Question

Let $p$ be an integer and let $1,\alpha ,\dots \dots {\alpha }^{n-1}$ be the $n^{\text{th}}$ roots of unity. Then $1+{\alpha }^p+({\alpha }^2{)}^p+\dots \dots +({\alpha }^{n-1}{)}^p=$

• A. $n$ if $p$ is a multiple of $n$
• B. $n$ if $p$ is not a multiple of $n$
• C. $0$ if $p$ is a multiple of $n$
• D. $0$ if $p$ is not a multiple of $n$

Answer 1

Answer: (A), (D)

$0$ if $p$ is not a multiple of $n$

$\because \alpha$ is $n^{\text{th}}$ root of unity
$\therefore {\alpha }^n=e^{2i\pi }=1$

Case $1:$ $p$ is a multiple of $n$.
Now let $p=nk$
$1+{\alpha }^{nk}+({\alpha }^2{)}^{nk}+\cdots ({\alpha }^{n-1}{)}^{nk}$
$=1+({\alpha }^n{)}^k+({\alpha }^n{)}^{2k}+\cdots +({\alpha }^n{)}^{(n-1)k}$
$=1+1+1+\cdots +1\text{(n times)}=n(\because {\alpha }^n=1)$

Case $2:$ $p$ is not a multiple of $n$.
Let $p=n\lambda +k,k=1,2,\cdots ,(n-1)[\lambda \in Z,n>1]$
and ${\alpha }^n=1$
Thus $1+{\alpha }^p+({\alpha }^2{)}^p+\cdots \cdots +({\alpha }^{n-1}{)}^p$ is
$=1+{\alpha }^{n\lambda +k}+({\alpha }^2{)}^{n\lambda +k}+\cdots +({\alpha }^{n-1}{)}^{n\lambda +k}$
$=1+{\alpha }^{n\lambda }\cdot {\alpha }^k+{\alpha }^{2n\lambda }\cdot {\alpha }^{2k}+\cdots +{\alpha }^{(n-1)n\lambda }\cdot {\alpha }^{k(n-1)}$
$=1+{\alpha }^k+{\alpha }^{2k}+\cdots +{\alpha }^{k(n-1)}[\because {\alpha }^n=1]$
$=\frac{{\alpha }^{nk}-1}{{\alpha }^k-1}=\frac{1-1}{{\alpha }^k-1}=0$