Solving for z in C: |z|^2 + |z| - 2 = 0 and z^8 + z^6 + z^4 + z^2

[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 can't find a convincing solution.

EDIT: sorry I forgot the = 0

 

[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.

 

[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?

 

[d_leet]

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.

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.

 

[arildno]

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

 

[interested_learner]

sorry arildno. I should have guessed something like that.

 

[arildno]

sorry arildno. I should have guessed something like that.

Punishment:
Make 50 genuflections&Ave Marias.