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Chapter 3: Problem 46

The locus of the points representing the complex numbers which satisfy \(|z|-2=0,|z-i|-|z+5 i|=0\) is: (A) a circle with centre at origin (B) a straight line passing through origin (C) the single point \((0,-2)\) (D) None of these

Short Answer

Expert verified
(C) The single point (0, -2).

Step by step solution

01

Understanding the First Equation

The first condition \(|z|-2=0\) implies \(|z|=2\). In the complex plane, this represents a circle with center at the origin \(0,0\) and radius equal to 2.
02

Analyzing the Second Equation

The second condition \(|z-i| - |z+5i| = 0\) means the distances from a point on the complex plane to \(0,1\) and \(0,-5\) are equal. This is the definition of the perpendicular bisector of the segment joining \(0,1\) and \(0,-5\).
03

Finding the Perpendicular Bisector

The perpendicular bisector of the line segment from \(0,1\) to \(0,-5\) is a horizontal line parallel to the real axis, equidistant from both points. Calculating the midpoint between these points gives \(0,-2\), this is the line y = -2.
04

Intersection of Circle and Line

Now we consider both the circle \(|z|=2\) and the line y = -2. To find their intersection, we check which points satisfy both conditions. At \(x, y = -2\), we calculate \(|z|=\sqrt{x^2 + (-2)^2}=2\). Substituting in the circle's equation gives \(|z|=\sqrt{x^2+4}=2\). Solving \(x^2 + 4 = 4\) leads to \(x^2=0\), hence \(x=0\).
05

Solution Interpretation

The only point that lies both on the circle and on the line is \(x, y = 0, -2\). Therefore, the locus is the single point \((0, -2)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Locus
In geometry and complex number theory, the term **locus** refers to the set of all points that satisfy a particular condition or equation. Here, the locus is utilized to find points on the complex plane that meet specific criteria. For this exercise, there are two conditions given:
  • The modulus of the complex number, \(|z|-2=0\), establishing a circle centered at the origin.
  • The equation \(|z-i| - |z+5i| = 0\), which defines a perpendicular bisector of the segment joining two points on the imaginary axis.
The locus, therefore, is the cross-section of these two conditions, leading to a unique solution in this problem. By finding a common point that lies on both formations, we can determine the set of all points that form the locus.
Circle Equation
Circles play a vital role in understanding geometric shapes using complex numbers. In the equation \(|z|-2=0\), the modulus \(|z|\) represents the distance of a point from the origin. Therefore, \(|z| = 2\) denotes a circle centered at the origin with a radius of 2.
  • Given a complex number \(z = x + yi\), the modulus \(|z|\) is calculated as \(\sqrt{x^2 + y^2}\).
  • Thus, the circle's equation can be written as \(x^2 + y^2 = 4\), where 4 is the square of the radius.
Being able to express complex properties through equations like this helps articulate the geometric structure and positions concerning the complex plane.
Perpendicular Bisector
To understand the concept of a **perpendicular bisector** in the complex plane, think of it as an invisible line that perfectly divides a segment into two equal parts, while being perpendicular to it. This is key when analyzing the equation \(|z-i|-|z+5i|=0\).
  • This equation describes a situation where any point \(z\) has an equal distance from both the points \(0,1\) (or \(i\)) and \(0,-5\) (or \(5i\)).
  • The perpendicular bisector of a segment connecting these points is derived by calculating their midpoint, here yielding the line \(y = -2\).
This line, therefore, represents a set of points equidistant from both references, central to finding the intersection with the given circle equation and locating the locus.
Complex Plane
The **complex plane** is a two-dimensional plane where each point corresponds to a complex number. With the horizontal axis representing real numbers and the vertical axis for the imaginary part, it is a major tool in visualizing and solving complex number problems.
  • A complex number \(z = x + yi\) places the real part \(x\) along the x-axis and the imaginary part \(y\) along the y-axis.
  • This plane becomes particularly useful when dealing with movements and transformations, as shown in our exercise's intersection problem.
Understanding how geometric shapes such as circles and lines manifest on this plane encourages deeper insight into their properties and interactions, helping reveal solutions to multiple constraint problems through visualization.

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Most popular questions from this chapter

If \(1, a_{1}, a_{2}, \ldots, a_{n-1}\) are the \(n, n\)th roots of unity, then \(\left(1-a_{1}\right)\left(1-a_{2}\right)\left(1-a_{3}\right) \ldots\left(1-a_{n-1}\right)=\) (A) \(n+1\) (B) \(n\) (C) \(n-1\) (D) None of these.

\(z_{1}=a+i b\) and \(z_{2}=c+i d\) are complex numbers such that \(\left|z_{1}\right|=\left|z_{2}\right|=1\) and \(\operatorname{Re}\left(z_{1} \bar{z}_{2}\right)=0 .\) If \(w_{1}=a+i c\) and \(w_{2}=b+i d(a, b, c, d \in R)\), then (A) \(\left|w_{1}\right|=1\) (B) \(\left|w_{2}\right|=1\) (C) \(\operatorname{Re}\left(w_{1} \bar{w}_{2}\right)=0\) (D) \(\operatorname{Re}\left(w_{1} \bar{w}_{2}\right)=1\)

The value of \(\sum_{k=1}^{10}\left(\sin \frac{2 k \pi}{11}+i \cos \frac{2 k \pi}{11}\right)\) is (A) \(i\) (B) 1 (C) \(-1\) (D) \(-i\)

Assertion: The locus of the points representing the complex numbers satisfying \(|z|-2=0,|z-i|-\) \(|z+5 i|=0\) is the single point \((0,-2)\) Reason: If \(z\) is a variable point and \(z_{1}, z_{2}\) are two fixed points in the argand plane, then \(\left|z-z_{1}\right|=\left|z-z_{2}\right| \Rightarrow\) locus of \(z\) is the perpendicular bisector of the line segment joining \(z_{1}\) and \(z_{2}\).

Let \(z, w\) be complex numbers such that \(\bar{z}+i \bar{w}=0\) and \(\arg z w=\pi\). Then \(\arg z\) equals \([2004]\) (A) \(\frac{\pi}{4}\) (B) \(\frac{5 \pi}{4}\) (C) \(\frac{3 \pi}{4}\) (D) \(\frac{\pi}{2}\)
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