Question

If $z$ is a complex number satisfying $z+z^{-1}=1$, then $z^n+z^{-n},n\in N$ has/have the value(s)

• A. $2(-1{)}^n$ when $n$ is a multiple of $3$
• B. $(-1{)}^{n+1}$ when $n$ is not a multiple of $3$
• C. $(-1{)}^{n+1},$ when $n$ is a multiple of $3$
• D. $0,$ when $n$ is not a multiple of $3$

Answer 1

Answer: (A), (B)

$(-1{)}^{n+1}$ when $n$ is not a multiple of $3$

$z+z^{-1}=1\Rightarrow z^2-z+1=0$
$\therefore z=\frac{1\pm \sqrt{3}i}{2}=-\omega ,-{\omega }^2$

$\therefore z^n+z^{-n}=(-\omega {)}^n+(-\omega {)}^{-n}\text{or}{(-{\omega }^2)}^n+{(-{\omega }^2)}^{-n}$
$=(-1{)}^n({\omega }^n+\frac{1}{{\omega }^n})\text{or}(-1{)}^n({\omega }^{2n}+\frac{1}{{\omega }^{2n}})$
$=(-1{)}^n\cdot ({\omega }^n+{\omega }^{2n})[\because {\omega }^{3n}=1]$

Now if $n$ is multiple of $3,$ then
$z^n+z^{-n}=(-1{)}^n\cdot (2)$

or if $n$ is not a multiple of $3,$ then
$z^n+z^{-n}=(-1{)}^n\cdot (-1)=(-1{)}^{n+1}$