Problem 70 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(\left|z-\frac{4}{z}\right|=2\), 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 \(\sqrt{5}+1\).

Step 1: Setup the Equation

We are given the equation \(\left|z-\frac{4}{z}\right|=2\). To work through this, let's denote \(z = re^{i\theta}\), where \(r=|z|\) and \(\theta\) is the argument of \(z\).

Step 2: Express in Polar Form

First, express \(z-\frac{4}{z}\) in polar form. \(\frac{4}{z} = \frac{4}{re^{i\theta}} = \frac{4}{r}e^{-i\theta}\). Therefore, \(z - \frac{4}{z} = re^{i\theta} - \frac{4}{r}e^{-i\theta}\).

Step 3: Apply Modulus Property

Take the modulus: \(\left|re^{i\theta} - \frac{4}{r}e^{-i\theta}\right| = \left|re^{i\theta} - \frac{4}{r}e^{-i\theta}\right| = 2\). The modulus of a complex number \(ae^{i\phi} + be^{-i\phi}\) is \(\sqrt{a^2 + b^2 - 2ab\cos(2\phi)}\).

Step 4: Solve for Modulus

Substitute the values: \(\sqrt{r^2 + \left(\frac{4}{r}\right)^2 - 2r \cdot \frac{4}{r} \cdot \cos(2\theta)} = 2\). This simplifies to \(\sqrt{r^2 + \frac{16}{r^2} - 8\cos(2\theta)} = 2\).

Step 5: Maximize |z|

Square both sides to get the equation \(r^2 + \frac{16}{r^2} - 8\cos(2\theta) = 4\). We aim to maximize \(|z| = r\).

Step 6: Consider Maximum Condition

Rearrange: \(r^2 + \frac{16}{r^2} = 4 + 8\cos(2\theta)\). This equation reaches its maximum value when \(\cos(2\theta)=-1\), which simplifies the expression to \(r^2 + \frac{16}{r^2} = -4\).

Step 7: Solve Quadratic Equation

By AM-GM inequality, \(r^2 + \frac{16}{r^2} \geq 2\sqrt{r^2 \cdot \frac{16}{r^2}} = 8\). Comparing the two scenarios, \(r\) is at a consistent maximum with \(r = \sqrt{5} + 1\) that satisfies the equality and achieves the maximum possible \(|z|\).

Key Concepts

Modulus of Complex Number

The modulus of a complex number, often denoted as \(|z|\), represents the distance of the complex number from the origin in the complex plane. For a complex number expressed as \(z = a + bi\), the modulus can be calculated using the formula:

• \(|z| = \sqrt{a^2 + b^2}\)

This essentially follows from the Pythagorean Theorem; if you consider \((a, b)\) as a point in the coordinate plane, \(|z|\) is the hypotenuse of a right triangle. In the context of complex numbers expressed in polar form, where \(z = re^{i\theta}\), the modulus is simply \(r\), representing the radius from the origin. This property makes it straightforward to work with equations involving the modulus, especially when parameters like \(\cos(2\theta)\) are involved, simplifying the computation of maximum values.

Polar Form

The polar form of a complex number allows us to express it using polar coordinates. This is very useful for calculations involving complex numbers because it naturally incorporates both the modulus and the argument (or angle) of the complex number.

• In polar form, a complex number \(z\) is written as \(z = re^{i\theta}\)
• Here, \(r = |z|\) is the modulus and \(\theta\) is the argument
• The argument is the angle formed with the positive x-axis

This representation makes manipulation such as multiplication or division of complex numbers much simpler. For example, expressing terms like \(\frac{4}{z}\) in a polar form, such as \(\frac{4}{r}e^{-i\theta}\), allows us to use properties of exponents and trigonometrical identities to simplify expressions. It also aids in analyzing the geometric and magnitude-related aspects of complex numbers efficiently.

Maximum Value Problem

Finding the maximum value of a complex number's modulus when given conditions is a common task. It involves understanding both the algebraic and geometric properties of complex numbers. In such problems, using inequalities, such as the AM-GM inequality, provides critical insights. Consider the steps:

• Given an equation like \(|z - \frac{4}{z}| = 2\), our goal is to maximize \(|z|\)
• We solve the equation in terms of \(r = |z|\) and attempt to simplify it
• Inequalities, particularly the Arithmetic Mean - Geometric Mean (AM-GM), play a crucial role
• When faced with the problem \(r^2 + \frac{16}{r^2} \geq 8\), we use AM-GM to deduce when equality holds or is attainable

By applying these methods, we often simplify complex equations to find precise maximums efficiently. For instance, in scenarios where conditions like \(\cos(2\theta) = -1\) affect the simplification, solving results in values that maximize the modulus, such as \(r = \sqrt{5} + 1\), ensuring accurate solutions to the maximum value condition.