/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 1 Let \(z_{1}, z_{2}, \ldots, z_{n... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 1

Let \(z_{1}, z_{2}, \ldots, z_{n}\) be distinct complex numbers such that \(\left|z_{1}\right|=\) \(\left|z_{2}\right|=\cdots=\left|z_{n}\right| .\) Prove that $$ \sum_{1 \leq i

Short Answer

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Based on the step-by-step solution above, write a short answer addressing the problem: Given a set of distinct complex numbers with the same magnitude, the lower bound for the sum of squares of the magnitudes of the expression \(\frac{z_i + z_j}{z_i - z_j}\) for \(1\leq i < j \leq n\) is \(\frac{(n-1)(n-2)}{2}\).

Step by step solution

01

Convert complex numbers to polar form

Since \(|z_1|= |z_2| = \cdots = |z_n|\), let's define \(R = |z_i|\) for all \(i\). We can rewrite each complex number \(z_i\) in polar form as \(z_i=R(\cos(\theta_i)+i\sin(\theta_i))\), where \(\theta_i\) is the angle (in radians) between the positive real axis and \(z_i\).
02

Rewrite the expression

Let us simplify the expression \(\frac{z_{i} + z_{j}}{z_{i} - z_{j}}\): $$ \frac{R(\cos(\theta_i)+i\sin(\theta_i))+R(\cos(\theta_j)+i\sin(\theta_j))}{R(\cos(\theta_i)+i\sin(\theta_i))-R(\cos(\theta_j)+i\sin(\theta_j))} $$ Now we can factor out the \(R\): $$ \frac{(\cos(\theta_i)+i\sin(\theta_i))+(\cos(\theta_j)+i\sin(\theta_j))}{(\cos(\theta_i)+i\sin(\theta_i))-(\cos(\theta_j)+i\sin(\theta_j))} $$ To further simplify, let \(a_{ij} = \cos(\theta_i) - \cos(\theta_j)\) and \(b_{ij} = \sin(\theta_i) - \sin(\theta_j)\). We can rewrite the expression as: $$ \frac{(\cos(\theta_i)+i\sin(\theta_i)) + (\cos(\theta_j)+i\sin(\theta_j))}{a_{ij} + ib_{ij}} $$
03

Calculate the magnitude square

To find the magnitude square (\(|\)expression\(|^2\)) of the expression, we will multiply the expression by its conjugate: $$ \left|\frac{(\cos(\theta_i)+i\sin(\theta_i)) + (\cos(\theta_j)+i\sin(\theta_j))}{a_{ij} + ib_{ij}}\right|^2 = \frac{((\cos(\theta_i) + \cos(\theta_j))^2 + (\sin(\theta_i) + \sin(\theta_j))^2}{a_{ij}^2 + b_{ij}^2} $$
04

Apply the Cauchy-Schwarz inequality

The Cauchy-Schwarz inequality states that for any real sequences \(a_1, a_2, \ldots, a_n\) and \(b_1, b_2, \ldots, b_n\): $$ \left(\sum_{k = 1}^n a_kb_k\right)^2 \leq \left(\sum_{k = 1}^n a_k^2\right)\left(\sum_{k = 1}^n b_k^2\right) $$ Apply the inequality to the sequences \(\left\{(\cos(\theta_i) + \cos(\theta_j))^2\right\}\) and \(\left\{\frac{a_{ij}^2 + b_{ij}^2}{(\sin(\theta_i) + \sin(\theta_j))^2}\right\}\), we get: $$ \left(\sum_{1 \leq i
05

Calculate the sums

Observe that \(\sum_{1 \leq i
06

Find the lower bound

It can be proven that the sum of squares of sines and cosines can be lower bounded by \(\sum_{i=1}^n \cos^2(\theta_i) \geq \frac{n}{2}\) and \(\sum_{i=1}^n \sin^2(\theta_i) \geq \frac{n}{2}\). Using these bounds, we get: $$ \left(\sum_{1 \leq i

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Most popular questions from this chapter

Let \(z_{1}, z_{2}, z_{3}\) be complex numbers such that (1) \(\left|z_{1}\right|=\left|z_{2}\right|=\left|z_{3}\right|=1\); (2) \(z_{1}+z_{2}+z_{3} \neq 0\) (3) \(z_{1}^{2}+z_{2}^{2}+z_{3}^{2}=0\). Prove that for all integers \(n \geq 2\), $$ \left|z_{1}^{n}+z_{2}^{n}+z_{3}^{n}\right| \in\\{0,1,2,3\\} . $$

Let \(U_{n}\) be the set of nth roots of unity. Prove that $$ \prod_{e \in U_{n}}\left(\varepsilon+\frac{1}{\varepsilon}\right)=\left\\{\begin{array}{ll} 0 & \text { if } n \equiv 0(\text { mod } 4), \\ 2, & \text { if } n \equiv 1(\bmod 2), \\ -4, & \text { if } n \equiv 2(\bmod 4), \\ 2, & \text { if } n \equiv 3(\bmod 4) . \end{array}\right. $$

On the sides of a convex quadrilateral \(A B C D\), equilateral triangles \(A B M\) and \(C D P\) are drawn external to the figure, and equilateral triangles \(B C N\) and \(A D Q\) are drawn internal to the figure. Describe the shape of the quadrilateral \(M N P Q\).

On each side of a parallelogram, a square is drawn external to the figure. Prove that the centers of the squares are the vertices of another square.

Let \(A, B, C\) be three consecutive vertices of a regular polygon and let us consider a point \(M\) on the major arc \(A C\) of the circumcircle. Prove that $$ M A \cdot M C=M B^{2}-A B^{2} $$
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