The least positive integer $n$ which will reduce ${{\left( \dfrac{i-1}{i+1} \right)}^{n}}$ to a real number is
A. 2
B. 3
C. 4
D. 5

Hint:We first rationalise the fraction form of $\dfrac{i-1}{i+1}$. We multiply $i-1$ to both the numerator and denominator of the fraction. We use the relations of imaginary numbers ${{i}^{2}}=-1,{{i}^{3}}=-i,{{i}^{4}}=1$. We replace the values to find the least positive integer $n$.

Complete step by step answer:
We first apply the rationalisation of complex numbers for $\dfrac{i-1}{i+1}$.
We multiply $i-1$ to both the numerator and denominator of the fraction.
So, \[\dfrac{i-1}{i+1}\times \dfrac{i-1}{i-1}=\dfrac{{{\left( i-1 \right)}^{2}}}{{{i}^{2}}-1}\].
We used the theorem of \[\left( a+b \right)\times \left( a-b \right)={{a}^{2}}-{{b}^{2}}\].
We now break the square using the theorem of \[{{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}\].
So, \[\dfrac{i-1}{i+1}=\dfrac{{{i}^{2}}-2i+1}{{{i}^{2}}-1}\].
We have the relations for imaginary $i$ where ${{i}^{2}}=-1,{{i}^{3}}=-i,{{i}^{4}}=1$. We place the values and get
\[\dfrac{i-1}{i+1}=\dfrac{{{i}^{2}}-2i+1}{{{i}^{2}}-1} \\
\Rightarrow \dfrac{i-1}{i+1}=\dfrac{-1-2i+1}{-1-1} \\
\Rightarrow \dfrac{i-1}{i+1}=i \\ \]
Therefore, ${{\left( \dfrac{i-1}{i+1} \right)}^{n}}={{i}^{n}}$. From the above relation relations ${{i}^{2}}=-1,{{i}^{3}}=-i,{{i}^{4}}=1$, we can see that the least positive integer $n$ which will reduce ${{\left( \dfrac{i-1}{i+1} \right)}^{n}}$ to a real number is 2.

Hence, the correct option is A.

Note:The integer power value of every eve number on $i$ will give the real number. Odd numbers of power value give back the imaginary part. The relation ${{i}^{2}}=-1,{{i}^{3}}=-i,{{i}^{4}}=1$ can also be represented as the unit circle on the complex plane.