1) If $z=\cos\frac{2\pi}{n}+i\sin\frac{2\pi}{n}$
2) Then apparently: $z^0+z^1+z^2...z^{n-1}=\frac{1-z^n}{1-z}=\frac{0}{1-z}=0$
Could someone please explain why: $z^0+z^1+z^2...z^{n-1}=\frac{1-z^n}{1-z}$?
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3It's the sum of a geometric series. – GoodDeeds Apr 05 '16 at 09:48
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1or $a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+...+ab^{n-2}+b^{n-1})$ – Roman83 Apr 05 '16 at 09:51
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Note that the $n$-th roots of unity are the roots of $X^n-1$. If you've learnt about polynomials, this tells you immediately that their sum is $0$ and their product is $(-1)^{n-1}$ – Gabriel Romon Apr 05 '16 at 09:53
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Let $z^0+z^1+z^2...z^{n-1}= p(z)$.
Then $z \ p(z)=z^1+z^2+z^3...z^{n}$.
Hence $p(z)-zp(z)=1-z^n$.
$p(z)(1-z)=1-z^n$.
$p(z)=\frac{1-z^n}{1-z}$
If $z=\cos\frac{2\pi}{n}+i\sin\frac{2\pi}{n}$. Then there is a formula (that is easy to prove) that states for any $i \in \mathbb{R}$, $z^i=\cos\frac{2i\pi}{n}+i\sin\frac{2i\pi}{n}$. Hence $z^n=\cos\frac{2n\pi}{n}+i\sin\frac{2n\pi}{n}=1$. Substitute that in the last expression of $p(z)$.
Miz
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Don't you mean $z^n = \cos \frac{2n\pi}{n} + i\sin \frac{2n\pi}{n} = 1$? – N. F. Taussig Apr 05 '16 at 10:17
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Sum of geometric series is given by $$\frac{a(1-r^n)}{1-r}$$ where $|r|\leq1$ so $a=z^0=1$ and $r$ is common ratio here its $z$ now we know z is nth root of unity so $z^n=1$ thus we get $1-z^n=1-1=0$ . also we have proved an important fact that sum of roots of unity is $0$.
Archis Welankar
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