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

Question

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

Answer 1

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 $$ \cis\frac{n(n+1)x}{2}=\cis0 $$