Question

If $i=\sqrt{-1},\omega$ = nonreal cube root of unity then $\frac{(1+i{)}^{2n}-(1-i{)}^{2n}}{(1+{\omega }^4-{\omega }^2)(1-{\omega }^4+{\omega }^2)}$ is equal to:

• A. $0$ if $n$ is even
• B. $0$ for all $nϵZ$
• C. $2^{n-1},i$ for all $nϵZ$
• D. None of these

Answer 1

The correct option is A $0$ if $n$ is even

$\frac{(1+i{)}^{2n}-(1-i{)}^{2n}}{(1+{\omega }^4-{\omega }^2)(1-{\omega }^4+{\omega }^2)}$

$H\omega +{\omega }^2=0$
${\omega }^2=\omega$
$\therefore H{\omega }^4=H\omega =-{\omega }^2$
$1+{\omega }^2=-\omega$
$\therefore \frac{(1+i{)}^{2n}-(1-i{)}^{2n}}{(-{2\omega }^2)(-\omega -\omega )}=\frac{(1+i{)}^{2n}-(1-i{)}^{2n}}{(-2{\omega }^2)(-2\omega )}$
= $\frac{(1+i{)}^{2n}-(1+i{)}^{2n}}{4}$
Even terms of ${(1+i)}^{2n}$ and ${(1-i)}^{2n}$ are same and will get cancelled
$\therefore (1+i{)}^{2n}-(1-i{)}^{2n}={[}^{2n}{C}_1\left(i{)}^{2n-1}{+}^{2n}{C}_3\right(i){\left(i\right)}^{2n-3}{+}^{2n}{C}_{2n-1}(i{)}^1$
$(i{)}^{2n-1}+(i)=0$ as $i^{2n-1}=-i$
$\therefore$ Expression $=0$