Problem 51 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

The largest value of for which the region represented by the set \(\\{\omega \in C|\omega-4-i| \leq r\\}\) is contained in the region represented by the set \((z \in c /|z-1| \leq|z+i|)\), is equal to: (a) \(\frac{5}{2} \sqrt{2}\) (b) \(2 \sqrt{2}\) (c) \(\frac{3}{2} \sqrt{2}\) (d) \(\sqrt{17}\)

Step-by-Step Solution

Answer

The largest value of \(r\) is \(2\sqrt{2}\).

Step 1: Understanding the Region Represented by the First Set

The set \( \{ \omega \in \mathbb{C} \mid |\omega - 4 - i| \leq r \} \) represents a circle centered at the point \((4, 1)\) in the complex plane with a radius of \(r\).

Step 2: Analyzing the Region Represented by the Second Set

The set \( \{ z \in \mathbb{C} \mid |z - 1| \leq |z + i| \} \) describes the region in the complex plane where points are closer to or equidistant from \(1\) on the real axis than from \(-i\) on the imaginary axis. This is the half-plane boundary defined by the perpendicular bisector of the line segment joining \(1\) and \(-i\).

Step 3: Finding the Perpendicular Bisector

The line segment joining \(1\) and \(-i\) has a midpoint at \((1, -0.5)\) in the complex plane. The perpendicular bisector will be the vertical line since the real and imaginary components are equally placed from the point \((1, -0.5)\). This vertical line is described by \(x = 1\). The line divides the plane into two regions.

Step 4: Determining Containment

For the circle centered at \((4, 1)\) to completely lie within the half-plane \(x \leq 1\), the center at \((4, 1)\) must be within the half-plane when considering the radius. The distance from the center of the circle \((4, 1)\) to the boundary line \(x = 1\) is \(4 - 1 = 3\). Thus, for the circle to be contained in the region, the radius \(r\) must be \leq 3.

Step 5: Identifying the Largest Possible Radius

Verification of options shows that only \(2\sqrt{2}\) (approximately \(2.828\)) is less than 3, which fits the condition for the largest \(r\). Hence, the largest value of \(r\) is \(2\sqrt{2}\).

Key Concepts

Geometry in the Complex Plane

In the complex plane, each point is linked to a complex number of the form \( z = x + yi \). The number \( x \) is the real part, and \( y \) is the imaginary part. This representation gives us a visual way to understand complex numbers, much like coordinates in the geometric plane.

When working with geometric shapes in the complex plane, circles and regions often come up. For example, the set \( \{ \omega \in \mathbb{C} \mid |\omega - 4 - i| \leq r \} \) creates a circle. Here, the point \( 4 + i \) acts as the center, and \( r \) is the radius.

Understanding geometry in the complex plane helps us solve problems involving distance and areas around points described by complex numbers. It's a crucial part of analyzing properties or transformations of these numbers.

Inequalities in Complex Numbers

Complex number inequalities are like sorting or understanding how complex numbers relate spatially. They show which numbers are closer or further from given points. For instance, \( |z - 1| \leq |z + i| \) tells us points \( z \) are closer to \( 1 \) on the real axis than to \( -i \) on the imaginary axis.

This inequality, therefore, describes a particular region. In this case, it's a half-plane in the complex plane. Such inequalities help set specific boundaries or conditions for these plane regions, leading to solutions involving intersections or containments of sets.

Perpendicular Bisector in Complex Plane

The concept of perpendicular bisectors helps to find regions equidistant from two points in geometry, and this translates to the complex plane. Suppose you have points \(1\) and \(-i\). The perpendicular bisector here finds the middle and forms a line of equidistance, without preference for either point.

The line segment from \(1\) to \(-i\) has its midpoint at \((1, -0.5)\). The perpendicular bisector is a vertical line, \(x = 1\), because the distance remains the same along this line horizontally. It's like cutting the plane into two halves, defining one side as closer to one point than the other. Such bisectors are helpful tools for determining containment, showing which side of the plane contains certain regions or shapes fully.