Problem 21 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(z=a+i b\) where \(a>0, b>0\), then (A) \(|z| \geq \frac{1}{\sqrt{2}}(a-b)\) (B) \(|z| \geq \frac{1}{\sqrt{2}}(a+b)\) (C) \(|z|<\frac{1}{\sqrt{2}}(a+b)\) (D) None of these

Step-by-Step Solution

Answer

The correct answer is D: None of these.

Step 1: Understanding the given problem

Given that \( z = a + ib \) where \( a > 0 \) and \( b > 0 \), we need to evaluate whether one of the provided mathematical statements involving the modulus \(|z|\) is true.

Step 2: Calculate the modulus of z

The modulus of a complex number \( z = a + ib \) is given by \(|z| = \sqrt{a^2 + b^2}\). This formula will be used to compare against the provided options.

Step 3: Analyze option A

Option A states that \(|z| \geq \frac{1}{\sqrt{2}}(a-b)\). Since \(a, b > 0\), \(a-b\) could be positive, negative, or zero, which makes this option ambiguous for comparison without additional information. Hence, it is difficult to establish if this inequality holds true.

Step 4: Analyze option B

Option B states that \(|z| \geq \frac{1}{\sqrt{2}}(a+b)\). Consider if this inequality can be proven:Substitute \(|z| = \sqrt{a^2 + b^2}\), and compare to \(\frac{1}{\sqrt{2}}(a+b)\). It can be rewritten as \(\sqrt{a^2+b^2} \geq \frac{1}{\sqrt{2}}(a+b)\). This inequality doesn't generally hold as \(|z|\) can be less than \(a+b\), so we conclude this isn't always true.

Step 5: Analyze option C

Option C states that \(|z| < \frac{1}{\sqrt{2}}(a+b)\). Since \(|z| = \sqrt{a^2 + b^2}\), and considering the sum \(a+b\), we check if \(|z|\) is less than this expression. Since \(|z|\) is actually often less than \(a+b\) but not necessarily less than \(\frac{1}{\sqrt{2}}(a+b)\), option C is also not suitable.

Step 6: Evaluate correct choice

Given that neither (A), (B), nor (C) can be conclusively asserted as true without further constraints on \(a\) and \(b\), and considering the analysis above, the correct choice for the question falls under (D) None of these.

Key Concepts

Modulus of a Complex Number

A complex number is typically expressed as \(z = a + ib\), where \(a\) and \(b\) are real numbers, and \(i\) is the imaginary unit with the property that \(i^2 = -1\). The modulus of a complex number, denoted as \(|z|\), gives us the magnitude or length of the vector represented by the complex number in the complex plane.
To find the modulus of a complex number \(z = a + ib\), you use the formula:

• Formula: \(|z| = \sqrt{a^2 + b^2}\)

This is similar to finding the distance of a point (\(a, b\)) from the origin in a Cartesian coordinate system. It uses the Pythagorean theorem, considering the real and imaginary parts as the legs of a right triangle.
Knowing how to find the modulus is crucial when comparing complex numbers and when working with inequalities involving them. It also has practical applications in fields like engineering, where complex numbers are used to describe wave functions, electrical circuits, and more.

Inequalities

Inequalities express a relationship between two values, indicating that one value is larger or smaller than the other. They play a crucial role in mathematics, helping us establish upper and lower bounds for different expressions. When dealing with complex numbers, it's common to compare the modulus of a complex number with other expressions.
In the given exercise, you encountered inequalities that compare the modulus \(|z| = \sqrt{a^2 + b^2}\) to certain expressions involving \(a\) and \(b\):

• Option A: \(|z| \geq \frac{1}{\sqrt{2}}(a-b)\)
• Option B: \(|z| \geq \frac{1}{\sqrt{2}}(a+b)\)
• Option C: \(|z| < \frac{1}{\sqrt{2}}(a+b)\)

Understanding how to analyze these inequalities can help determine the relative sizes of these expressions. This involves comparing squares of both sides or applying algebraic manipulations. Recognizing when an inequality holds, or does not, is a critical skill for solving mathematical problems effectively. It involves a keen understanding of properties like symmetry, positivity, and other mathematical theorems that might be relevant.

JEE Mathematics

The Joint Entrance Examination (JEE) is one of the most challenging entrance exams in India for students aspiring to pursue engineering. It requires a deep understanding of various mathematical concepts, including complex numbers and inequalities, among others.
To excel in JEE Mathematics, students must not only be able to perform standard calculations but also apply creative problem-solving skills to unfamiliar problems. This involves:

• Grasping complex numbers and their properties, such as modulus and arguments.
• Understanding and solving inequalities, ensuring a firm grasp of algebraic and trigonometric manipulations.
• Applying mathematical concepts in innovative ways to solve rigorous questions.

By mastering these areas, students can enhance their ability to tackle JEE Mathematics problems efficiently. The modular approach to these concepts can often aid in simplifying the seemingly complex problems encountered in the exam. Therefore, regular practice and familiarity with a variety of problems is key to success.