Question

The complex numbers $\frac{2^n}{(1+i{)}^{2n}}+\frac{(1+i{)}^{2n}}{2^n}$, $nϵZ$, is equal to

• A. 0
• B. 2
• C. $\{1+(-1{)}^n\}.i^{-n}$
• D. None of these

Answer 1

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

Simplifying, we get
$e^{i\frac{-2n\pi }{4}}+e^{i\frac{2n\pi }{4}}$
$=e^{i\frac{-n\pi }{2}}+e^{i\frac{n\pi }{2}}$
$=\frac{1+e^{i\frac{2n\pi }{2}}}{e^{i\frac{n\pi }{2}}}$
$=\frac{1+e^{in\pi }}{e^{i\frac{n\pi }{2}}}$
$=i^{-n}[1+(-1{)}^n]$
Hence, option 'C' is correct.