Let $A=\left \{ \left | z^{n}+\frac{1}{z^{n}} \right | |n \epsilon \mathbb {N} ,z\epsilon \mathbb{C},z^{4}+z^{3}+z^{2}+z^{1}+1=0\right \}$

Question

Let $A=\left \{ \left | z^{n}+\frac{1}{z^{n}} \right| \mid n \in \mathbb {N} ,z\in \mathbb{C},z^{4}+z^{3}+z^{2}+z^{1}+1=0\right \}$ Calculate the sum of the squares of the elements of A.

I know that you start from $z^{5}-1=0$, but I can't understand what $A$ means or which are the elements, can someone please explain the exercise

Welcome to MSE. Please read this text about how to ask a good question. — José Carlos Santos, Dec 27 '19 at 11:16
With $z=e^{ix}$ you get $z^n+z^{-n}=2\cos x$. There are only three elements in $A$. — Jyrki Lahtonen, Dec 27 '19 at 11:23
So i have to find every root with euler and then do the sum? sorry, I don't know that much — Not important, Dec 27 '19 at 11:28
Oh it's ok, but how do I apply the formula here, sorry, I don't know how to do by this method, can you please show me a little more detailed? — Not important, Dec 27 '19 at 12:38

lab bhattacharjee · Accepted Answer · 2019-12-27T11:56:47.910

2

As $z^5=1,z\ne1, $

so, the unique members of $A$ are $f(n)=z^n+\dfrac1{z^n}; n=1,2$ as $f(1)=f(5-1), f(2)=f(5-2)$

Like Quadratic substitution question: applying substitution $p=x+\frac1x$ to $2x^4+x^3-6x^2+x+2=0$ divide both sides by $z^2$

$$z^2+\dfrac1{z^2}+z+\dfrac1z+1=0$$

$$\left(z+\dfrac1z\right)^2+z+\dfrac1z-1=0$$

We need to find $$f^2(1)+f^2(2)$$ $$=z^2+\dfrac1{z^2}+2+z^4+\dfrac1{z^4}+2$$ $$=z^2+\dfrac1{z^2}+z+\dfrac1{z}+4=?$$ as $z^4=\dfrac1z$

edited Dec 27 '19 at 11:56

answered Dec 27 '19 at 11:42

lab bhattacharjee

274,582

So I don't understand why you took only n=1,2, since there are 4 solutions. Is it because polynomials have symetrical complex roots? – Not important Dec 27 '19 at 11:51
And I can't figure out what do you mean by f(1)=f(5-1), and f(2), why is it like that? – Not important Dec 27 '19 at 11:52
@Notimportant, As $z^4=\dfrac{z^5}z=\dfrac1z\implies f(4)=f(1)$ right? – lab bhattacharjee Dec 27 '19 at 11:56
Why isn't it f(2)=f(5-2)? – Not important Dec 27 '19 at 11:56
Yeah, i got the first part why f(1) – Not important Dec 27 '19 at 11:56
@Notimportant, Typo rectifed – lab bhattacharjee Dec 27 '19 at 11:56
Oh ok sorry, thanks for clearing that out, one more thing is why don't we take f(3) squared and f(4) also? – Not important Dec 27 '19 at 11:57
@Notimportant, I said "unique members". If you allow repetition, then we need to include them and consequently the resultant sum will be doubled, right? – lab bhattacharjee Dec 27 '19 at 11:59
So after that we use the equation above like t^2 + t- 1=0 and put it in t^2 +t +2 and we get 3? – Not important Dec 27 '19 at 12:04
@Notimportant, Yes, if you need the sum of unique elements – lab bhattacharjee Dec 27 '19 at 12:11
So if we get for all the elements, do we get 2*3+1? The answer is 7 so I suppose it is the 4 solutions and 1? – Not important Dec 27 '19 at 12:13
Oh the thing is it is modulus, like the sum of the modulus – Not important Dec 27 '19 at 12:21
Basically for the answer they took f(0) aswell – Not important Dec 27 '19 at 12:23
@Notimportant, But $$z=1$$ is not a root of the given equation, right? – lab bhattacharjee Dec 27 '19 at 12:24
Well yes, that is true, but i don't know why in my book they took f(0) as a solution too – Not important Dec 27 '19 at 12:26
like f(0)^2 +f(1)^2 + f(2)^2 – Not important Dec 27 '19 at 12:26
They also took f(0)^2= (1+1)^2 =4 and added to the rest – Not important Dec 27 '19 at 12:27

Let $A=\left \{ \left | z^{n}+\frac{1}{z^{n}} \right | |n \epsilon \mathbb {N} ,z\epsilon \mathbb{C},z^{4}+z^{3}+z^{2}+z^{1}+1=0\right \}$

1 Answers1