Question

Find the multiplicative inverse of $2 - 3i$.

Answer 1

Here, we need to find the multiplicative inverse of $2 - 3i$. The multiplicative inverse of a number is its reciprocal. We will first find the reciprocal of the given number. Then, we will use rationalisation and algebraic identities to simplify the expression, and get the required value of the multiplicative inverse.

Formula Used: The product of the sum and difference of two numbers is given by the algebraic identity $\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}$.

Complete step-by-step answer:
If two numbers $x$ and $y$ are such that their product is equal to 1, then $x$ and $y$ are called the multiplicative inverse of each other.
For example, the multiplicative inverse of 2 is $\dfrac{1}{2}$.
Now, we will find the multiplicative inverse of $2 - 3i$.
The multiplicative inverse of $2 - 3i$ is the reciprocal of $2 - 3i$.
Therefore, we get
Multiplicative inverse of $2 - 3i$ is $\dfrac{1}{{2 - 3i}}$ .
We will simplify the expression to get the required answer.
Rationalising the denominator by multiplying and dividing by $2 + 3i$, we get
$\begin{array}{l} \Rightarrow \dfrac{1}{{2 - 3i}} = \dfrac{1}{{2 - 3i}} \times \dfrac{{2 + 3i}}{{2 + 3i}}\\\ \Rightarrow \dfrac{1}{{2 - 3i}} = \dfrac{{2 + 3i}}{{\left( {2 - 3i} \right)\left( {2 + 3i} \right)}}\end{array}$
We know that the product of the sum and difference of two numbers is given by the algebraic identity $\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}$.
Substituting $a = 2$ and $b = 3i$ in the algebraic identity, we get
$\Rightarrow \left( {2 - 3i} \right)\left( {2 + 3i} \right) = {2^2} - {\left( {3i} \right)^2}$
Simplifying the expression, we get
$\Rightarrow \left( {2 - 3i} \right)\left( {2 + 3i} \right) = 4 - 9{i^2}$
We know that ${i^2}$ is equal to $- 1$.
Therefore, substituting in the expression, we get
$\Rightarrow \left( {2 - 3i} \right)\left( {2 + 3i} \right) = 4 - 9\left( { - 1} \right)$
Multiplying the terms of the expression, we get
$\Rightarrow \left( {2 - 3i} \right)\left( {2 + 3i} \right) = 4 + 9$
Adding 4 and 9, we get
$\Rightarrow \left( {2 - 3i} \right)\left( {2 + 3i} \right) = 13$
Now, substituting $\left( {2 - 3i} \right)\left( {2 + 3i} \right) = 13$ in the equation $\dfrac{1}{{2 - 3i}} = \dfrac{{2 + 3i}}{{\left( {2 - 3i} \right)\left( {2 + 3i} \right)}}$, we get
$\Rightarrow \dfrac{1}{{2 - 3i}} = \dfrac{{2 + 3i}}{{13}}$
Rewriting the expression in the form $a + bi$, we get
$\Rightarrow \dfrac{1}{{2 - 3i}} = \dfrac{2}{{13}} + \dfrac{3}{{13}}i$
$\therefore$ We get the multiplicative inverse of the number $2 - 3i$ as $\dfrac{2}{{13}} + \dfrac{3}{{13}}i$.

Note: The number $2 - 3i$ is a complex number. A complex number is a number which can be written in the form $a + bi$, where $a$ and $b$ are real numbers, and $i$ is the imaginary unit. Here, $i = \sqrt { - 1}$, which is not real. This is why we substituted ${i^2} = - 1$.