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

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

• For constant $a$, $\cos a x + i \sin a x = e^{iax}$ – Kevin Feb 24 '16 at 10:22
• Oh I didn't know that. I'll try. Thanks – Karen Feb 24 '16 at 10:23
• Also, another hint would be to know that $\sum_{i=1}^{n}k=n(n+1)/2$ – Kevin Feb 24 '16 at 10:24
• I got: $ix(1+n)(n/2)=1$. So I need to square both sides? – Karen Feb 24 '16 at 10:29

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 $$\cis\frac{n(n+1)x}{2}=\cis0$$
• But then: $n(n+1)=0, x=0$ but isn't a period for $cis$? – Karen Feb 24 '16 at 10:35
• @Karen $\cis a=\cis b$ means $a=b+2k\pi$ for some integer $k$. – egreg Feb 24 '16 at 10:35
• Got it. So $x=4\pi k/n(n+1)$, and $k=0,1,...,n$? – Karen Feb 24 '16 at 10:37
• @Karen Almost. ;-) Check the $2$'s – egreg Feb 24 '16 at 10:38