Question

If $w$ be complex $n^{th}$ root of unity and $r$ is an integer not divisible by $n$, then the sum of the $r$th powers of the nth roots of unity is

• A. $0$
• B. $1$
• C. $w$
• D. $n$

Answer 1

The correct option is A $0$

We have, for $n^{th}$ root of unity denoted by $\alpha i$

$1+{\alpha }_1+{\alpha }_2+{\alpha }_3---------{\alpha }_{n-1}=\frac{1(1-(\alpha {)}^2)}{1-\alpha }$ [where ${\alpha }_i=e^{i\frac{2\pi k}{n}}$ Thus, ${\alpha }_1=e^{i\frac{2\pi }{n}}$ ${\alpha }_1=e^{i\frac{4\pi }{n}}$---------------form a G.P. with ratio $\alpha =e^{i2\pi /2}$]

for power r, expression changes to,

$1+{\alpha }_1^r+{\alpha }_2^r+{\alpha }_3^r---------{\alpha }_{n-1}^r=\frac{1-({\alpha }^r{)}^2}{}$ [Given $r\ne kn$ $\Rightarrow {\alpha }^r\ne 1$]

$=\frac{1-({\alpha }^n{)}^r}{1-{\alpha }^r}$

But ${\alpha }^2=1$

Thus $\frac{1-1}{1-{\alpha }^r}=0$