The least positive integer n for which ${\left( {\dfrac{{1 + i\sqrt 3 }}{{1 - i\sqrt 3 }}} \right)^n} = 1$, is
A) 2
B) 6
C) 5
D) 3

Hint: According to given in the question we have to determine the least positive integer n for which ${\left( {\dfrac{{1 + i\sqrt 3 }}{{1 - i\sqrt 3 }}} \right)^n} = 1$. So, first of all we have to multiply the reciprocal of the term in the denominator with the numerator and denominators of the given expression.
Now, we have to multiply the terms in the denominator with the numerator and denominator to obtain the values of n.
To solve the obtained expression we have to apply the formulas as mentioned below:

Formula used: $
   \Rightarrow {(a + b)^2} = {a^2} + {b^2} + 2ab..................(A) \\
   \Rightarrow ({a^2} - {b^2}) = (a + b)(a - b)................(B) \\
   \Rightarrow {(a - b)^2} = {a^2} + {b^2} - 2ab..................(C)
 $
Now, we have to find the cube of the given expression and after substituting all the values we can determine n.

Complete step-by-step answer:
Step 1: First of all we have to determine the reciprocal or inverse of the term given in the denominator which is $1 - i\sqrt 3 $ and its reciprocal is $1 + i\sqrt 3 $
Step 2: Now, we have to multiply the obtained reciprocal with the numerator and denominator of the expression given. Hence,
\[
   \Rightarrow \left( {\dfrac{{1 + i\sqrt 3 }}{{1 - i\sqrt 3 }}} \right) \times \left( {\dfrac{{1 + i\sqrt 3 }}{{1 + i\sqrt 3 }}} \right) \\
   \Rightarrow \dfrac{{{{(1 + i\sqrt 3 )}^2}}}{{{1^2} - {{(i\sqrt 3 )}^2}}}
 \]
Step 3: Now, to solve the expression as obtained in the solution step 2 we have to apply the formula (A) and formula (B) as mentioned in the solution hint. Hence,
$
   = \left( {\dfrac{{{1^2} + {{(i\sqrt 3 )}^2} + 2 \times 1 \times (i\sqrt 3 )}}{{1 - 3{i^2}}}} \right) \\
   = \dfrac{{1 + 3{i^2} + 2i\sqrt 3 }}{{1 - 3{i^2}}}
 $
Step 4: Now, to solve the expression as obtained in the solution step 3 we have to substitute the value of${i^2} = - 1$. Hence,
$
   = \dfrac{{1 - 3 + 2i\sqrt 3 }}{{1 - 3( - 1)}} \\
   = \dfrac{{ - 2 + 2i\sqrt 3 }}{{1 + 3}} \\
   = \dfrac{{1 - i\sqrt 3 }}{{ - 2}}..................(1)
 $
Step 5: Now, we have to multiply the denominator with the expression to numerator and denominator. Hence,
\[
   \Rightarrow \left( {\dfrac{{1 + i\sqrt 3 }}{{1 - i\sqrt 3 }}} \right) \times \left( {\dfrac{{1 - i\sqrt 3 }}{{1 - i\sqrt 3 }}} \right) \\
   \Rightarrow \dfrac{{{1^2} - {{(i\sqrt 3 )}^2}}}{{{{(1 - i\sqrt 3 )}^2}}}
 \]
Step 6: Now, to solve the expression as obtained in the solution step 5 we have to apply the formula (C) as mentioned in the solution hint. Hence,
$ = \dfrac{{1 - 3{i^2}}}{{{1^2} + {{(i\sqrt 3 )}^2} - 2 \times 1 \times (i\sqrt 3 )}}$
Now, to solve the expression obtained just above, we have to substitute the value of ${i^2} = - 1$
\[
   = \dfrac{{1 - 3( - 1)}}{{1 + 3{i^2} - 2i\sqrt 3 }} \\
   = \dfrac{{1 + 3}}{{1 - 3 - 2i\sqrt 3 }} \\
   = \dfrac{4}{{ - 2 - 2i\sqrt 3 }} \\
   = \dfrac{{ - 2}}{{1 + i\sqrt 3 }}.....................(2)
 \]
Step 7: Now, we have to find the cube of the given expression, $\left( {\dfrac{{1 + i\sqrt 3 }}{{1 - i\sqrt 3 }}} \right)$. Hence,
$
   \Rightarrow {\left( {\dfrac{{1 + i\sqrt 3 }}{{1 - i\sqrt 3 }}} \right)^3} = \left( {\dfrac{{1 + i\sqrt 3 }}{{1 - i\sqrt 3 }}} \right) \times \left( {\dfrac{{1 + i\sqrt 3 }}{{1 - i\sqrt 3 }}} \right) \times \left( {\dfrac{{1 + i\sqrt 3 }}{{1 - i\sqrt 3 }}} \right) \\
   \Rightarrow {\left( {\dfrac{{1 + i\sqrt 3 }}{{1 - i\sqrt 3 }}} \right)^3} = {\left( {\dfrac{{1 + i\sqrt 3 }}{{1 - i\sqrt 3 }}} \right)^2} \times \left( {\dfrac{{1 + i\sqrt 3 }}{{1 - i\sqrt 3 }}} \right)
 $
Now, with the help of the expression (1) and expression (2),
$
   \Rightarrow {\left( {\dfrac{{1 + i\sqrt 3 }}{{1 - i\sqrt 3 }}} \right)^3} = \left( {\dfrac{{1 + i\sqrt 3 }}{{1 - i\sqrt 3 }}} \right) \times \left( {\dfrac{{ - 2}}{{1 + i\sqrt 3 }}} \right) \times \left( {\dfrac{{1 - i\sqrt 3 }}{{ - 2}}} \right) \\
   \Rightarrow {\left( {\dfrac{{1 + i\sqrt 3 }}{{1 - i\sqrt 3 }}} \right)^3} = 1
 $

Hence, with the help of formula (A), (B), and (C) we have obtained the value of n which is n = 3.

Note: In the given expression i is an imaginary term and on multiplying i with i or we can say on apply the square to the imaginary term i we will obtain ${i^2} = - 1$
To find the inverse or a reciprocal of a given term or expression we have to convert the given sign means if the signs between the two terms is positive then we have to convert it to negative or vice-versa.