# Find all z

1. Jul 21, 2006

### kezman

find all z in C that verify simultaneously:

$$|z|^2 + |z| - 2 = 0$$

$$z^8 + z^6 + z^4 + z^2 = 0$$

Of the first equation I know that 1 is a root.
And I think i is a solution for the second one.
But I cant find a convincing solution.

EDIT: sorry I forgot the = 0

Last edited: Jul 21, 2006
2. Jul 21, 2006

### arildno

Solve the first one as a quadratic equation in the REAL number |z|.
The solutions are |z|=-2, and |z|=1
Since the modulus cannot be negative, it follows that the solutions of eq.1 constitute the unit circle in the complex plane.

Your second "equation" is not an equation as it stands; something is missing.

Last edited: Jul 21, 2006
3. Jul 22, 2006

### interested_learner

It was a good hint that arildno gave you. But I have to respectfully disagree with his statement that the second equation is not an equation. It is too an equation. As much as the first.

A couple more hints. You are correct that i is a solution of the second equation. Can there be any real solutions? Is the solution set of the first equation real or imaginary or complex?

4. Jul 22, 2006

### d_leet

It wasn't an equation when Arildno posted that, the original poster initially forgot an equals sign and I presume edited it when Arildno pointed this out.

5. Jul 22, 2006

### arildno

We have:
$$z^{8}+z^{6}+z^{4}+z^{2}=z^{2}(z^{6}+z^{4}+z^{2}+1)=z^{2}(z^{4}+1)(z^{2}+1)$$

Last edited: Jul 22, 2006
6. Jul 22, 2006

### interested_learner

sorry arildno. I should have guessed something like that.

7. Jul 22, 2006

### arildno

Punishment:
Make 50 genuflections&Ave Marias.