Problem 29 - A Complete Resource Book in Mathematics for JEE Main

Question

The greatest value of the moduli of complex numbres \(z\) satisfying the equation \(\left|z-\frac{4}{z}\right|=2\) is (A) \(\sqrt{5}\) (C) \(\sqrt{5}+1\) (B) \(\sqrt{5}-1\) (D) None of these

Step-by-Step Solution

Answer

(C) \(\sqrt{5} + 1\)

Step 1: Interpret the Equation with Modulus

The given equation is \( \left| z - \frac{4}{z} \right| = 2 \). We are asked to find the greatest modulus of the complex numbers \( z \) that satisfy this condition. Recall that the modulus of a complex number \( z \) is \( |z| \). The task is to find the maximum value of \( |z| \).

Step 2: Substitute and Simplify

Let \( z = re^{i\theta} \) where \( r = |z| \) and \( e^{i\theta} \) describes the angle of \( z \. \) The expression \( \frac{4}{z} \) becomes \( \frac{4}{re^{i\theta}} = \frac{4}{r}e^{-i\theta} \). Substitute these into the given equation: \[\left| re^{i\theta} - \frac{4}{r}e^{-i\theta} \right| = 2\] This simplifies to \[\left| r e^{i\theta} - \frac{4}{r} e^{-i\theta} \right| = 2 \text{ or } \sqrt{ \left( r - \frac{4}{r} \right)^2 + 0 } = 2\] Essentially, this reduces to \( \left| r - \frac{4}{r} \right| = 2 \).

Step 3: Solve for r

Now we solve \( \left| r - \frac{4}{r} \right| = 2 \). This implies two cases: 1. \( r - \frac{4}{r} = 2 \)2. \( r - \frac{4}{r} = -2 \) **Case 1:** \[r - \frac{4}{r} = 2 \Multiply by r: \quad r^2 - 4 = 2r \Rightarrow r^2 - 2r - 4 = 0 \]The solutions to this quadratic equation are:\[r = \frac{2 \pm \sqrt{(2)^2 + 4 \times 4}}{2} = \frac{2 \pm \sqrt{20}}{2} = 1 \pm \sqrt{5}\]

Step 4: Analyze the Results

From the solution of the quadratic, we find \( r = 1 + \sqrt{5} \) and \( r = 1 - \sqrt{5} \). Since \( r \) represents the modulus and must be positive, the valid solution from this case is \( r = 1 + \sqrt{5} \).**Case 2:**\[r - \frac{4}{r} = -2 \Multiply by r: \quad r^2 + 2r - 4 = 0 \]Solving this quadratic gives:\[r = \frac{-2 \pm \sqrt{4 + 16}}{2} = \frac{-2 \pm \sqrt{20}}{2} = -1 \pm \sqrt{5}\]Here, the positive solution is \( r = \sqrt{5} - 1 \).

Step 5: Determine the Maximum Modulus

Comparing the positive values obtained, \( r = 1 + \sqrt{5} \) and \( r = \sqrt{5} - 1 \), the greatest value is clearly \( 1 + \sqrt{5} \) since \( 1 + \sqrt{5} > \sqrt{5} \).

Key Concepts

Modulus of Complex Numbers

Understanding the modulus of a complex number is crucial when working with complex numbers. The modulus, denoted as \(|z|\), represents the distance from the origin to the point \(z\) in the complex plane. If you have a complex number represented as \(z = a + bi\), where \(a\) is the real part and \(b\) is the imaginary part, the modulus can be calculated using the formula:

• \(|z| = \sqrt{a^2 + b^2}\)

Finding the modulus gives the "magnitude" or "absolute value" of the complex number. In geometric terms, it shows how far away the complex number is from the zero point or origin.

For example, let's say you have \(z = 3 + 4i\). The modulus would be:

• \(|z| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5\)

In this way, you can always quantify the "size" of the complex number. Knowing the modulus is particularly important in problems where you're asked to compare the size of complex numbers or solve equations involving complex numbers.

Quadratic Equations

Quadratic equations are mathematical expressions where the highest power of the unknown variable is two. The standard form of a quadratic equation is \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants. Solving quadratic equations often involves finding the values of \(x\) that satisfy the equation.

There are several methods to solve quadratic equations, including:

• Factoring: Finding two numbers that multiply to \(c\) and add up to \(b\).
• Using the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
• Completing the square: Transforming the equation into a perfect square trinomial.

The quadratic formula is particularly useful because it provides a straightforward means to find solutions, even when the equation cannot be easily factored.

Let’s consider the quadratic equation from our solution: \(r^2 - 2r - 4 = 0\). By applying the quadratic formula, we obtained the possible values for \(r\) as \(1 \pm \sqrt{5}\). This indicates how the quadratic formula helps in determining possible moduli in such problems.

Polar Form of Complex Numbers

Complex numbers can also be represented in polar form, which is often helpful in simplifying multiplication and division of complex numbers. The polar form expresses a complex number \(z\) as the product of its modulus \(r\) and a complex exponential.

The formula for the polar form is:

• \(z = re^{i\theta}\)

Here, \(r\) is the modulus (magnitude of the complex number) and \(\theta\) represents the argument or angle of the complex number, measured from the positive real axis in the counterclockwise direction. This format leverages Euler's formula, \(e^{i\theta} = \cos(\theta) + i\sin(\theta)\).

Using polar form allows us to easily express and manipulate complex numbers in rotational (angle-based) contexts, such as when performing multiplication or division. For instance, in our solution, substituting \(z = re^{i\theta}\) simplified the expression for solving the modulus relations. Converting complex numbers to polar form can often simplify complex mathematical operations, making solutions more intuitive.