# Adding primitive roots of unity

1. Aug 6, 2013

### Artusartos

1. The problem statement, all variables and given/known data

I was trying to figure out whether or not $\zeta_5 + \zeta_5^2$ and $\zeta_5^2 + \zeta_5^3$ were complex (where $\zeta_5$ is the fifth primitive root of unity).

2. Relevant equations

3. The attempt at a solution

$\zeta_5 + \zeta_5^2 = \cos(2\pi/5) + i\sin(2\pi/5) + (\cos(2\pi/5) + i\sin(2\pi/5))^2 = \cos(2\pi/5) + i\sin(2\pi/5) + \cos(4\pi/5) + i\sin(4\pi/5)$.

Since $i\sin(2\pi/5)$ and $i\sin(4\pi/5)$ do not cancel out each other, $\zeta_5 + \zeta_5^2$ must be complex, right?

$\zeta_5^2 + \zeta_5^3 = (\cos(2\pi/5) + i\sin(2\pi/5))^2 + (\cos(2\pi/5) + i\sin(2\pi/5))^3 = \cos(4\pi/5) + i\sin(4\pi/5) + \cos(6\pi/5) + i\sin(6\pi/5)$

But again, the complex numbers don't cancel out each other, right?

2. Aug 6, 2013

### tiny-tim

draw an argand diagram!!

(ie mark them out as vectors on a unit circle)

3. Aug 6, 2013

### haruspex

$\bar{\zeta_n}\zeta_n = 1$, so $\bar{\zeta_n} = 1/\zeta_n$. Suppose $\zeta_n^r+\zeta_n^s$ is real. What can you deduce about r+s?

4. Aug 7, 2013

### Artusartos

When I draw a diagram, I need to go three times the angle of $\zeta_5$ for $\zeta_5+\zeta_5^2$, and so I end up in quadrant three, right? So it is not real.

But for $\zeta_5^2 + \zeta_5^3$ or $\zeta_5+\zeta_5^4$ we need to go 2+3=1+4=5 times the angle of $\zeta_5$, and so we will be going all the way and end up back in 1, right? So it must be real.

Do you think that's correct?

Last edited: Aug 7, 2013
5. Aug 7, 2013

### tiny-tim

Hi Artusartos!

I'm not sure what you're doing there (looks more like multiplication than addition ).

Draw the vectors OB and OC (for $\zeta_5^2$ and $\zeta_5^3$), and then use the parallelogram law to add them …

what do you see?

6. Aug 7, 2013

### Artusartos

Well, that looks like it's in the second quaderant, right?

7. Aug 7, 2013

### tiny-tim

do you know the parallelogram law?

you should be seeing two vectors OB and OC, and a parallelogram with BOC on three sides​

8. Aug 7, 2013

### Artusartos

Oh I'm sorry. I was looking at $\zeta_5$ instead of $\zeta_5^2$ and $\zeta_5^2$ instead of $\zeta_5^3$...

So if I look at $\zeta_5^2 + \zeta_5^3$, then it lies on the x-axis and so the sine is zero and it must be real, right?

9. Aug 7, 2013

### tiny-tim

yup!

(though it worries me that you add "the sine is zero" …

i see what you mean, but it really has nothing to do with the problem, does it? )​

10. Aug 7, 2013

### Artusartos

Thanks a lot.

But I'm not sure why it doesn't have to do with this problem. We want sine to be zero, because it is the complex term in $\cos(2\pi/n) + i\sin(2\pi/n)$, right?

11. Aug 7, 2013

### tiny-tim

First, you need an "r" there: $r[\cos(2\pi/n) + i\sin(2\pi/n)]$.

Second, it's much simpler to say that a real number is a complex number x + iy with y = 0 …

and the x + iy form is perfect for addition (of two complex numbers), while the re (polar) form is pretty much useless!

12. Aug 7, 2013

### haruspex

Seems you didn't like my hint, so I'll take it a bit further:

If $\zeta_n^r+\zeta_n^s$ is real then $\zeta_n^r+\zeta_n^s = \bar{\zeta_n^r}+\bar{\zeta_n^s} = 1/\zeta_n^r+1/\zeta_n^s$. Multiplying out, $\zeta_n^{2r+s}+\zeta_n^{r+2s} = \zeta_n^r+\zeta_n^s = \zeta_n^{r+s}(\zeta_n^r+\zeta_n^s)$. Can you take it from there?

13. Aug 7, 2013

### Artusartos

Thanks. So if $\zeta_n^r+\zeta_n^s$ is real, then $r+s$ must be a multiple of $n$.

So now I should prove that $\zeta_n^r+\zeta_n^s = \bar{\zeta_n^r}+\bar{\zeta_n^s}$, right?

But shouldn't we be proving the converse instead? That if $r+s$ is a multiple of $n$, then $\zeta_n^r+\zeta_n^s$ must be real?

14. Aug 7, 2013

### haruspex

There is another solution to the equation.
Not sure what you mean. If the sum is real then $\zeta_n^r+\zeta_n^s = \overline{\zeta_n^r+\zeta_n^s}$, from which the above follows quickly.
It depends. You asked whether a particular sum of powers is real. We've shown that if it is then a certain relationship holds. If the powers do not satisfy that relationship then the sum is not real. If the powers do satisfy it then, yes, there's a bit more work to do, but it's basically running the same argument backwards.