How do you simplify $\dfrac{{5 - 2i}}{{3 + 2i}}$?

Hint: First, multiply the numerator and denominator of $\dfrac{{5 - 2i}}{{3 + 2i}}$ by the conjugate of $3 + 2i$ to make the denominator real. Next, simplify the numerator by expanding $\left( {5 - 2i} \right)\left( {3 - 2i} \right)$ using the FOIL Method. Next, simplify the denominator by expanding $\left( {3 + 2i} \right)\left( {3 - 2i} \right)$ using algebraic identity. Next, Split the fraction obtained into two fractions.

Formula used:
${i^2} = - 1$
${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$

Complete step by step answer:
Given: $\dfrac{{5 - 2i}}{{3 + 2i}}$
First, multiply the numerator and denominator of $\dfrac{{5 - 2i}}{{3 + 2i}}$ by the conjugate of $3 + 2i$ to make the denominator real.
i.e., Multiply numerator and denominator of $\dfrac{{5 - 2i}}{{3 + 2i}}$ by $3 - 2i$.
$ \Rightarrow \dfrac{{5 - 2i}}{{3 + 2i}} \times \dfrac{{3 - 2i}}{{3 - 2i}}$
Combine.
$ \Rightarrow \dfrac{{\left( {5 - 2i} \right)\left( {3 - 2i} \right)}}{{\left( {3 + 2i} \right)\left( {3 - 2i} \right)}}$
Simplify the numerator.
Expand $\left( {5 - 2i} \right)\left( {3 - 2i} \right)$ using the FOIL Method.
Apply the distributive property.
$ \Rightarrow \dfrac{{5\left( {3 - 2i} \right) - 2i\left( {3 - 2i} \right)}}{{\left( {3 + 2i} \right)\left( {3 - 2i} \right)}}$
Apply the distributive property.
$ \Rightarrow \dfrac{{15 - 10i - 6i + 4{i^2}}}{{\left( {3 + 2i} \right)\left( {3 - 2i} \right)}}$
Put ${i^2} = - 1$.
$ \Rightarrow \dfrac{{15 - 10i - 6i + 4\left( { - 1} \right)}}{{\left( {3 + 2i} \right)\left( {3 - 2i} \right)}}$
Simplify and combine like terms.
$ \Rightarrow \dfrac{{11 - 16i}}{{\left( {3 + 2i} \right)\left( {3 - 2i} \right)}}$
Simplify the denominator.
Expand $\left( {3 + 2i} \right)\left( {3 - 2i} \right)$ using algebraic identity ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$.
$ \Rightarrow \dfrac{{11 - 16i}}{{{3^2} - {{\left( {2i} \right)}^2}}}$
Put ${i^2} = - 1$.
$ \Rightarrow \dfrac{{11 - 16i}}{{9 + 4}}$
Add $9$ and $4$, we get
$ \Rightarrow \dfrac{{11 - 16i}}{{13}}$
Split the fraction $\dfrac{{11 - 16i}}{{13}}$ into two fractions.
i.e., Separate pure and imaginary parts.
$ \Rightarrow \dfrac{{11}}{{13}} - \dfrac{{16}}{{13}}i$

Therefore, simplified version of $\dfrac{{5 - 2i}}{{3 + 2i}}$ is $\dfrac{{11}}{{13}} - \dfrac{{16}}{{13}}i$.

Additional information:
FOIL Method:
FOIL (the acronym for first, outer, inner and last) method is an efficient way of remembering how to multiply two binomials in a very organized manner.
To put this in perspective, suppose we want to multiply two arbitrary binomials, $\left( {a + b} \right)\left( {c + d} \right)$.
The first means that we multiply the terms which occur in the first position of each binomial.
The outer means that we multiply the terms which are located in both ends (outermost) of the two binomials when written side-by-side.
The inner means that we multiply the middle two terms of the binomials when written side-by-side.
The last means that we multiply the terms which occur in the last position of each binomial.
After obtaining the four partial products coming from the first, outer, inner and last, we simply add them together to get the final answer.$\left( {a + b} \right)\left( {c + d} \right) = a \cdot c + a \cdot d + b \cdot c + b \cdot d$
$ \Rightarrow \left( {a + b} \right)\left( {c + d} \right) = ac + ad + bc + bd$

Note: If $a$, $b$ are two real numbers, then a number of the form $a + ib$ is called a complex number.
If $z = a + ib$ is a complex number, then ‘$a$’ is called the real part of $z$ and ‘$b$’ is known as the imaginary part of $z$.
The square root of a negative real number is called an imaginary quantity or an imaginary number.
For any positive real number $a$, we have $\sqrt { - a} = \sqrt { - 1 \times a} = \sqrt { - 1} \times \sqrt a = i\sqrt a $.
If ${z_1} = a + bi$ and ${z_2} = c + di$ be two complex numbers. Then the multiplication of ${z_1}$with ${z_2}$ is denoted by ${z_1} \cdot {z_2}$ and is defined as the complex number $\left( {ac - bd} \right) + \left( {ad + bc} \right)i$.