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Suppose $a$ is a complex number such that $$a^2+a+\frac 1a +\frac 1{a^2} +1 =0$$ If $m$ is a positive integer find value of $$a^{2m}+a^m+\frac 1{a^m} + \frac 1{a^{2m}}$$

Setting $a=|a|e^{i\theta}$ we get $$2|a| \cos \theta + 2 |a|^2 \cos 2\theta+1=0$$ I could not get anything from this. Any ideas? Thanks.

Navin
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1 Answers1

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Here's another method that avoids (explicit) trigonometry:

Hint Multiplying the equation through by $a^2$ gives $$a^4 + a^3 + a^2 + a + 1 = 0 ,$$ which we can rewrite as $$\frac{a^5 - 1}{a - 1} = 0.$$

Travis Willse
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  • Another version of this is to multiply both sides of the original equation by $a-1$ and note that it simplifies to $a^3-a^{-2}=0$. – Semiclassical May 10 '17 at 20:49
  • @Travis Perhaps a silly question but what is the reasoning that allows the quartic function to simplify to that fraction? I've tried looking it up but I'm having no luck. – WaveX May 10 '17 at 20:56
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    This is usual formula for the sum of a geometric sequence. To see how it works in this case, multiply $a^4 + \cdots + a + 1$ by $a - 1$ and see what happens. – Travis Willse May 10 '17 at 21:03
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    Ahh. Okay. Now I understand. :) – WaveX May 10 '17 at 21:06