Problem 134
Question
If \(z_{1}, z_{2}, z_{3}\) are non-zero, non-collinear complex numbers such that \(\frac{2}{z_{1}}=\frac{1}{z_{2}}+\frac{1}{z_{3}}\), then the points \(z_{1}, z_{2}, z_{3}\) lie (A) in the interior of a circle (B) on a circle passing through origin (C) in the exterior of a circle (D) None of these
Step-by-Step Solution
Verified Answer
(B) on a circle passing through origin
1Step 1: Understanding the Equation
We are given \( \frac{2}{z_{1}} = \frac{1}{z_{2}} + \frac{1}{z_{3}} \). This can be rewritten as \( \frac{2}{z_{1}} - \frac{1}{z_{2}} - \frac{1}{z_{3}} = 0 \). By finding a common denominator, this equation becomes \( \frac{2z_2z_3 - z_1z_3 - z_1z_2}{z_1z_2z_3} = 0 \), which simplifies to \( 2z_2z_3 = z_1z_3 + z_1z_2 \).
2Step 2: Reorganizing Terms
Let's rearrange the equation \(2z_2z_3 = z_1z_3 + z_1z_2\) as \( 2z_2z_3 - z_1z_3 - z_1z_2 = 0 \) or \( z_2(2z_3 - z_1) - z_1z_3 = 0 \). Factor out \(z_3\) to get \( (2z_2 - z_1)z_3 = z_1z_2 \).
3Step 3: Analyzing Vector Representation
To interpret this geometrically, consider representing these complex numbers as vectors in the complex plane. If we take \(z_1, z_2, z_3\) as vector endpoints, the equation becomes relevant to Apollonius Circle Theorem in geometry, which shows the ratios of distances from a point to two fixed points form a circle.
4Step 4: Interpreting the Geometric Condition
The condition \( \frac{2}{z_1} = \frac{1}{z_2} + \frac{1}{z_3} \) implies a harmonic division. In the complex plane, this means the points \( z_1, z_2, \) and \( z_3 \) lie on an Apollonius circle of which the origin is on the circle.
Key Concepts
Apollonius Circle TheoremVector Representation in Complex PlaneHarmonic Division in Complex Plane
Apollonius Circle Theorem
The Apollonius Circle Theorem is a fascinating concept in geometry that relates to the position of points in a plane. It's named after the Greek mathematician Apollonius of Perga. This theorem states that, for any given point, there is a circle such that the distance of any point on this circle from two fixed points is in a constant ratio.
To put it simply, imagine you have two fixed points, say A and B, and you pick a point C somewhere on the plane. The Apollonius circle would include all points, like C, where the distance to A and B maintains a specific ratio. In the context of complex numbers, this can translate into equations involving those numbers.
To put it simply, imagine you have two fixed points, say A and B, and you pick a point C somewhere on the plane. The Apollonius circle would include all points, like C, where the distance to A and B maintains a specific ratio. In the context of complex numbers, this can translate into equations involving those numbers.
- The Apollonius circle offers insight into how these complex numbers relate to each other spatially.
- It provides a geometric interpretation, helping us visualize equations as circular paths.
Vector Representation in Complex Plane
The vector representation in the complex plane is a powerful way to visualize and solve equations involving complex numbers. Each complex number can be seen as a vector originating from the origin of the plane, with its endpoint at the coordinates defined by the real and imaginary parts.
In simple terms, it's like viewing each complex number as an arrow that points somewhere on the plane. This arrow's length and direction are determined by the number's magnitude and argument.
In simple terms, it's like viewing each complex number as an arrow that points somewhere on the plane. This arrow's length and direction are determined by the number's magnitude and argument.
- Vector representation helps to uncover the spatial relationships among complex numbers.
- It allows us to see geometric patterns, like those formed by circles or lines connected by these vectors.
Harmonic Division in Complex Plane
Harmonic division is an intriguing concept, particularly in the field of projective geometry, which extends into the complex plane. In essence, a harmonic division is a partitioning of a line segment into four points \(A, B, C, D\) such that the cross-ratio \( (A, B; C, D) \) equals -1. This special condition relates the perpendicular distances or vectors stemming from these points, even if they are complex values.
In the equation \( \frac{2}{z_1} = \frac{1}{z_2} + \frac{1}{z_3} \), the concept of harmonic division aids in understanding how these points are positioned to form a balanced or "harmonic" layout around a given circle or line.
In the equation \( \frac{2}{z_1} = \frac{1}{z_2} + \frac{1}{z_3} \), the concept of harmonic division aids in understanding how these points are positioned to form a balanced or "harmonic" layout around a given circle or line.
- Harmonic division in the complex plane highlights a specific symmetry and division of areas.
- This adds depth to interpreting equations involving more than one point or line segment.
Other exercises in this chapter
Problem 129
If \(P(x)\) and \(Q(x)\) are two polynomials such that \(f(x)=\) \(P\left(x^{3}\right)+x Q\left(x^{3}\right)\) is divisible by \(x^{2}+x+1\), then (A) \(P(x)\)
View solution Problem 132
\(e^{2 m i \cot ^{-1} p} \cdot\left(\frac{p i+1}{p i-1}\right)^{m}=\) (A) 0 (B) 1 (C) \(-1\) (D) None of these
View solution Problem 135
If \(|z-25 i| \leq 15\), then the least positive value of arg \(z\) is \(\begin{array}{ll}\text { (A) } \pi-\tan ^{-1} \frac{4}{3} & \text { (B) } \tan ^{-1} \f
View solution Problem 136
If \(|z-4+3 i| \leq 2\), then the least and the greatest values of \(|z|\) are (A) 3,7 (B) 4,7 (C) 3,9 (D) None of these
View solution