Question

Which among the following expressions are equivalent (where $n\in I$)

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

Answer 1

Answer: (B), (D)

The correct options are
B $\frac{(1+i{)}^{2n}}{2^n}+\frac{(1-i{)}^{2n}}{2^n}$
D $\frac{2^n}{(1+i{)}^{2n}}+\frac{2^n}{(1-i{)}^{2n}}$

$(1-i{)}^{2n}=((1-i{)}^2{)}^n=(1-1-2i{)}^n=(-2i{)}^n=2^n\times (-i{)}^n$
$(1+i{)}^{2n}=((1+i{)}^2{)}^n=(1-1+2i{)}^n=(2i{)}^n=2^n\times (i{)}^n$
Also, $-i=\frac{-i.i}{i}=\frac{1}{i}=i^{-1}$

Now from options,
$\left(A\right)\frac{2^n}{(1-i{)}^{2n}}+\frac{(1+i{)}^{2n}}{2^n}$
$=\frac{1}{(-i{)}^n}+(i{)}^n$
$=i^n+i^n=2i^n$


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

$=i^n+(-i{)}^n=i^n+i^{-n}$

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

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