( r^2 + 2r - 4 \geq 0 \).This is the same quadratic as in Step 4, solved for \( r = 1 + \sqrt{5} \) or \( r = 1 - \sqrt{5} \), but since \( r \geq 0 \), \( r = 1 + \sqrt{5} \) is valid.

Step 6: Determine the maximum value

Since \( 1 + \sqrt{5} \) is the only non-negative solution to both inequalities, it is the maximum value of \(|z|\) that satisfies the initial condition.

Key Concepts

Triangle Inequality

The Triangle Inequality is an essential principle in mathematics, particularly when dealing with complex numbers. It gives us insight into the relationship between the absolute values (or magnitudes) of sums or differences of complex numbers. For any two complex numbers, say \( z_1 \) and \( z_2 \), the Triangle Inequality is expressed as:

• \( |z_1 + z_2| \leq |z_1| + |z_2| \)
• \( |z_1 - z_2| \leq ||z_1| - |z_2|| \)

This inequality helps in visualizing the 'stretch' or 'shrink' of complex numbers on the complex plane.
In our exercise, we applied the Triangle Inequality to assess the equation \( |z - \frac{4}{z}| = 2 \).
By using this property, we transformed it to derive more manageable inequalities, such as \( ||z| - \frac{4}{|z|}| \leq 2 \).
This approach allows us to isolate the terms involving magnitudes, leading to simpler expressions and eventually finding the maximum value of \(|z|\).

Magnitude of Complex Numbers

Understanding the magnitude of complex numbers is crucial in many areas of mathematics. The magnitude, also known as the absolute value, of a complex number \( z = a + bi \) is given by the formula:

• \( |z| = \sqrt{a^2 + b^2} \)

This value represents the distance of the point \((a, b)\) from the origin on the complex plane.
In the context of our problem, we express the complex number in polar form, \( z = re^{i\theta} \), where \( r = |z| \) is the magnitude.
This allows us to apply trigonometric principles and equations like the Triangle Inequality efficiently.
Understanding polar representations not only helps in calculations but also provides a clear geometric interpretation of the magnitude.
In this exercise, finding the maximum magnitude \(|z|\) needed transforming and solving inequalities arising from its structure.

Quadratic Equations

Quadratic equations are a fundamental part of algebra, characterized by the standard form \( ax^2 + bx + c = 0 \). The solutions, or roots, of these equations, can be found using the quadratic formula:

• \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

In our exercise, solving the inequalities required transforming them into quadratic equations.
For instance, we derived \( r^2 - 2r - 4 = 0 \) while solving for the magnitude \(|z|\).
Using the quadratic formula, we found solutions to this equation revealing values \( r = 1 + \sqrt{5} \) and \( r = 1 - \sqrt{5} \).
Since \( r \) represents a magnitude, only non-negative solutions are considered valid, like \( r = 1 + \sqrt{5} \).
Understanding how to manipulate and solve quadratic equations is crucial in isolating mathematical values and finding solutions to more complex problems.