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Each of the $m$ individual roots of complex numbers can be multiplied with itself $m$ times, where $m =$ number of roots, to yield the original number. But, does this have any geometric significance. I am having a simple example of $z^6 = 1$, which has 6 roots as below:
(i) $1$
(ii) $\frac{1}{2} + i\frac{\sqrt{3}}{2}$
(iii) $-\frac{1}{2} + i\frac{\sqrt{3}}{2}$
(iv) $- 1$
(v) $\frac{1}{2} - i\frac{\sqrt{3}}{2}$
(vi) $-\frac{1}{2} - i\frac{\sqrt{3}}{2}$

Each of the $6$ roots can be multiplied by itself $6$ times to yield the original complex number. Let the 3rd root be taken and binomial expansion applied as: $(-\frac{1}{2} + i\frac{\sqrt{3}}{2})^6$ leading to following terms:
$(-\frac{1}{2})^6, 6.(-\frac{1}{2})^5.(\frac{\sqrt{3}}{2}i)^1, 15.(-\frac{1}{2})^4.(\frac{\sqrt{3}}{2}i)^2, 20.(-\frac{1}{2})^3.(\frac{\sqrt{3}}{2}i)^3, 15.(-\frac{1}{2})^2.(\frac{\sqrt{3}}{2}i)^4, 6.(-\frac{1}{2})^1.(\frac{\sqrt{3}}{2}i)^5, (\frac{\sqrt{3}}{2}i)^6$

Taking the real terms (i.e. alternate, or those having even power of imaginary part) only as the sum of real terms should be $=1$, while the imaginary terms should cancel out each other :

$\implies \frac{1}{64}, -\frac{45}{64}, \frac{135}{64}, -\frac{27}{64} $

But, the sum of real terms is not $=1$ (as it should be), and the geometrical significance part is also there to be asked for.

Addendum Sorry, the sum is $=1$. Please help in the geometrical significance part.

jiten
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    The sixth roots of 1 form a regular hexagon centered at 0. In general the $n$th roots of a general complex number form a regular $n$-gon centered at 0. The size and rotation of the $n$-gon depend on the location of the original complex number. – jgon Jan 11 '18 at 06:01
  • @jgon Thanks for the geometrical significance. – jiten Jan 11 '18 at 06:03
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    I'm confused, the fractions you pulled out are the real terms, and they sum to 1 as they should. – jgon Jan 11 '18 at 06:08
  • @jgon Thanks, I have added OP to beg sorry for the haste in putting the question. – jiten Jan 11 '18 at 06:09
  • @jgon Can you please elaborate the 2 phrases : "location of the original complex number" "affecting the size and rotation" of the hexagon. I am unclear about the first, as well as the second phrases. – jiten Jan 11 '18 at 06:14
  • @jgon Please elaborate if possible. – jiten Jan 11 '18 at 06:21
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    The google search complex number roots geometric and the corresponding google image search will lead you to a lot of different explanations. – Dave L. Renfro Jan 11 '18 at 06:26
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    Try plotting some examples in Wolfram Alpha to give you an idea of what I mean. https://www.wolframalpha.com/input/?i=roots+of+x%5E5-(1%2Bi) Then note that if $\alpha^n=a$, then the other $n$th roots of $a$ are $\alpha\omega_n^i$ where $\omega_n$ is a primitive $n$th root of unity and $0\le i\le n-1$. Since multiplication of complex numbers corresponds to scaling and rotating, (see https://math.stackexchange.com/q/6916/90543), and the $n$th roots of unity form a regular $n$-gon of radius 1 with a vertex at 1, $\alpha$ determines the scaling and rotation. Note the scale is $\sqrt[n]{|a|}$. – jgon Jan 11 '18 at 06:26

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