Problem 207 - A Complete Resource Book in Mathematics for JEE Main

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 2, corresponding to option (C).

Step 1: Understand the Given Equation

We are given that \( |Z - \frac{4}{Z}| = 2 \), where \(Z\) is a complex number. The expression represents the distance between the complex number \(Z\) and \(\frac{4}{Z}\) on the complex plane being equal to 2.

Step 2: Rewrite the Equation in Terms of Real and Imaginary Parts

Let \( Z = x + yi \), where \( x \) and \( y \) are real numbers and \( i \) is the imaginary unit. Then \( \frac{4}{Z} = \frac{4}{x + yi} \). Express \( \frac{4}{Z} \) in real and imaginary terms by multiplying the numerator and denominator by the conjugate: \( \frac{4(x-yi)}{x^2 + y^2} \). Thus:\[ Z - \frac{4}{Z} = (x + yi) - \left( \frac{4x}{x^2+y^2} \right) - \left( \frac{4y}{x^2+y^2} \right)i \].

Step 3: Solve for Modulus

The modulus \( |Z - \frac{4}{Z}| = 2 \) can now be expressed using real components: \\[ \left( x - \frac{4x}{x^2+y^2} \right)^2 + \left( y - \frac{4y}{x^2+y^2} \right)^2 = 4 \].

Step 4: Analyze for Maximum |Z| by Substitution

Try substitutions or geometrical interpretations to simplify calculations. Assume \( |Z| = r \), where \( r = \sqrt{x^2+y^2} \), and substitute to check consistency: \[ |Z|^2 + \left| \frac{4}{Z} \right|^2 - 2\frac{4}{|Z|} = 4 \] implies \[ r^2 + \frac{16}{r^2} - 8 = 4 \]. Solve for \( r \).

Step 5: Solve Quadratice Equation

Simplify the equation derived in Step 4: \[ r^4 - 8r^2 + 16 = 0 \].\Use substitution \( u = r^2 \) leading to \( u^2 - 8u + 16 = 0 \). Solve for \( u \). By factorizing, \( (u-4)^2 = 0 \) giving \( u = 4 \). Then \( r = \sqrt{4} = 2 \).

Step 6: Calculate |Z| and Cross Check Options

The value obtained for \(|Z|\) is 2. Cross-check against given options (A-D). The maximum value 2 matches option (C).

Key Concepts

Modulus of a Complex Number

The modulus of a complex number is a fundamental concept. It measures how far a point, representing the complex number, is from the origin in the complex plane. For a complex number expressed as \( Z = x + yi \), where \( x \) is the real part and \( y \) is the imaginary part, the modulus of \( Z \) is calculated using the formula:

• \( |Z| = \sqrt{x^2 + y^2} \)

This modulus represents the "size" or "magnitude" of the complex number. In our specific exercise, the given condition involves the modulus, which signifies a specific distance constraint between two points on the complex plane: \( Z \) and \( \frac{4}{Z} \). The provided constraint \( |Z - \frac{4}{Z}| = 2 \) ensures that the distance between these points is exactly 2 units, serving as the key to solving the problem.

Distance in the Complex Plane

Understanding distance in the complex plane aids in visualizing and solving problems involving complex numbers. The complex plane is similar to a Cartesian coordinate system, with the real and imaginary parts of a complex number acting like the x and y coordinates. The distance between two complex numbers \( Z_1 = x_1 + y_1i \) and \( Z_2 = x_2 + y_2i \) is calculated using:

• \( |Z_1 - Z_2| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)

In the specific problem, the expression \( |Z - \frac{4}{Z}| = 2 \) represents the Euclidean distance formula applied in the complex plane. This formula becomes a guide in transitioning from abstract complex representations to genuine geometric interpretations, allowing us to translate the algebraic condition into a comprehensible geometric visualization. This visualization forms the foundation in assessing potential choices for the modulus \(|Z|\).

Solving Quadratic Equations

Quadratic equations often appear in problems involving complex numbers due to their squared terms. In the exercise, after substituting and simplifying, we encountered a quadratic equation of the form \( u^2 - 8u + 16 = 0 \). Here, \( u \) represents \( r^2 \), which comes from expressing the modulus \(|Z|\) in terms of real numbers.To solve quadratic equations:

• Utilize the general formula for roots: \( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
• In this case, the equation \((u - 4)^2 = 0\) was solved by recognizing its perfect square form, directly providing \( u = 4 \).

The intersection of algebraic manipulation and root-solving skills leads us to determine that the maximum value of \(|Z|\) is 2. This value satisfies the quadratic equation and the initial condition given, demonstrating that mastering quadratic solutions is key to unlocking distance and modulus constraints in complex number problems.