The nth roots of a complex number

Question

Let's consider this following equation in $ \mathbb C $ :

$$ (E) : z^6 + z^5 + z^4 + z^3 + z^2 + z +1=0 $$ for every $ K \in \{0,1,2,3,4,5,6\} $ : $$Z_k= e^{i \frac {2k\pi}7 } $$

1) Show that: $ z_1,z_2 ,...$ and $z_6 $ are solutions of (E)

now I tried to answer this question by demonstrating in every situation that it is equal to Z I tried to show that $z^1 = Z_1$ and $z^2 = Z_2$and $z^3 = Z_3$ ....furthermore but I really don't think my first method is right. Unfortunately, I failed to recognize the answer and I'm not asking for it please don't downvote at least I tried I just need an idea where to begin since the method I thought of is false.

your polynomial E, what happens when you multiply that by $(z-1) ; ?$ — Will Jagy, Dec 30 '17 at 23:18
You should know that $Z_k$ is a specific root of a common number — Triatticus, Dec 30 '17 at 23:24
Geometric sum: $E=\frac{z^7-1}{z-1}$. $E=0$ iff $z^7=1$ and $z\ne 1$. — A.Γ., Dec 30 '17 at 23:33
$Z_k$ is a 7th root of unity, so $Z_k^7=1$ and as such $Z_k^7-1=0$ — Triatticus, Dec 30 '17 at 23:35
A follow up, since we know that Z is a 7th root of unity and $Z \neq 1$ for $k\neq 0$. Since $Z_k^7-1=(Z_k-1)\cdot E=0$ what can you conclude about E for the stated values of $k$ — Triatticus, Dec 31 '17 at 00:46

score 2 · Accepted Answer · answered Dec 31 '17 at 01:36

Hint: Consider multiplying both sides of $(E)$ by $(z-1)$ and how $Z_k$ might satisfy the new equation

Solution:

$$z^6 + z^5 + z^4 + z^3 + z^2 +z +1=0$$ $$(z-1)(z^6 + z^5 + z^4 + z^3 + z^2 +z +1)=0 \space \text{ (1)}$$ $$z^7-1=0 \space \text{(2)}$$ Substituting $z = Z_k = e^{i \frac{2k \pi}{7}}$ $$ \left( e^{i \frac{2k \pi}{7}} \right)^7 -1 = e^{i 2 k \pi} - 1 = 1 - 1 = 0 \space \text{(3)}$$ Hence, as $Z_k$ satisfies $z^7 - 1=0$, and does not satisfy $z-1=0$ (which we multiplied the original equation by), it thus is a solution to $(E)$, for $k \in \{0,1,2,3,4,5,6\}$.

Explanation:

$(1)$ Multiplied both sides by $(z-1)$

$(2)$ This is a common result progressing from the second to third line

$(3)$ As $e^{i2k\pi} = 1$ for $k \in \mathbb{Z}$

Yes thank you so much for the explanation I was already in result (2) but that's where I got stuck. — Manuela NIEVES, Dec 31 '17 at 19:09

The nth roots of a complex number

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