Question

The value of $i^n+i^{-n}$, for $i=\sqrt{-1}$ and $n\in 1$ is

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

Answer 1

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

$(1+i{)}^2=1+i^2+2i=1-1+2i=2i$

$⟹(1+i{)}^{2n}=2^n{i}^n$

$⟹\frac{(1+i{)}^{2n}}{2^n}=i^n$

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

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

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