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Question: Let $z=\cos(2\pi /n)+i\sin(2\pi /n)$ for an integer $n\geq2$. Show that $1+z+z^2+...+z^{n-1}=0$.

Work thus far: My initial idea is was to use euler's formula, but this leads to a sum of exponentials that can't be simplified easily. If I can separate the sum into two sums (one real, one imaginary) which both add to zero then I can show that the equality holds. Using DeMoivre's formula $z^a=\cos(a2\pi /n)+i\sin(a2\pi /n)$. So we get $$1+\cos(2\pi /n)+\cos(4\pi /n)+...+\cos(2n\pi/n)$$ $$i\sin(2\pi /n)+i\sin(4\pi /n)+...+i\sin(2n\pi/n)$$ From there however I do not know how to find the sum of both series.

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    I would draw the numbers up in the plane, and argue by symmetry. – 2017-01-15
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    Use the fact that your sum is a geometric series to simplify the sum. – 2017-01-15
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    Recall that $1+z +\cdots + z^{n-1} = \frac{1-z^n}{1-z}$. – 2017-01-15
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    Find a monic polynomial that has exactly those roots, and study the coefficient of the term with degree one less than that of your polynomial. – 2017-01-15

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Let $z = e^{2\pi i/n}$. Now, note that: $$\sum_{i = 0}^{n-1} z^i = \frac{z^{n}-1}{z-1}$$ (by the finite geometric sum formula). Putting in $e^{2\pi i/n}$, we get that: $$\frac{e^{2\pi i/n\times n}-1}{e^{2\pi i/n}-1} = \frac{0}{e^{2\pi i/n}-1}$$ So, we have that the sum is zero.

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Note that $\{1,z,z^2,...,z^{n-1}\}$ are all of the $n$-th roots of unity. Using this fact, the multiplication with $z$ is a bijection on $\{1,z,z^2,...,z^{n-1}\}$, thus $$1+z+...+z^{n-1} = z+z^2+...+z^n = z(1+...+z^{n-1}).$$ Since $z\neq 1$, we can conclude that $1+...+z^{n-1} = 0$.

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The sum is a finite geometric series and hence is equal to $$ \frac{1-z^n}{1-z} $$ Substitute $z=e^{2\pi i/n}$.