# What does the graph of the seventh root of unity look like on a unit circle like the one in the pictured attached?

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Alan N. Share
May 21, 2018

Another view
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#### Explanation:

It is interesting to note that the ${n}^{t h}$ roots of unity lie on the unit circle on the complex plane and form a regular polygon of with n sides.

The 7 roots of ${z}^{7} = 1 : z \in \mathbb{C}$ are shown below.

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Dean R. Share
May 21, 2018

#### Answer:

There are seven seventh roots of unity, ${e}^{\frac{2 \pi k i}{7}}$, all on the unit circle, $r = 1$ above. The first one is at $\theta = \frac{2 \pi}{7} = {360}^{\circ} / 7 = 51 \frac{3}{7} {\setminus}^{\circ}$, and there are others at $\frac{4 \pi}{7} , \frac{6 \pi}{7} , \frac{8 \pi}{7} , \frac{10 \pi}{7} , \frac{12 \pi}{7}$ and of course at $0$ radians, i.e. unity itself.

#### Explanation:

Euler's Identity to an even integer power of $2 k$ tell us

${\left({e}^{i \pi}\right)}^{2 k} = {\left(- 1\right)}^{2 k}$

${e}^{2 \pi k i} = 1$

Now we see

${1}^{\frac{1}{7}} = {\left({e}^{2 \pi k i}\right)}^{\frac{1}{7}} = {e}^{\frac{2 \pi k i}{7}}$

That's seven distinct seventh roots, given by any seven consecutive $k$s. (After that they repeat.)

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