Problem 127 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(z\) satisfies \(|z+1|<|z-2|\), and \(\omega=3 z+2+i\), then (A) \(|\omega+1|<|\omega-8|\) (B) \(|\omega+1|<|\omega-7|\) (C) \(\omega+\bar{\omega}>7\) (D) \(|\omega+5|<|\omega-4|\)If \(z\) satisfies \(|z+1|<|z-2|\), and \(\omega=3 z+2+i\), then (A) \(|\omega+1|<|\omega-8|\) (B) \(|\omega+1|<|\omega-7|\) (C) \(\omega+\bar{\omega}>7\) (D) \(|\omega+5|<|\omega-4|\)

Step-by-Step Solution

Answer

(B) \(|\omega+1| < |\omega-7|\)

Step 1: Interpret the condition on z

The inequality \(|z+1|<|z-2|\) describes a set of points in the complex plane. Geometrically, this means that the point \(z\) is closer to \(-1\) than to \(2\). This is the region between the perpendicular bisector of the line segment joining \(-1\) and \(2\) and the point \(-1\).

Step 2: Locate the region described by |z+1| < |z-2|

The line \(x=\frac{1}{2}\) is the perpendicular bisector of the segment connecting \(-1\) and \(2\). The region satisfying \(|z+1| < |z-2|\) is the left half-plane defined by \(x < \frac{1}{2}\).

Step 3: Express \\omega in terms of z.

Given \(\omega = 3z + 2 + i\), express \(\omega\) in terms of the real part \(x\) and imaginary part \(y\) of \(z\) as \(z = x + yi\). So, \(\omega = 3x + 2 + (3y + 1)i\).

Step 4: Transform the region for \\omega using the mapping

Since \(z\) must satisfy \(x < \frac{1}{2}\), we substitute \(x\) in \(\omega = 3x + 2 + (3y + 1)i\). Therefore, \(3x + 2 < \frac{3}{2} + 2 = \frac{5}{2}\). \(\omega\) lies in the left half of the complex plane with the real part less than \(\frac{5}{2}\).

Step 5: Evaluate each given condition and find the correct answer

- Check \(|\omega + 1| < |\omega - 8|\). The real part comparison is \(2 < 8\), which holds. - Check \(|\omega + 1| < |\omega - 7|\). It checks out as \(2 < 7\).- Check if \(\omega + \overline{\omega} > 7\). Notice that \(\overline{\omega} = 3x + 2 - (3y + 1)i\), so \(\omega + \overline{\omega} = 6x + 4\). For \(6x + 4 > 7\), we require \(x > \frac{1}{6}\), which may not always be true since \(x < \frac{1}{2}\).- Check \(|\omega + 5| < |\omega - 4|\). Real part comparison gives \(7 < 4\). This isn't correct.

Key Concepts

Inequalities in Complex Numbers

Understanding inequalities in complex numbers is similar to understanding them in real numbers, with the added complexity of dealing with the imaginary part. However, rather than thinking about 'greater' or 'lesser,' we focus on the distance from the origin or between points.
In complex number terms, the inequality \(|z+1| < |z-2|\) tells us something about the relative positioning of the complex number \(z\) concerning the points \(-1\) and \(2\) on the complex plane. This specific inequality means that the point \(z\) is closer to \(-1\) than to \(2\). When interpreting this geometrically, we imagine drawing circles around each point, \(-1\) and \(2\), and noting where these circles intersect or overlap.

These sorts of inequalities can be visualized as half-planes, where the number lies either inside or outside certain boundaries determined by the values in the inequality. This division helps in locating where our complex number \(z\) resides concerning other points. The understanding of these regions enables us to make further calculations when transforming or mapping these numbers.

Geometric Interpretation of Complex Numbers

Geometrically, complex numbers are represented on a plane similar to the Cartesian coordinate system, known as the complex plane or Argand plane. Here, any complex number \(z = x + yi\) can be plotted as a point, where \(x\) refers to the position on the horizontal axis (real part), and \(y\) on the vertical axis (imaginary part).

For the inequality \(|z+1|<|z-2|\), its geometric interpretation is about measuring distances. Specifically, it asks us to consider the locus of points for which the distance from \(z\) to \(-1\) is less than the distance to \(2\).
This results in the perpendicular bisector of that line segment, which forms a straight line as a part of the larger region on the complex plane. For our problem, the line is \(x = \frac{1}{2}\), dividing the complex plane, and telling us that the point \(z\) exists to the left of this bisector. This visual stance helps in understanding why certain inequalities hold or do not hold based on where the points are located in terms of each other.

Transformation of Complex Regions

Transforming complex numbers involves using mappings—functions that take each input to exactly one output. In this exercise, we explore a transformation through mapping. The given mapping is \(\omega = 3z + 2 + i\).

The transformation is achieved by replacing \(z\) with an expression in terms of \(\omega\), creating a new region where \(\omega\) satisfies a similar kind of inequality. Concretely, since \(|z+1|<|z-2|\) places constraints on where \(z\) can reside in the complex plane, the transformation reshapes this region similarly for \(\omega\).

• We know \(x < \frac{1}{2}\) for \(z\), thus when translating to \(\omega\), the real part of \(\omega\) due to the transformation \( \omega = 3x + 2 + (3y + 1)i \) becomes \(3x + 2 < \frac{5}{2}\).
• Essentially, the inequality is \(\omega\) must lie to the left half of the complex plane where the real part is less than \(\frac{5}{2}\).

Understanding the transformation of these complex regions helps in evaluating how specific mappings affect locations on the complex plane, and which inequality conditions correspondingly hold true.