An Interesting Complex Number Problem

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Discussion Overview

The discussion revolves around a complex number problem from the 2013 Panhellenic exams, specifically focusing on the properties of complex numbers and inequalities. Participants share their thoughts on the difficulty of the problem and the implications of solving it within the context of an exam setting.

Discussion Character

  • Homework-related
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Michael presents the problem and expresses concern about its difficulty and the overall exam structure.
  • Some participants argue that the problem is not particularly hard, suggesting that finding the points satisfying the first equation leads to a simple inequality.
  • One participant mentions a theorem that provides a weaker but sufficient result related to the inequality.
  • Another participant clarifies that the inequality is in the complex number system, not just real numbers, and discusses the geometric interpretation of the points satisfying the equation.
  • Concerns are raised about the overall exam workload, given that there are multiple complex problems to solve in a limited time.

Areas of Agreement / Disagreement

Participants express differing views on the difficulty of the problem, with some finding it manageable while others share concerns about its complexity and the exam's demands. The discussion remains unresolved regarding the overall assessment of the problem's difficulty.

Contextual Notes

There are limitations regarding the assumptions made about the complexity of the problem and the definitions of the terms involved, particularly in the context of complex numbers and inequalities.

Mkbul
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Hey all.

I am giving Panhellenic exams this year, and part of my preparation comes from solving previous' years questions. This problem was a complex number based one, in the 2013 Panhellenic exams. It goes as follows:

Consider 3 complex numbers, α01 and α2 which belong in the line given by the following equation:
(α-2)(α'-2)+|α-2|=2 , where α∈ℂ and α' is α's conjugate

If complex number v satisfies the following expression:

v32v21v+α0=0 , Prove that:

|v|<4

It is said that it was the hardest complex number problem they ever put in the exam, and it was totally against exam "spirit". By the way the exams are done in the last year of Senior High School.

I really want to hear your opinions on the matter. Thanks for reading.
Michael.

EDIT: I apologize if this is the wrong forum for this question. I've been a member for quite a while now but i haven't visited in many years. I didn't post this problem here to get it solved. I just wanted people to discuss it and give their thoughts about the exams. If i broke any guidelines please move my thread, i didn't mean to cause harm.
 
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I don't think it is that hard. You have to find the points that satisfy the first equation, then the rest is a simple inequality in the real numbers. There is also a theorem that gives a slightly weaker (but still sufficient) result compared to the inequality.
 
mfb said:
I don't think it is that hard. You have to find the points that satisfy the first equation, then the rest is a simple inequality in the real numbers.

Thanks for the reply, mfb. I did the first step and solved the complex equation, and the points that satisfy it belong in a circle with center K(2,0) and radius r=1.
Unfortunately that was the easy part. The inequality is not in the real numbers but in the complex number system. V a0, a1 and a2 are all complex numbers. I will try to solve this one now.

What worries me is that this was only one exercise, and the exam had 4. The proof of the fundamental theorem of calculus, this one, and two calculus problems with fuction-integrals. It seems quite a lot to solve in 3 hours.

By the way, i love your avatar. Isn't this an artists concept of that newly discovered planet with a massive ring system?
 
Mkbul said:
Thanks for the reply, mfb. I did the first step and solved the complex equation, and the points that satisfy it belong in a circle with center K(2,0) and radius r=1.
Right.
Mkbul said:
Unfortunately that was the easy part. The inequality is not in the real numbers but in the complex number system. V a0, a1 and a2 are all complex numbers.
Sure, but it is sufficient to consider their magnitude. For large v, the first term is very large - |v3| is at least 64 in the region where you have to exclude zeros...
Mkbul said:
By the way, i love your avatar. Isn't this an artists concept of that newly discovered planet with a massive ring system?
Yes it is :). I scaled it a bit to fit.
 
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