# Homework Help: Complex number equation

1. Sep 17, 2010

### zcserei

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

I've been recapitulating some lessons we learned in high school 2 years ago for the exams I need to take this year. There was this exercise I couldn't solve in a nice way.

z^2005+(1/z^2005) if we know that z^2+z+1=0

2. Relevant equations

I couldn't came up with a good solution, so I looked at the key at the end of the book, and it said that from z^2+z+1=0 => z^3=1, and then you do z^2005=(z^3)^668+z=1^668+z=1+z

3. The attempt at a solution

I calculated the root for the equation, and I found that it is (-1±i√3)/2. Now, that complex number definitely isn't 1 on the third power, and I can't take (1+z)+1/(1+z) too far either.
Where did I mess up? Is the key in my book correct?

Thank you for your answers.

2. Sep 17, 2010

### hunt_mat

Okay, they get z^{3} from:
$$z^{2}+z+1=0\Rightarrow z^{3}+z^{2}+z=0\Rightarrow z^{3}-1=0\Rightarrow z^{3}=1$$
Then
$$z^{2004}=(z^{3})^{668}=1^{668}=1$$
From here it is clear what to do.

3. Sep 17, 2010

### statdad

"Now, that complex number definitely isn't 1 on the third power"

Are you sure about that? Check it again.

As a different approach: remember that

$$A^3 -1 = (A-1)(A^2 + A + 1)$$

so, given $z^2 + z + 1 = 0$

\begin{align*} z^2 + z + 1 & = 0 \\ \frac{z^3 - 1}{z-1} & = 0 \\ z^3 - 1 & = 0 \\ z^3 & = 1 \end{align*}

I can eliminate the denominator since it is obvious that z is not equal to 1. This gets you to the same statement about z as direct calculation, but without having to work with complex numbers (not that that is a huge problem). Once here, the rest of the solution goes as you note.

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