Finding two possible values of $z^2 + z + 1$ given that $z$ is one of the three cube roots of unity.

Question

Factorise $z^3 - 1 $. If $z$ is one of the three cube roots of unity, find the two possible values of $z^2 + z + 1$.

Factorising gives you :

$(z - 1)(z^2 + z + 1) = 0$ since $z$ is one of the three cube roots of unity.

z is complex so $z \neq 1$ so $z-1 \neq 0$

Hence, $z^2 + z + 1 = 0$

Where do I go from here?

Any help is appreciated!!

Dietrich Burde · Answer 1 · 2017-01-19T20:07:50.330

3

You are done. The two possible values of $z^2+z+1$ are $0$ for $z$ a third root of unity different from $1$, as you have computed, and $1+1+1=3$ for $z=1$.

edited Jan 19 '17 at 20:07

answered Jan 19 '17 at 19:55

Dietrich Burde

130,978

Why $z$ = 1???? – TripleA Jan 19 '17 at 19:56
1

Because $z=1$ also satisfies $z^3=1$, hence is a third root of unity. We have 3 different third roots of unity, namely $z=1,e^{2\pi i/3},e^{4\pi i/3}$. – Dietrich Burde Jan 19 '17 at 19:57
I really don't get it – TripleA Jan 19 '17 at 20:04
I don't understand why $z=1$ is able to be substituted in, for what reason? Also the answer below gives you a complex root? – TripleA Jan 19 '17 at 20:12
The questions says." If $z$ is one of the three cube roots of unity (so $z=1$ for example), find the two possible values of $z^2 + z + 1$." Now do this for $z=1$. – Dietrich Burde Jan 19 '17 at 20:14
Why $z = 1$ though? Why not take $z = 2$ or something? This may seem really easy, but I'm quite slow, sorry. – TripleA Jan 19 '17 at 20:16
Dear TripleA: $z=2$ is not a third cube of unity, because its third power is not $1$. Rather we have $2^3=8\neq 1$. A cube root of unity is a number $z$ with $z^3=1$. – Dietrich Burde Jan 19 '17 at 20:17
Then what are the 3 cubes of unity? – TripleA Jan 19 '17 at 20:18
Read the line after your "Why $z=1$?????" – Dietrich Burde Jan 19 '17 at 20:18
Ahhhhh so why aren't we subbing in the other two possible values? – TripleA Jan 19 '17 at 20:20
@TripleA You don't need to substitute, because you know that $(z-1)(z^2+z+1)=0$ for the three roots. Therefore either $z-1=0$ or $z^2+z+1=0$. In other words, the two roots that are not $z=1$ must satisfy $z^2+z+1=0$. – Théophile Jan 19 '17 at 20:41

mvw · Answer 2 · 2017-01-19T20:09:56.403

0

$$ 0 = z^2 + z + 1 = (z+1/2)^2 - (1/2)^2 + 1 = (z+1/2)^2 + 3/4 \iff \\ z + \frac{1}{2} = \pm i \frac{\sqrt{3}}{2} \iff \\ z = \frac{-1 \pm i \sqrt{3}}{2} $$ By the way: $$ 1 + z + z^2 = \frac{z^3-1}{z-1} $$ as a finite term geometric series.

edited Jan 19 '17 at 20:09

answered Jan 19 '17 at 20:01

mvw

34,562

I'm getting different answers..? Apparently it's 3? – TripleA Jan 19 '17 at 20:05
The third root, $z=1$, you already factored off. – mvw Jan 19 '17 at 20:06

mathisfun · Answer 3 · 2017-01-19T20:35:53.823

0

More generally, you have $$(z-1)\sum_{j=0}^{N-1} z^j = z^N-1,$$ so for $z=1$ you have $\sum_{j=0}^{N-1} z^j = N$ and if $z\neq 1$ you get $$\sum_{j=0}^{N-1} z^j = \frac{z^N-1}{z-1}.$$ Hence, if $z$ is any $N^{th}$ root of unity not equal to 1, we have $$\sum_{j=0}^{N-1} z^j = 0.$$

edited Jan 19 '17 at 20:35

answered Jan 19 '17 at 20:21

mathisfun

376

Finding two possible values of $z^2 + z + 1$ given that $z$ is one of the three cube roots of unity.

3 Answers3