Problem 31 - A Complete Resource Book in Mathematics for JEE Main

Question

\(\frac{z^{2}}{z-1}\) is always real, then (A) \(z\) lies only on a circle (B) \(z\) lies only on the real axis (C) \(z\) lies either on the real axis or on a circle (D) None of these

Step-by-Step Solution

Answer

(C) \(z\) lies either on the real axis or on a circle.

Step 1: Express z in terms of x and y

To handle this problem, let the complex number be expressed as \( z = x + yi \) where \( x \) and \( y \) are real numbers, and \( i \) is the imaginary unit.

Step 2: Substitute z in the expression and simplify

Substitute \( z = x + yi \) into the given expression \( \frac{z^2}{z-1} \). First, calculate \( z^2 = (x + yi)^2 = x^2 - y^2 + 2xyi \). Now, substitute into the expression: \[\frac{z^2}{z-1} = \frac{x^2 - y^2 + 2xyi}{x + yi - 1}.\]

Step 3: Handle the denominator

The denominator \( x + yi - 1 \) can be simplified to \( (x-1) + yi \). Multiply the numerator and the denominator of the expression by the conjugate of the denominator, \( (x-1) - yi \). The expression becomes:\[\frac{(x^2 - y^2 + 2xyi)((x-1) - yi)}{((x-1) + yi)((x-1) - yi)}.\]

Step 4: Evaluate the denominator

Calculate the denominator: \[((x-1) + yi)((x-1) - yi) = (x-1)^2 + y^2.\] This is derived from summing the square of the real and imaginary parts.

Step 5: Expand and combine like terms in the numerator

Expand the numerator:\[(x^2 - y^2 + 2xyi)((x-1) - yi) = (x^2 - y^2)(x-1) - (x^2 - y^2)yi + 2xyi(x-1) - 2xyi(yi)\]Combine like terms:\[[(x^2 - y^2)(x-1) + 2xyi(x-1)] - [(x^2 - y^2) + 2x(y^2)]i.\]

Step 6: Set conditions for a real value

For \( \frac{z^2}{z-1} \) to be real, the imaginary part of the numerator must be zero:\[-(x^2 - y^2) - 2x(y^2) = 0\]Solving this equation gives:\[x^2 = y^2 + 2xy^2\]Simplify:\[(x - y)(x + y) = 2xy^2\]

Step 7: Evaluate possible cases

We find solutions for when the expression is real:1. \(y = 0\) (which means \(z\) is on the real axis), or2. \(x = \frac{y}{y + 1}\) (which forms a circle when squared on the complex plane). **(This implies possible circle solutions.)**

Step 8: Conclude based on the conditions

From the analysis above, \(z\) either lies on the real axis (when \(y = 0\)) or is constrained to a circle (from manipulation of the circular form when \(x = \frac{y}{y + 1}\)). Thus, \(z\) can lie on either the real axis or on a specific circle.

Key Concepts

real axis

The real axis is a fundamental part of the complex plane, representing all the real numbers. When handling complex numbers, the real axis provides a reference for determining the position of these numbers in a two-dimensional space. For any complex number, supposed as \( z = x + yi \), the part \( x \) corresponds to its position on the real axis. Here, the imaginary part \( y \) equals zero, making the number purely real.

Understanding when \( z \) lies on the real axis is crucial for determining solutions in complex number problems. If the imaginary component of a formula or equation equates to zero, it means the number lies solely on this axis.

In our specific problem, when \( y = 0 \), \( z = x \), indicating that the solution lies on the real axis. This becomes significant as it helps simplify and solve complex equations by reducing them to real number problems.

complex plane

The complex plane is a two-dimensional space essential for visualizing and working with complex numbers. It consists of a horizontal axis for real numbers and a vertical axis for imaginary numbers. Each complex number \( z = x + yi \) is represented as a point in this plane, combining its real part \( x \) and imaginary part \( yi \).

When solving problems involving complex numbers, such as our given expression \( \frac{z^2}{z-1} \), it's crucial to understand how these numbers interact on the complex plane. The complex plane allows for geometric interpretations of complex operations, making it easier to analyze relationships, like whether \( z \) lies on a circle or line.

In the context of our exercise, exploring where \( z \) lies helps us understand potential solutions like points on the real axis or configurations forming a circle. These interpretations are essential to graphically visualizing solutions and simplifying complex mathematical expressions.

imaginary unit

The imaginary unit, denoted as \( i \), is a foundational element in the study of complex numbers. It is defined as the square root of \(-1\), forming the basis of all imaginary numbers. The imaginary unit facilitates defining complex numbers as \( z = x + yi \), where \( yi \) incorporates \( i \) to introduce imaginary value.

To solve problems like our example, breaking down expressions into real and imaginary components using the imaginary unit is essential. By expressing \( z = x + yi \), we manage both components distinctly, enabling us to apply algebraic techniques to simplify and resolve the equation. In particular, verifying that the imaginary part equals zero ensures that the complex expression yields a real result.

Mastery of the imaginary unit allows for seamless manipulation of complex numbers, easing their inclusion in real-world problems and theoretical math. Understanding its role and effects in operations like division and multiplication is key to solving and interpreting complex expressions.