Question

The smallest positive integer n for which $(1+i{)}^{2n}=(1-i{)}^{2n}$ is :

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (B)

Given: ${\left[\frac{\left(1+i\right)}{\left(1-i\right)}\right]}^{2n}=1...\left(i\right)$
Let $A=\frac{1+i}{1-i}...\left(ii\right)$
So $A=\frac{1+i}{1-i}\times \frac{1+i}{1+i}=\frac{2i}{1-i^2}$
Now $i^2=-1$ and $i^4=1$
$\therefore A=\frac{2i}{2}=i$
So, from equations $\left(i\right)$ and $\left(ii\right)$,
${\left(i\right)}^{2n}=1=i^4$
$\Rightarrow 2n=4\therefore n=2.$