Find the multiplicative inverse of the complex number given in the following: $4-3i$.

Hint: The multiplicative inverse of complex number$z=\dfrac{1}{z}$. We will rationalise the complex number first and then we will try to solve it further by using the formula, ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$ to find the inverse of the complex number $z$.

Complete step-by-step solution -
In the question we have been asked to find the multiplicative inverse of complex number $z$. Before we proceed with the question, let us find out about complex numbers and the meaning of multiplicative inverse of complex numbers. Complex numbers are those numbers that can be expressed in the form of $a+bi$. The terms $a$ and $b$ represent real numbers. $i$ is the solution of the equation ${{x}^{2}}=-1$. Since there are no real numbers that satisfy this condition, $i$ is termed as an imaginary number. Multiplicative inverse of complex numbers is simply the reciprocal of the number. In the question here we have $z=4-3i$. The multiplicative inverse of $z$ is given by $\dfrac{1}{z}$, so we get, $4-3i=\dfrac{1}{4-3i}$.
Rationalizing $\dfrac{1}{\left( 4-3i \right)}$ by multiplying by, $\dfrac{4+3i}{\left( 4-3i \right)}$, we get,
$\begin{align}
  & =\dfrac{1}{\left( 4-3i \right)}\times \dfrac{4+3i}{\left( 4-3i \right)} \\
 & =\dfrac{\left( 4+3i \right)}{\left( 4-3i \right)\left( 4+3i \right)} \\
\end{align}$
We know that $\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$, so in the denominator, we get,
$\begin{align}
  & =\dfrac{\left( 4+3i \right)}{{{\left( 4 \right)}^{2}}-{{\left( 3i \right)}^{2}}} \\
 & =\dfrac{4+3i}{16-9{{i}^{2}}} \\
\end{align}$
We know that ${{i}^{2}}=-1$ from basic complex numbers, so we get,
$\begin{align}
  & =\dfrac{4+3i}{16-9\left( -1 \right)} \\
 & =\dfrac{4+3i}{16+9} \\
 & =\dfrac{4+3i}{25} \\
\end{align}$
Splitting into the real and complex numbers, we get,
$=\dfrac{4}{25}+\dfrac{3i}{25}$
Therefore, the multiplicative inverse of the given complex number $\left( 4-3i \right)=\left( \dfrac{4}{25}+\dfrac{3i}{25} \right)$.

Note: Alternatively, we can also find the multiplicative inverse of any complex number $z$ by using the direct formula of ${{z}^{-1}}=\dfrac{{\bar{z}}}{{{\left| z \right|}^{2}}}$. We have $z=3-4i$, and we also have $\bar{z}=3+4i$ and ${{\left| z \right|}^{2}}={{4}^{2}}+{{\left( -3 \right)}^{2}}=16+9=25$. Substituting the values, we get, ${{z}^{-1}}=\dfrac{4+3i}{25}=\dfrac{4}{25}+\dfrac{3i}{25}$.