Question

$i=\sqrt{-1}$, $\omega =$ nonreal cube root 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 $nϵN$
• D. None of these

Answer 1

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

Since $w$ is the cube root of unity.$\therefore w^3=1\&1+w+w^2=0$where $w=\frac{-1+i\sqrt{3}}{2}\&w^2=\frac{-1-i\sqrt{3}}{2}$
$z=\frac{{(1+i)}^{2n}{-(1-i)}^{2n}}{\left({1+w}^4{-w}^2\right)\left({1-w}^4{+w}^2\right)}=\frac{2^n\left[{\left(\frac{1+i}{2}\right)}^{2n}{-\left(\frac{1-i}{2}\right)}^{2n}\right]}{\left({1+w}^4{-w}^2\right)\left({1-w}^4{+w}^2\right)}$
$\Rightarrow z=\frac{2^n\left[{\left(cis\frac{\pi }{4}\right)}^{2n}{-\left(cis(-\frac{\pi }{4})\right)}^{2n}\right]}{\left({1+w}^4{-w}^2\right)\left({1-w}^4{+w}^2\right)}=\frac{2^n\left[\left(cis\frac{n\pi }{2}\right)-cis(-\frac{n\pi }{2})\right]}{\left({1+w}^4{-w}^2\right)\left({1-w}^4{+w}^2\right)}$
$\Rightarrow z=\frac{2^{n+1}i\sin \frac{n\pi }{2}}{\left({1+w}^4{-w}^2\right)\left({1-w}^4{+w}^2\right)}$
If n is even.
$z=0$
Ans: A