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Given the equation:

$$\operatorname{cis} x \operatorname{cis} 2x \operatorname{cis} 3x \dots \operatorname{cis} nx=1$$

How can I solve it?

I know that $\operatorname{cis} x=\cos x+i\sin x$, but I can't see how to proceed from there.

Thank you

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  • $\begingroup$ For constant $a$, $\cos a x + i \sin a x = e^{iax}$ $\endgroup$ – Kevin Feb 24 '16 at 10:22
  • $\begingroup$ Oh I didn't know that. I'll try. Thanks $\endgroup$ – Karen Feb 24 '16 at 10:23
  • $\begingroup$ Also, another hint would be to know that $\sum_{i=1}^{n}k=n(n+1)/2$ $\endgroup$ – Kevin Feb 24 '16 at 10:24
  • $\begingroup$ I got: $ix(1+n)(n/2)=1$. So I need to square both sides? $\endgroup$ – Karen Feb 24 '16 at 10:29
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The $\def\cis{\operatorname{cis}}\cis$ function obeys a very simple rule: $$ \cis a\cis b=\cis(a+b) $$ Just apply the decomposition $\cis a=\cos a+i\sin a$, in order to prove it. Then you have $$ \cis(x+2x+\dots+nx)=1 $$ and now it should be easy.

Hint: $x+2x+\dots+nx=x\frac{n(n+1)}{2}$, so you have $$ \cis\frac{n(n+1)x}{2}=\cis0 $$

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  • $\begingroup$ But then: $n(n+1)=0, x=0$ but isn't a period for $cis$? $\endgroup$ – Karen Feb 24 '16 at 10:35
  • $\begingroup$ @Karen $\cis a=\cis b$ means $a=b+2k\pi$ for some integer $k$. $\endgroup$ – egreg Feb 24 '16 at 10:35
  • $\begingroup$ Got it. So $x=4\pi k/n(n+1)$, and $k=0,1,...,n$? $\endgroup$ – Karen Feb 24 '16 at 10:37
  • $\begingroup$ @Karen Almost. ;-) Check the $2$'s $\endgroup$ – egreg Feb 24 '16 at 10:38
  • $\begingroup$ I'm not sure about the k's. $\endgroup$ – Karen Feb 24 '16 at 10:39

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