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This problem has been treated several times on the site, but there's a minor (and possibly stupid) aspect I fail to understand.

The general statement is "The sum of n-th degree complex-roots of 1 is 0".

How I fail to understand it:

(1) The equation $z^n=1$ has n roots.

(2) If $n=1$ the equation is $z=1$ has 1 root and is the pair $1+0i$, and since it is the only root it does not add up to zero.

A related question: Prove that sum of n-th degree roots of complex number is 0

Cure
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    You're perfectly correct, the assertion is only correct for $n>1$. – paul garrett Oct 04 '20 at 23:33
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    Look at the proof of the statement, it uses the fact that sum of roots is the minus coeffcient at $z^{n-1}$, which is $0$ for polynomial $z^n-1$ with $n>1$, and it is $1$ for $n=1$. – Sil Oct 04 '20 at 23:38

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The roots of $z^n-1 = 0$ (with multiplicity, in theory) should add up to the negative of the coefficient of $z^{n-1}$. This is a general rule for any polynomial.

For $n>1$, that coefficient is $0$, and so the roots add up to $0$. For $n=1$, however, that coefficient is $-1$, and so the roots add up to $1$.

Misha Lavrov
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The $n$-th roots of unity, $n\gt1$, the roots of $z^n-1$, form a cyclic group.

Since they are all powers of a primitive root, say $e^{2πi/n}$, the sum of them is actually a geometric series: $$\sum_{k=0}^{n-1}e^{2πik/n}=\dfrac{1-e^{2πi}}{1-e^{2πi/n}}=0$$.

When $n=1$, the sum is of course just $1$.

Alternatively, the sum of the roots of a polynomial of degree $n$ is the first elementary symmetric polynomial in those roots. That's the negative of the coeff of the $z^{n-1}$ term.

Thirdly, this is a special case of Vieta's formulas.

  • I did the very same thing to show it is zero. But I still feel uneasy about the proof, because I don't completely understand why should be possible in this case to add complex numbers "just like that", without doing what is normally (I think) done, which is converting them back to binomial form to add them. – Cure Oct 05 '20 at 02:54
  • The geometric series "telescopes", whether you have real or complex numbers. So the proof goes through. All you need is that a complex number plus its opposite is zero. –  Oct 05 '20 at 03:03
  • What I mean is: $(1-z)(1+z+z^2+\dots+z^{n-1})=1-z^n$. Thus the $k$-th partial sum of the geometric series is $\dfrac{1-z^{k+1}}{1-z}$. –  Oct 05 '20 at 03:43