Question

If $z^2-z+1=0,$ then possible value(s) of $z^n-z^{-n},$ where $n$ is even number

• A. $\sqrt{3}i$
• B. $0$
• C. $-\sqrt{3}i$
• D. $-2$

Answer 1

Answer: (A), (B), (C)

$-\sqrt{3}i$

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

If $z=-\omega ,$ then
$z^n-z^{-n}=(-\omega {)}^n-(-\omega {)}^{-n}$
$=(-1{)}^n{\omega }^n-\frac{(-1{)}^{-n}}{{\omega }^n}$
$={\omega }^n-{\omega }^{2n}$
Here, there are three cases possible
Case $1:n$ is multiple of $3$
$={\omega }^n-{\omega }^{2n}=1-1=0$
Case $2:n=3m+1,m\in Z^{+}$
$={\omega }^{3m+1}-{\omega }^{2(3m+1)}={\omega }^{3m}\omega -{\omega }^{6m}{\omega }^2$
$=\omega -{\omega }^2=\sqrt{3}i$
Case $3:n=3m+2,m\in Z^{+}$
$={\omega }^{3m+2}-{\omega }^{2(3m+2)}={\omega }^{3m}{\omega }^2-{\omega }^{6m}{\omega }^4$
$={\omega }^2-\omega =-\sqrt{3}i$

If $z=-{\omega }^2,$ then
$z^n-z^{-n}=(-{\omega }^2{)}^n-(-{\omega }^2{)}^{-n}$
$=(-1{)}^n{\omega }^{2n}-\frac{(-1{)}^{-n}}{{\omega }^{2n}}$
$={\omega }^{2n}-{\omega }^n$
Similarly as above we can do.