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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!!

3 Answers 3

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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 $z=1$.

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    Why $z$ = 1???? – 2017-01-19
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    Because $z=1$ also satisfies $z^3=1$, hence is a third root of unity. We have 3 different third roots of unity, namely $z=1,e^{2\pi i/3},e^{4\pi i/3}$. – 2017-01-19
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    I really don't get it – 2017-01-19
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    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? – 2017-01-19
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    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 $z=1$. – 2017-01-19
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    Why $z = 1$ though? Why not take $z = 2$ or something? This may seem really easy, but I'm quite slow, sorry. – 2017-01-19
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    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 $z^3=1$. – 2017-01-19
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    Then what are the 3 cubes of unity? – 2017-01-19
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    Read the line after your "Why $z=1$?????" – 2017-01-19
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    Ahhhhh so why aren't we subbing in the other two possible values? – 2017-01-19
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    @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 $z^2+z+1=0$. – 2017-01-19
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$$ 0 = z^2 + z + 1 = (z+1/2)^2 - (1/2)^2 + 1 = (z+1/2)^2 + 3/4 \iff \\ z + \frac{1}{2} = \pm i \frac{\sqrt{3}}{2} \iff \\ z = \frac{-1 \pm i \sqrt{3}}{2} $$ By the way: $$ 1 + z + z^2 = \frac{z^3-1}{z-1} $$ as a finite term geometric series.

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    I'm getting different answers..? Apparently it's 3? – 2017-01-19
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    The third root, $z=1$, you already factored off. – 2017-01-19
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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 $$\sum_{j=0}^{N-1} z^j = 0.$$