How do you simplify $\dfrac{{2 - 3i}}{{3 + 4i}}$?

Hint: To solve this complex number, take complex conjugate and proceed with normal simplification. And remember some basic formula in the topic complex number like ${i^2} = - 1$. This will help you to solve this problem.

Complete step by step answer:
Let us consider the given solution,
$ \Rightarrow \dfrac{{2 - 3i}}{{3 + 4i}}$
Let us take complex conjugate for the above question, to take complex conjugate for a given number, we have to consider the denominator part and have to change the sign near the imaginary part and multiply and divide in the above question we get,
$
\Rightarrow \dfrac{{2 - 3i}}{{3 + 4i}} = \dfrac{{2 - 3i}}{{3 + 4i}} \times \dfrac{{3 - 4i}}{{3 - 4i}} \\
    \\
 $
Now use the formula ${a^2} - {b^2} = (a + b)(a - b)$ in the above equation and we get,
$ \Rightarrow \dfrac{{2 - 3i}}{{3 + 4i}} = \dfrac{{6 - 8i - 9i + 12{i^2}}}{{9 - 12i + 12i - 16{i^2}}}$
We can perform operation real part with real part and imaginary part with imaginary part only i.e., we cannot add real number with imaginary number We know that ${i^2} = - 1$, substituting the value we get,
$ \Rightarrow \dfrac{{2 - 3i}}{{3 + 4i}} = \dfrac{{6 - 17i - 12}}{{9 + 16}} = \dfrac{{ - 6 - 17i}}{{25}} = \dfrac{{ - 6}}{{25}} + \dfrac{{ - 17}}{{25}}$
This is our required solution.

Additional information: Some of the properties of complex numbers are
1) If we add two complex conjugate numbers, it will give a real number.
2) And also if we multiply two complex conjugate numbers, it will give a real number.
3) The complex conjugate for $z = a + ib$ will be \[\bar z = a - ib\].
4) It obeys both addition and multiplication commutative property.

Note: The complex number is expressed in terms of $a + ib$, where $i$ is the imaginary number and the other two variables are the real number. When we come up with the solution $x = \sqrt { - 1} $, there is no solution for this in real numbers and here we consider the complex number as $x = i$. Complex numbers will always have real and imaginary parts.