/*! 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

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

Let \(A B C\) be a triangle such that \(\widehat{A C B}=2 \widehat{A B C} .\) Let \(D\) be the point on the side \(B C\) such that \(C D=2 B D .\) The segment \(A D\) is extended to \(E\) so that \(A D=D E .\) Prove that $$ \widehat{E C B}+180^{\circ}=2 \widehat{E B C} $$

Let \(A B C\) be a triangle, \(H\) its orthocenter, \(O\) its circumcenter, and \(R\) its circumradius. Let \(D\) be the reflection of \(A\) across \(B C\), let \(E\) be that of \(B\) across \(C A\), and \(F\) that of \(C\) across \(A B .\) Prove that \(D, E\), and \(F\) are collinear if and only if \(O H=2 R\). (39th IMO - Shortlist)

Let \(A B C D\) be a square with center \(O\) and let \(M, N\) be the midpoints of segments \(B O, C D\) respectively. Prove that triangle AMN is an isosceles right triangle.

Let \(z_{1}, z_{2}, \ldots, z_{n}\) be distinct complex numbers with the same modulus such that $$ z_{3} z_{4} \ldots z_{n-1} z_{n}+z_{1} z_{4} \ldots z_{n-1} z_{n}+\cdots+z_{1} z_{2} \ldots z_{n-2}=0 $$ Prove that $$ z_{1} z_{2}+z_{2} z_{3}+\cdots+z_{n-1} z_{n}=0 . $$
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