# How do you simplify 4/(2-3i)?

May 7, 2018

Via "razionalization"

#### Explanation:

When dividing by imaginari numbers, you can do something similar to rationalization, when you need to get rid of roots in the denominator.

Using the identity $\left(a - b\right) \left(a + b\right) = {a}^{2} - {b}^{2}$, we have

$\setminus \frac{4}{2 - 3 i} = \setminus \frac{4}{2 - 3 i} \setminus \frac{2 + 3 i}{2 + 3 i} = \setminus \frac{4 \left(2 + 3 i\right)}{\left(2 + 3 i\right) \left(2 - 3 i\right)}$

The numerator is a simple product:

$4 \left(2 + 3 i\right) = 8 + 12 i$

At the denominator we have the identity discussed before:

$\left(2 + 3 i\right) \left(2 - 3 i\right) = {2}^{2} - {\left(3 i\right)}^{2} = 4 - 9 {i}^{2} = 4 + 9 = 13$

Now the denominator is a simple real number, and we have

$\setminus \frac{4}{2 - 3 i} = \setminus \frac{8 + 12 i}{13} = \setminus \frac{8}{13} + \setminus \frac{12}{13} i$