Question

The least positive integer $n$,which will reduce ${\left[\frac{\left(1+i\right)}{\left(1-i\right)}\right]}^n$to a real number is

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

The correct option is C

$4$

Solve the given expression

Given expression, ${\left[\frac{\left(1+i\right)}{\left(1-i\right)}\right]}^n$

$\therefore Z={\left[\frac{\left(1+i\right)}{\left(1-i\right)}\right]}^n$

Now, multiply and divide the above expression by $\left(1+i\right)$ we get,

$$\begin{gathered} Z={\left[\frac{\left(1+i\right)\left(1+i\right)}{\left(1-i\right)\left(1+i\right)}\right]}^n \\ \Rightarrow Z={\left[\frac{{\left(1+i\right)}^2}{1-i^2}\right]}^n \\ \Rightarrow Z={\left[\frac{1+i^2+2i}{1-\left(-1\right)}\right]}^n \\ \Rightarrow Z={\left[\frac{1-1+2i}{2}\right]}^n \\ \Rightarrow Z={\left[\frac{2i}{2}\right]}^n \\ \Rightarrow Z=i^n \\ \Rightarrow Z=1 \end{gathered}$$

Where $i=1$ is given (which is not real)

Now, for $n=2$ we have

,$$\begin{gathered} Z=i^2 \\ \therefore Z=-1 \end{gathered}$$

So, the least positive integer is $n=4$ $[\because Z=i^4=1]$

Hence, option C is correct.