How do you use Demoivre’s theorem to simplify ${\left( { - \sqrt 3 - i} \right)^4}$ ?

Hint: In the given problem, we are required to simplify the given expression involving the complex numbers using the Demoivre’s theorem. Demoivre’s theorem is of vital importance when any expression involving the complex number set is raised to a specific power. For applying the Demoivre’s theorem, we need to represent the expressions involving the complex number set in polar form.

Complete step by step answer:
So, we need to simplify the expression ${\left( { - \sqrt 3 - i} \right)^4}$ using the Demoivre’s theorem.
Firstly, we need to convert the expression involving complex numbers into polar form.
So, let us assume $Z = \left( { - \sqrt 3 - i} \right)$.
Then, $\left| Z \right| = \sqrt {{{\left( { - \sqrt 3 } \right)}^2} + {{\left( { - 1} \right)}^2}} = \sqrt {3 + 1} = 2$
So, Modulus of the complex number $Z = \left( { - \sqrt 3 - i} \right)$ is $\left| Z \right| = 2$.
Now, Converting the complex number into polar form, we get,
 $Z = 2\left( {\dfrac{{ - \sqrt 3 }}{2} - \dfrac{i}{2}} \right)$
Now, we have to figure out the principal argument of the complex number. So, we need to compare the complex number with the polar representation of the same.
Comparing $2\left( {\dfrac{{ - \sqrt 3 }}{2} - \dfrac{i}{2}} \right)$ with the standard form of complex number $r\left( {\cos \theta + i\sin \theta } \right)$, we get $\cos \theta = - \dfrac{{\sqrt 3 }}{2}$ and $\sin \theta = \dfrac{{ - 1}}{2}$ . So, if the complex number is represented on an Argand plane, it will lie in the fourth quadrant. So, also by dividing sine and cosine, we get $\tan \theta = \dfrac{1}{{\sqrt 3 }}$. So, the principal argument of the complex number should be $\dfrac{{11\pi }}{6}$ .
$Z = 2\left( {\cos \left( {\dfrac{{11\pi }}{6}} \right) + i\sin \left( {\dfrac{{11\pi }}{6}} \right)} \right)$
So, the modulus of the complex number is $2$ and the argument of the complex number is $\left( {\dfrac{{11\pi }}{6}} \right)$ .
Now, ${Z^4} = {\left[ {2\left( {\cos \left( {\dfrac{{11\pi }}{6}} \right) + i\sin \left( {\dfrac{{11\pi }}{6}} \right)} \right)} \right]^4}$
Distributing the power, we get,
$ = {2^4}{\left( {\cos \left( {\dfrac{{11\pi }}{6}} \right) + i\sin \left( {\dfrac{{11\pi }}{6}} \right)} \right)^4}$
$ = 16{\left( {\cos \left( {\dfrac{{11\pi }}{6}} \right) + i\sin \left( {\dfrac{{11\pi }}{6}} \right)} \right)^4}$
 Now using the Demoivre’s Theorem, ${\left( {\cos \theta + i\sin \theta } \right)^n} = \left( {\cos \left( {n\theta } \right) + i\sin \left( {n\theta } \right)} \right)$ ,
$ = 16\left( {\cos \left( {4 \times \dfrac{{11\pi }}{6}} \right) + i\sin \left( {4 \times \dfrac{{11\pi }}{6}} \right)} \right)$
Further simplifying, we get,
$ = 16\left( {\cos \left( {\dfrac{{22\pi }}{3}} \right) + i\sin \left( {\dfrac{{22\pi }}{3}} \right)} \right)$
Sine and cosine are periodic functions with fundamental period as $2\pi $.
$ = 16\left( {\cos \left( {\dfrac{{4\pi }}{3}} \right) + i\sin \left( {\dfrac{{4\pi }}{3}} \right)} \right)$
$ = 16\left( { - \dfrac{1}{2} - \dfrac{{\sqrt 3 }}{2}i} \right)$
$ = - 8 - 8\sqrt 3 i$

So, Using Demoivre’s theorem, we get the value of ${\left( { - \sqrt 3 - i} \right)^4}$ as $ - 8 - 8\sqrt 3 i$

Note: Demoivre’s Theorem can be used to evaluate expressions where the complex numbers are raised to powers. The argument of the resultant complex number is obtained by simply multiplying the principal argument of the complex number by the power and the modulus is obtained by raising it to the same power.