Simplify ${\left( { - 16} \right)^{\dfrac{{ - 1}}{2}}}$?

Hint:In the given problem, we need to evaluate the inverse of square root of a given number. The given question requires knowledge of the concepts of complex numbers and square roots. The square root of a negative number is always a complex number. Hence, we must have in mind the definition of complex numbers and their basic properties.

Complete step by step answer:
In the question, we need to evaluate the value of ${\left( { - 16} \right)^{\dfrac{{ - 1}}{2}}}$ . For evaluating the inverse of the square root of $16$, we need to have an idea of complex numbers, their properties and how to do basic operations on the complex numbers set.

So, ${\left( { - 16} \right)^{\dfrac{{ - 1}}{2}}}$

$ = $${\left( { - \dfrac{1}{{16}}} \right)^{\dfrac{1}{2}}}$

${\left( { - \dfrac{1}{{16}}} \right)^{\dfrac{1}{2}}}$ can be represented as a product of two entities: $\left( { - 1} \right)$ and $\left( {\dfrac{1}{{16}}} \right)$. While splitting ${\left( { - \dfrac{1}{{16}}} \right)^{\dfrac{1}{2}}}$ into two entities, we need to distribute the power to both of them.

$ = $${\left( { - 1} \right)^{\dfrac{1}{2}}}{\left( {\dfrac{1}{{16}}} \right)^{\dfrac{1}{2}}}$
Now, we know that the square root of $16$ is $4$ since square root of any number is always positive.

$ = $${\left( { - 1} \right)^{\dfrac{1}{2}}}\left( {\dfrac{1}{4}} \right)$

Now, we know that ${\left( { - 1} \right)^{\dfrac{1}{2}}} = \sqrt { - 1} $ is known as iota and is represented as $i$. So, the given expression can be represented as:

$ = $ $i\left( {\dfrac{1}{4}} \right) = \dfrac{i}{4}$

So, ${\left( { - 16} \right)^{\dfrac{{ - 1}}{2}}}$ can be simplified as $\left( {\dfrac{i}{4}} \right)$ .

Note: The given question involves solving the square root of a negative number and that’s where the set of complex numbers comes into picture and plays a crucial role in mathematics.
The answer can be verified by working the solution backwards and observing that square of $\left( {\dfrac{i}{4}} \right)$ is $\left( { - 16} \right)$