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I'm trying to prove that if $z=\operatorname{cis}(2\pi/n) = \cos(2\pi/n) + i\sin(2\pi/n)$, that is, $z$ is a primitive $n$-th root of unity, for any integer $n\geq 2$, $1+z+z^2+\cdots+z^{n-1}=0$. I've already come across a nice and concise proof here, and that same link also has a comment pointing out that it's just a geometric sum which can be expressed as $\dfrac{1-\operatorname{cis}^n(2\pi/n)}{1-\operatorname{cis}(2\pi/n)}$ which is just $0$ in the numerator. However, I was wondering if I could do it just using trig functions. It's an inefficient way of proving it, but I was fixated on this approach for so long I was wondering if someone knew how to do it.

Proving that the imaginary part is $0$ is easy- you just use the identity $\sin(a)+\sin(b)=2\sin(\frac{a+b}{2})\sin(\frac{a-b}{2})$ and for each integer $j$ where $0< j<n$, pair $i\sin(2\pi j/n)$ with $i\sin(2\pi (n-j)/n)$ to get $0$. (If $n$ is even, $i\sin(\pi)$ can't be paired, but that's of course $0$ as well.)

This same approach doesn't work for the real part- using the identity $\cos(a)+ \cos(b) =2\cos(\frac{a+b}{2})\cos(\frac{a-b}{2})$, and adding the same pairs gets $2\cos(2\pi)\cos(2\pi(n-2j)/n)=2\cos(2\pi(n-2j)/n)$ so this gets $1+2\sum_{j=1}^{\lfloor n/2 \rfloor}\cos(2\pi(n-2j)/n)$ with $\cos(\pi/n)=-1$ added if $n$ is even. Then I need to show that that sum is $0$ if $n$ is even and $-1/2$ if $n$ is odd. Is there a clean way of doing this? The only thing I can think to do is repeat the sum of $\cos$ identity, and that doesn't seem too helpful.

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Use the identity $$\displaystyle\sum\limits_{m=0}^{n-1} \cos(mx+y)=\frac{\cos\left(\dfrac{n-1}{2}x+y\right)\sin\left(\dfrac{n}{2}\, x\right)}{\sin\left(\dfrac{x}{2}\right)}$$

and evaluate where $x=2\pi/n$ and $y=0$ to deduce that the real part is zero.

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  • $\begingroup$ Thanks for the answer. lab pointed out a proof to this in the comments, but there's a slight difference between them. Should that be $\operatorname{cos}$ and not $\operatorname{sin}$ as the first factor in the numerator? $\endgroup$ – Kevin Long May 13 '16 at 15:42
  • $\begingroup$ Yes It should be. Thanks for pointing out the error. $\endgroup$ – Foobaz John May 13 '16 at 17:11

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