Find the sum of the squares of the elements of A

[CristianTheodor]

I have no idea how to start .


12. Fie A = {|z^n + 1/z^n| / n ∈ ℕ, z ∈ ℂ, z^4+z^3+z^2+z+1=0}. Să se determine suma pătratelor elementelor mulţimii A. (6 pct.)

a) 1; b) 9; c) 4; d) 10; e) 7; f) 5.

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12. Let A = {| z ^ n + 1 / z ^ n | / n ∈ ℕ, z ∈ ℂ, z ^ 4 + z ^ 3 + z ^ 2 + z + 1 = 0}. Determine the sum of the squares of the elements of the set A. (6 points)

a) 1; b) 9; c) 4; d) 10; e) 7; f) 5.

 

[Dr.Peterson]

Please at least translate the question into English [that is, don't just hide it in the title!], and tell us what you have learned that might be relevant.

Even when you "have no idea how to start", there is something you can try just to see if it will help at all. Please show us that work, which will help us see where you are starting from!

What do you know about the roots of a fourth-degree equation? Is there any way you can rewrite z^n + 1/z^n that might be related to that?

 

[CristianTheodor]

The problem asks for the sum of the squares of the elements of A.
I have found this

So the solutions for

are the same complex solutions for

,

We will have

and

but that's not very helpful

 

[Dr.Peterson]

That's a very interesting observation, and may well turn out to be very helpful! So the values of z are four of the five fifth roots of 1.

(By the way, I haven't worked out an answer yet; I'm just thinking along with you.)

I will tell you my next thought. Reading the definition of A carefully, I realize that, on the surface, it is an infinite set, since n can be any natural number; so how can you sum the squares of all these elements? That leads me to check whether {z^n | n in N, z^4 + z^3 + z^2 + z + 1 = 0} might be a nicer set than it appears to be. What do you find that this set is?

While we discuss this, it may also be helpful if you tell me the context of the question. I'm imagining that it may be more like a contest question than a problem from a course you are taking. Am I right?

 

[CristianTheodor]

Yes , this is a question from a contest . The admission contest for the University Politehnica of Bucharest .
Maybe it would be useful to write z^n in the trigonometric form .

 

[Dr.Peterson]

That's possible. Or you might see the idea I have in mind just because you have previously seen the trigonometric form! Give it a try.

Notice that "try" will be a key word throughout this process; it's an intentionally unfamiliar problem, to get you to put ideas together, and you can expect not to know ahead of time what will work best.

 

[CristianTheodor]

But if z^5=1 , i guess r = 1 so

 

[Dr.Peterson]

Okay. But you not only know what [MATH]r[/MATH] is, you know what [MATH]\theta[/MATH] is (up to a point).

 

[CristianTheodor]

 

[CristianTheodor]

there should be 4 solutions for theta

 

[pka]

there should be 4 solutions for theta

\(\displaystyle \theta=\left[\dfrac{k\pi i}{5}\right],~k=1,2,3,4\)

Above equation for \(\displaystyle \theta\) is incorrect (small typo). It should have been:

\(\displaystyle \theta=\left[\dfrac{k\pi}{5}\right],~k=1,2,3,4\)

 

[CristianTheodor]

\(\displaystyle \theta=\left[\dfrac{k\pi i}{5}\right],~k=1,2,3,4\)

oh , if θ = 0 , z = 1 , which we excluded . But what does i mean in your notation

 

[pka]

oh , if θ = 0 , z = 1 , which we excluded . But what does i mean in your notation

I am not at all sure about your notation either. Here \(\displaystyle i\) is the imaginary unit or \(\displaystyle i^2=-1\). You may use \(\displaystyle j\) instead.

P.S.It just occurred to me that you may not know that \(\displaystyle \exp\left(\frac{\pi i }{n}\right)=\cos\left(\frac{\pi }{n}\right)+i\sin\left(\frac{\pi }{n}\right)\)

 

[Dr.Peterson]

\(\displaystyle \theta=\left[\dfrac{k\pi i}{5}\right],~k=1,2,3,4\)

I'm not familiar with this notation meaning an angle, either! I would have just said [MATH]\theta=\dfrac{k\pi}{5},~k=1,2,3,4[/MATH].

 

[pka]

View attachment 15999View attachment 15999

I'm not familiar with this notation meaning an angle, either! I would have just said [MATH]\theta=\dfrac{k\pi}{5},~k=1,2,3,4[/MATH].

Prof Peterson, I am so use to writing complex numbers as \(\displaystyle z=r\exp(i\theta)\) that I automatically posted wth an \(\displaystyle i\) in the argument.
Here is a question for you: Do you understand the notation in the question as written? I have no idea why the \(\displaystyle n^{th}\) powers.
I think that the question implies some very interesting possibles, I just do not understand the Hungarian notation.
It seems that as fifth roots of unity, \(\displaystyle |z|=1\) then \(\displaystyle z=\exp\left(\frac{k\cdot i\cdot\pi}{5}\right),~k=0,1,2,3,4\)
I think that the question means to exclude \(\displaystyle k=0\).
Recall that \(\displaystyle \frac{1}{z}=\frac{\overline{~z~}}{|z|^2}\) Now if \(\displaystyle \bf\large |z|=1\) then \(\displaystyle \left|z^n+\frac{1}{z^n}\right|=\left|z^n+z^{-n}\right|=2\left|\cos(n\theta)\right|\) where \(\displaystyle \arg(z)=\theta\)

 

[Steven G]

there should be 4 solutions for theta

[MATH]5\theta[/MATH] does not only equal 0! [MATH]5\theta[/MATH] can equal 0, 2pi, 4pi, 6pi and 8pi. Then [MATH]\theta[/MATH] can equal 0, 2pi/5, 4pi/5, 6pi/5 and 8pi/5

 

[Deleted member 4993]

[MATH]5\theta[/MATH] does not only equal 0! [MATH]5\theta[/MATH] can equal 0, 2pi, 4pi, 6pi and 8pi. Then [MATH]\theta[/MATH] can equal 0, 2pi/5, 4pi/5, 6pi/5 and 8pi/5

For this problem, 0 is excluded.

Corner.... Corner.....

 

[Dr.Peterson]

I see no notation issues, but the problem is intentionally hard to think about.

As I see it, z itself can be one of four fifth roots; but z^n can be any of the fifth roots. Then 1/z^n is its conjugate. This gives only three numbers in the set A. We just have to add them up.

 

[Steven G]

For this problem, 0 is excluded.

Corner.... Corner.....

Yes, I knew that. I was just answering \(\displaystyle \sin(5\theta) = 1\). I wanted the OP to remove the zero root.