Question

If $i=\sqrt{-1}$ and $\omega$ is non real 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. $0,$ when $n$ is even
• B. $2^{n-1}\cdot i^n,$ when $n$ is odd
• C. $0\forall n\in I$
• D. $2^{n-1}\cdot i^n,$ when $n$ is even

Answer 1

Answer: (A), (B)

The correct options are
A $0,$ when $n$ is even
B $2^{n-1}\cdot i^n,$ when $n$ is odd

$\frac{(1+i{)}^{2n}-(1-i{)}^{2n}}{(1+{\omega }^4-{\omega }^2)(1-{\omega }^4+{\omega }^2)}=\frac{\left(\right(1+i{)}^2{)}^n-\left(\right(1-i{)}^2{)}^n}{(1+\omega \cdot {\omega }^3-{\omega }^2)(1-\omega \cdot {\omega }^3+{\omega }^2)}$ $=\frac{(1+i^2+2i{)}^n-(1+i^2-2i{)}^n}{(1+\omega -{\omega }^2)(1-\omega +{\omega }^2)}$ $=\frac{(2i{)}^n-(-2i{)}^n}{(-2{\omega }^2)(-2\omega )}$ $=\frac{2^n{i}^n[1-(-1{)}^n]}{4}$ $=\{\begin{array}{c}0,\text{when}n\text{is even}\\ \frac{2^{n+1}\cdot i^n}{2^2}=2^{n-1}\cdot i^n,\text{when}n\text{is odd}\end{array}$