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If \(z_{1} \neq-z_{2}\) and \(\left|z_{1}+z_{2}\right|=\left|\frac{1}{z_{1}}+\frac{1}{z_{2}}\right|\), then (A) at least one of \(z_{1}, z_{2}\) is unimodular (B) \(z_{1} \times z_{2}\) is unimodular (C) both \(z_{1}, z_{2}\) are unimodular (D) None of these

Short Answer

Expert verified
Option (B): \(z_{1} \times z_{2}\) is unimodular.

Step by step solution

01

Understand Given Equation

We start with the equation \( \left|z_{1}+z_{2}\right|=\left|\frac{1}{z_{1}}+\frac{1}{z_{2}}\right| \). This can be rewritten as \( \left| z_1 + z_2 \right| = \left| \frac{z_1 + z_2}{z_1 z_2} \right| \).
02

Simplify the Absolute Value Equation

By simplifying \( \left| z_1 + z_2 \right| = \frac{\left| z_1 + z_2 \right|}{\left| z_1 z_2 \right|} \), we see that it simplifies to \( \left| z_1 z_2 \right| = 1 \). This implies that \( z_1 \times z_2 \) is unimodular.

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Key Concepts

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

Unimodular Complex Numbers
Unimodular complex numbers are an interesting type of complex number. They are defined as numbers whose absolute value is equal to 1. In simpler terms, if we take a complex number, say \( z \), and calculate its absolute value which is represented as \( |z| \), a unimodular complex number satisfies the equation \( |z| = 1 \).
Examples of unimodular complex numbers include \( 1, -1, i, \text{and} -i \). These are points located on the unit circle of the complex plane, meaning they are a specific distance (1 unit) away from the origin.
Unimodular numbers have a special property in multiplication. When two complex numbers multiply together to form a unimodular number, it usually hints at a unique and symmetrical relationship between the numbers, as seen in problems involving complex solutions.
Absolute Value of Complex Numbers
The absolute value of a complex number provides a measure of its magnitude or distance from the origin on the complex plane. For a complex number \( z = a + bi \), where \( a \) and \( b \) are real numbers, the absolute value \( |z| \) is calculated using the formula: \[ |z| = \sqrt{a^2 + b^2} \]
This is similar to the Pythagorean theorem where \( a \) and \( b \) are sides of a right triangle, and \( |z| \) is the hypotenuse.
Knowing the absolute value helps determine how far a number is from zero and can be useful in many operations, such as when deciding if two sum or product results in unimodular behavior.
In the given exercise, the condition \( \left|z_{1} + z_{2}\right| = \left| \frac{1}{z_{1}} + \frac{1}{z_{2}} \right| \) becomes simplified to find \( |z_1 z_2| = 1 \). This indicates that the product \( z_1 \times z_2 \) is unimodular, deepening the importance of understanding absolute values in these problems.
Complex Number Multiplication
Complex number multiplication involves both magnitude and angle considerations. When two complex numbers \( z_1 \) and \( z_2 \) are multiplied, their magnitudes multiply, and their angles add up.
If \( z_1 = a + bi \) and \( z_2 = c + di \), their product is:\[ z_1 \times z_2 = (ac - bd) + (ad + bc)i \]
This operation not only combines the real and imaginary parts separately but also impacts the final result's magnitude and angle.
In the context of unimodular numbers, this product becomes particularly interesting. If the product \( |z_1 z_2| = 1 \), it implies that they have perfect balancing magnitudes that together, their unit-circle properties bring about a simplification feature, reflecting the unit modulus trait.
Understanding complex multiplication is crucial in identifying harmonious relationships in complex number operations which are frequent in many mathematical and engineering fields.

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Most popular questions from this chapter

If \(f(x)\) and \(g(x)\) are two polynomials such that the polynomial \(h(x)=x f\left(x^{3}\right)+x^{2} g\left(x^{6}\right)\) is divisible by \(x^{2}+x+1\), then (A) \(f(1)=g(1)\) (B) \(f(1)=-g(1)\) (C) \(h(1)=0\) (D) \(h(-1)=0\)

If \(1, \alpha, \alpha^{2}, \ldots, \alpha^{n}{ }^{1}\) are the \(n n\)th roots of unity then \(\sum_{i=1}^{n-1} \frac{1}{2-\alpha^{i}}\) is equal to (A) \(\frac{(n-2) 2^{n-1}+1}{2^{n}-1}\) (B) \((n-2) \times 2^{n}\) (C) \(\frac{(n-2) \cdot 2^{n-1}}{2^{n}-1}\) (D) None of these

If \(z\) is a complex number of unit modulus and argument \(\theta\), then \(\left(\frac{1+z}{1+\bar{z}}\right)\) equals \(\quad\) [2013] (A) \(\frac{\pi}{2}-\theta\) (B) \(\theta\) (C) \(\pi-\theta\) (D) \(-\theta\)

Assertion: If the area of the triangle on the argand plane formed by the complex numbers \(-z, i z, z-i z\) is 600 square units, then \(|z|=20\) Reason: Area of the triangle on the argand plane formed by the complex numbers \(-z, i z, z-i z\) is \(\frac{3}{2}\left|z^{2}\right|\)

If \(1, \omega, \omega^{2}, \ldots, \omega^{n-1}\) are the \(n, n\)th roots of unity, then \((1-\omega)(1-\omega)^{2} \ldots\left(1-\omega^{n-1}\right)\) is equal to (A) 0 (B) 1 (C) \(n\) (D) \(n^{2}\)Passage 3 Solution of Equations Certain types of algebraic equations can be solved with the help of De'Moivre's theorem Equations of the type \(p z^{n}+q=0:\) If \(p z^{n}+q=0\), where \(p\) and \(q\) are complex numbers, and \(p \neq 0\), then $$ z^{n}=-q / p $$ The roots of the given equation are, therefore, the \(n\) values of \((-q / p)^{1 / n}\). For example, consider the equation \(z^{7}+1=0\).\(z^{7}+1=0 \Rightarrow z^{7}=-1=\) cis \((2 p+1) \pi\), where \(p\) is an integer. Therefore, \(z=\operatorname{cis}[(2 p+1) \pi 7], p=0,1, \ldots, 6\) On putting \(p=0,1,2,3,4,5,6\), the roots are seen to be \(\cos (\pi 7) \pm i \sin (\pi / 7), \cos (3 \pi / 7) \pm i \sin \left(3 \pi^{\prime} 7\right), \cos\) \((5 \pi / 7) \pm i \sin (5 \pi / 7),-1 .\) Equations of the type \(p z^{2 n}+q z^{n}+r=0\), where \(p, q\) and \(r\) are complex numbers and \(p \neq 0\). $$ z^{n}=\frac{-q \pm \sqrt{q^{2}-4 p r}}{2 p} $$ Denoting these values of \(z^{n}\) by \(\alpha\) and \(\beta\), we have two equations \(z^{n}=\alpha\) and \(z^{n}=\beta\), each of which can be solved by the method given in the above example. Equations of the type \(a(p z+q)^{n}+b(r z+s)^{n}=0:\) The substitution \(\frac{p z+q}{n+s}=w\) reduces the given equation to the form $$ a w^{n}+b=0 $$ which can be solved by the method given above. If \(w_{k}\) be a root of the equation (i), the corresponding root \(z_{k}\) of the given equation is obtained by solving the equation $$ \frac{p z_{k}+q}{r z_{k}+s}=w_{k} $$

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