Question

The complex number $\frac{2^n}{(1+i{)}^{2n}}+\frac{(1+i{)}^{2n}}{2^n},n\in I$ is equal to

• A. $0$
• B. $2$
• C. $\{1+(-1{)}^n\}\cdot i^n$
• D. $\{1-(-1{)}^n\}\cdot i^n$

Answer 1

The correct option is C $\{1+(-1{)}^n\}\cdot i^n$

Here, $\frac{2^n}{(1+i{)}^{2n}}+\frac{(1+i{)}^{2n}}{2^n}$
$=\frac{2^n}{(1+i^2+2i{)}^n}+\frac{(1+i^2+2i{)}^n}{2^n}$
$=\frac{2^n}{(2i{)}^n}+\frac{(2i{)}^n}{2^n}$
$={\left(\frac{1}{i}\right)}^n+(i{)}^n$
$={\left(\frac{i}{i^2}\right)}^n+(i{)}^n$
$=(-i{)}^n+(i{)}^n$
$=(i{)}^n\{(-1{)}^n+1\}$