How can I find fifth root of unity?

Question

I have no idea to do this question, how can I find the fifth root of unity?

Question :

Find all the distinct fifth root of unity. Let $\alpha$ be a fifth root of unity such that $\alpha \ne 1$.

Prove that $1 + \alpha^2 + \alpha^3 + \alpha^4 = 0$.

Your support is much appreciated! thank you.

Do you know how to solve an equation of the form $z^n=r cis(\theta)$ or $z^n=r e^{i\theta}$? — Galc127, Oct 22 '14 at 12:04
By definition - if $z$ is a n-th root of unity, then $z^n=1$. Now, you should know that the solution of an equation of the form $z^n=re^{i \theta}$ is $\displaystyle z_k=\sqrt[n]{r}e^{\frac{i\theta+2\pi k}{n}}$. Good luck. — Galc127, Oct 22 '14 at 12:13
You're gooing to have a hard time proving $1+a^2+a^3+a^4=0$. You'll have better luck with $1+a+a^2+a^3+a^4=0$. — Gerry Myerson, Oct 22 '14 at 12:16

score 2 · Answer 1 · answered Oct 22 '14 at 12:10

2

Here's one: $$ \alpha = \cos(2\pi/5) + \mathbf i \sin (2\pi/5). $$

Now that you know one, can you find a second one, and indeed, the rest of them?

answered Oct 22 '14 at 12:10

John Hughes

1

It is a more interesting exercise, perhaps, to find exact values of the components. – Travis Willse Oct 22 '14 at 12:12
Agreed. (Of course, $\cos(2\pi/5)$ is an exact value.) :) Also: this person seems to be wondering about complex numbers. His/her trig skills might be sufficient to determine the value of the trig functions in, say, surd form. I didn't want to presume that they were not. – John Hughes Oct 22 '14 at 12:14
(Yes, of course I mean radical expressions not involving trigonometric functions.) And yes, I agree, I meant not to suggest you'd misinterpreted the thrust of the problem, but just to suggest an interesting follow-up (in fact, one of my favorite precalculus problems) for the inclined reader. – Travis Willse Oct 22 '14 at 12:17
Ah...nice point! – John Hughes Oct 22 '14 at 12:18
1

As vertices of a regular pentagon, the fifth roots of unity in the complex plane can be found by straight-edge and compass construction. – hardmath Oct 22 '14 at 12:23

score 0 · Answer 2 · answered Jul 18 '21 at 17:29

0

If you can construct, which isn't that difficult with Pythagoras:

$$\cos(36°) = \frac{\sqrt{5} +1}{4}$$

You are done, since $2*36° = 72°$.

answered Jul 18 '21 at 17:29

score 0 · Answer 3 · answered Oct 22 '14 at 12:12

0

Hint: $~a^5=1\iff a^5-1=0.~$ Now, what formula do you know for $a^n-1$ ? :-$)$

answered Oct 22 '14 at 12:12

Lucian

3 Answers3