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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

Analyzing the equation |z| - 2 = 0

The equation \(|z| - 2 = 0\) means that the modulus of the complex number \(z\) is 2. Therefore, the locus described by this equation is a circle centered at the origin with a radius of 2.
02

Analyzing the equation |z-i| - |z+5i| = 0

The equation \(|z - i| - |z + 5i| = 0\) can be rewritten as \(|z - i| = |z + 5i|\). This describes the locus of points equidistant from the complex numbers \(i\) and \(-5i\). Such a locus is a perpendicular bisector of the line segment joining \(i (0,1)\) and \(-5i (0,-5)\) in the complex plane.
03

Determine the perpendicular bisector

To find the perpendicular bisector of the points \((0, 1)\) and \((0, -5)\), we first find the midpoint of the segment joining these two points: \((0, (1 + (-5))/2) = (0, -2)\). The perpendicular bisector is a horizontal line passing through this point, \(y = -2\).
04

Intersecting the loci

The intersection of the circle \(|z| = 2\) and the line \(y = -2\) must be found. The points on the circle have coordinates \((x, y)\) such that \(x^2 + y^2 = 4\). Setting \(y = -2\) gives \(x^2 + (-2)^2 = 4\), or \(x^2 + 4 = 4\), leading to \(x^2 = 0\), so \(x = 0\). Thus, the point \((0, -2)\) is the sole solution.

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Key Concepts

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

Locus
The term "locus" signifies the set of points satisfying a particular condition or a collection of rules. In complex numbers, the locus is a collection of points in the complex plane that these numbers correspond to. If you're working with an algebraic equation that involves complex numbers, the solution often describes a geometric shape or a path, known as the locus.
In our given exercise, there are two main equations that describe loci:
  • The first equation \(|z| - 2 = 0\) represents a circle's circumference centered at the origin.
  • The second equation \(|z-i| - |z+5i| = 0\) describes a line on the complex plane that is equidistant from the points corresponding to \(+i\) and \(-5i\).
Understanding loci is crucial as it allows for visualizing and conceptualizing complex numbers through simple geometric shapes or lines.
Modulus
In the context of complex numbers, the modulus of a complex number is its distance from the origin of the complex plane. If a point is labeled as \(z = x + yi\), where \(x\) is the real part and \(y\) is the imaginary part, the modulus is represented as \(|z| = \sqrt{x^2 + y^2}\).
  • The modulus essentially converts a complex number into a positive real number, showing its magnitude in the complex plane.
  • In our example, \(|z| - 2 = 0\), suggests that the distance (modulus) of complex number \(z\) from the origin is \(2\), characterizing a circle around the origin.
The modulus plays an integral role when defining distances between complex points, enabling the transformation of algebraic conditions into geometric interpretations on the complex plane.
Perpendicular Bisector
A perpendicular bisector, in its simplest form, is a line that divides another line into two equal parts at a 90-degree angle. When dealing with complex numbers, we often see perpendicular bisectors show up as a result of equations that state one point is equidistant from two fixed points.
For the equation \(|z - i| = |z + 5i|\), think of it as a scenario where each point \(z\) on this line is equidistant to \(i (0, 1)\) and \(-5i (0, -5)\). Here, the perpendicular bisector runs horizontally through the midpoint calculated as \((0, -2)\).
  • The presence of a perpendicular bisector indicates symmetry on the complex plane, reflecting equal distances from two points.
  • It's crucial for geometrically dividing sections of circles, such as in circle intersections.
Understanding perpendicular bisectors in the complex plane connects algebraic solutions to visible geometric phenomena.
Circle Equation
The classic circle equation in the complex plane is derived from the general equation \(x^2 + y^2 = r^2\), where \(r\) is the radius, and the center lies on the origin, represented by the equation \(|z| = r\). Circles described in this manner are quite common in complex numbers problems, showcasing loci centered around specific points.
In our example from the exercise, \(|z| - 2 = 0\) means that every point \(z\) on the circle is \(2\) units away from the origin.
  • Such equations help easily identify circles in the complex plane, providing the benefit of recognizing and calculating intersections with other lines, such as that of a perpendicular bisector.
  • Understanding circle equations is fundamental to grasp geometric implications of algebraic conditions on complex numbers.
The use of circle equations in problems involving complex numbers reveals the spatial relationship and symmetry present within the solution set of such equations.

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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.

If \(|z-i|=1\) and \(\arg (z)=\theta, \theta \in\left(0, \frac{\pi}{2}\right)\), then the value of \(\cot \theta-\frac{2}{z}\) is equal to (A) 0 (B) \(i\) (C) \(-i\) (D) 1

If \(k=\frac{3 n}{2}\), where \(n\) is an even positive integer, then \(\sum_{r=1}^{k}(-3)^{r-1} \cdot{ }^{3 n} C_{2 r-1}=\) (A) 0 (B) 1 (C) \(-1\) (D) None of these

If \(n>1\), then the roots of \(z^{n}=(z+1)^{n}\) lie on a (A) circle (B) straight line (C) parabola (D) None of these

If \(z^{2}+z+1=0\), where \(z\) is a complex number, then the value of \(\left(z+\frac{1}{z}\right)^{2}+\left(z^{2}+\frac{1}{z^{2}}\right)^{2}+\left(z^{3}+\frac{1}{z^{3}}\right)^{2}+\ldots+\left(z^{6}+\frac{1}{z^{6}}\right)^{2}\) is \([\mathbf{2 0 0 6}]\) (A) 18 (B) 54 (C) 6 (D) 12
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