For a complex number ${\left( {1 + i\sqrt 3 } \right)^2}$, find the multiplicative inverse of the above complex number:
A) $ - \dfrac{1}{8} - i\dfrac{{\sqrt 3 }}{8}$
B) $ - \dfrac{1}{8} + i\dfrac{{\sqrt 3 }}{8}$
C) $\dfrac{1}{8} - i\dfrac{{\sqrt 3 }}{8}$
D) None of the above.

Hint: We take the given complex number and expand it using the complex number expansions. Multiplicative inverse is simply the reciprocal, so the complex number is in the denominator. We rationalize the number to form the multiplicative inverse and then rearrange it in the form of the given options.

Complete step-by-step solution:
Let us consider the given complex number to be $z$.
Given,
$z = {\left( {1 + i\sqrt 3 } \right)^2}$
By using the formula: ${\left( {a + ib} \right)^2} = {a^2} - {b^2} + 2ab$
From the given complex number, equating the complex number to its general form $a + bi$, we get,
$a = 1$ and $b = \sqrt 3 $
Substituting the values into the formula,
$ \Rightarrow z = {\left( 1 \right)^2} - {\left( {\sqrt 3 } \right)^2} + 2\left( 1 \right)\left( {i\sqrt 3 } \right)$
Simplifying we get,
$ \Rightarrow z = 1 - 3 + 2\sqrt 3 i$
Subtracting we get,
$ \Rightarrow z = - 2 + 2\sqrt 3 i$
Multiplicative inverse of the number is basically the reciprocal of the given number. That means the multiplicative inverse of $z$ is ${z^{ - 1}}$, that is $\dfrac{1}{z}$.
Now,
$\dfrac{1}{z} = \dfrac{1}{{ - 2 + 2\sqrt 3 i}}$
Rationalizing the given complex number,
Rationalizing is the condition where the given fraction’s denominator is additively inversed and multiplied both on numerator and denominator.
$\dfrac{1}{z} = \dfrac{1}{{ - 2 + 2\sqrt 3 i}} \times \dfrac{{ - 2 - 2\sqrt 3 i}}{{ - 2 - 2\sqrt 3 i}}$
Multiplying the terms,
$ \Rightarrow \dfrac{1}{z} = \dfrac{{ - 2 - 2\sqrt 3 i}}{{( - 2 + 2\sqrt 3 i)( - 2 - 2\sqrt 3 i)}}$
By using the formula $\left( {{a^2} + {b^2}} \right) = \left( {a + b} \right)\left( {a - b} \right)$,
$ \Rightarrow \dfrac{1}{z} = \dfrac{{ - 2 - 2\sqrt 3 i}}{{{{( - 2)}^2} - {{(2\sqrt 3 i)}^2}}}$
Squaring the terms,
$ \Rightarrow \dfrac{1}{z} = \dfrac{{ - 2 - 2\sqrt 3 i}}{{4 - 12{i^2}}}$
Since we know that, ${i^2} = - 1$,
$ \Rightarrow \dfrac{1}{z} = \dfrac{{ - 2 - 2\sqrt 3 i}}{{4 + 12}}$
Adding the terms,
$ \Rightarrow \dfrac{1}{z} = \dfrac{{ - 2 - 2\sqrt 3 i}}{{16}}$
Taking the common terms out in the numerator
$ \Rightarrow \dfrac{1}{z} = \dfrac{{2( - 1 - \sqrt 3 i)}}{{16}}$
Dividing the terms we get,
$ \Rightarrow \dfrac{1}{z} = \dfrac{{( - 1 - \sqrt 3 i)}}{8}$
Splitting the numerator into two terms, we get:
$ \Rightarrow \dfrac{1}{z} = \dfrac{{ - 1}}{8} - \dfrac{{\sqrt 3 }}{8}i$

$\therefore $ The correct option is A.

Note: A complex number of the form $a + bi$, where $a$ and $b$ are real numbers and I is an indeterminate satisfying ${i^2} = - 1$. The real number $a$ is called the real part of the complex number $a + bi$ the real number $b$ is called its imaginary part. To emphasize, the imaginary part does not include $a$ factor $i$, that is the imaginary part $b$, not $bi$.
A complex number $z$ can thus be identified with an ordered pair (Re(z), Im(z)) of real numbers, which in turn may be interpreted as coordinates of a point in a two-dimensional space.
The additive inverse is simply the reverse of the given sign. For example, the additive inverse of subtraction is addition and vice versa. The terms at the end must be split in order to get the exact solution as given in the options of the questions.