Question

If $\omega \ne 1$, is $n$th root of unity and
${(\omega +x)}^n=1+\overline{\omega }+\frac{n-1}{2n}{\overline{\omega }}^2+....$
then value of $x$ is

• A. $1/n$
• B. $2/n$
• C. $1/2$
• D. $1/4$

Answer 1

The correct option is A $1/n$

$(w+x{)}^n$ $=w^n+nx(w{)}^{n-1}+\frac{n(n-1)}{2}{x}^2{w}^{n-2}+...$
$=1+\left(\overline{w}\right)+\frac{(n-1)}{2n}\left({\overline{w}}^2\right)+...$
$w^n=1$
Hence $n=3r$ where r is a positive real number.
Hence $w^{n-1}=\overline{w}$ ...
Thus by comparing coefficients, we get
$nx=1$
$x=\frac{1}{n}$