Question

How do you simplify $\dfrac{1}{\left( 2i \right)^{3}}$ ?

Answer 1

In this question, we need to simplify and find the value of $\dfrac{1}{\left( 2i \right)^{3}}$ .The given expression consists of complex numbers. Mathematically, complex numbers are represented as $x\ + \ iy$ where $x$ and $y$ are the real numbers and here $i$ is an imaginary number. Firstly, we need to expand the denominator of the given expression. Then we can substitute the value of the imaginary number . The value of the imaginary number $i^{2}$ equals the minus of $1$ . Then we need to multiply both the numerator and denominator by $i$, then again by substituting the value of the imaginary number , we can simplify the given expression.

Complete step by step answer:
Given, $\dfrac{1}{\left( 2i \right)^{3}}$
On simplifying the denominator ,
We get,
$\Rightarrow \dfrac{1}{\left( 2 \right)^{3}\left( i \right)^{3}}$
Here $i^{3}$ can be written as $i^{2} \times i$ ,
$\Rightarrow \dfrac{1}{8 \times \left( i \right)^{2} \times (i)}$
We know that $i^{2} = - 1$
Thus we get,
$\Rightarrow \dfrac{1}{8 \times ( - 1) \times (i)}$
On further simplifying,
We get,
$\dfrac{1}{\left( 2i \right)^{3}} = - \dfrac{1}{8i}$

On multiplying both the numerator and denominator by $i$ ,
We get,
$\Rightarrow-\dfrac{1}{8i}\times\dfrac{i}{i}$
On simplifying,
We get,
$\dfrac{-i}{8i^{2}}$
We know that $i^{2}=-1$ ,
$\dfrac{-i}{8(-1)}$
On simplifying,
We get,
$\dfrac{1}{(2i)^{3}} =\dfrac{i}{8}$
We also know that the value of $i=\sqrt-1$
Thus we get,
$\dfrac{1}{(2i)^{3}} =\dfrac{\sqrt-1}{8}$

Thus the value of $\dfrac{1}{(2i)^{3}}$ is equal to $\dfrac{\sqrt-1}{8}$.

Note: The set of complex numbers is basically denoted by $C$. To solve these types of questions we need to know the value of the imaginary number. The term $i$ is known as an imaginary number and it is called iota as it has the value of $\sqrt-1$ . We also know the value of $i^{2} = - 1$ . Example for complex numbers is $(2 + 3i)$ . Complex number consists of two parts namely the real part and the imaginary part. Imaginary part is denoted by Im(z) and the real part is denoted by Re(z). It is the sum of real numbers and imaginary numbers. It also helps to find the square root of negative numbers. And the imaginary number $i$ leads to another topic that is the complex plane.