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

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Answer 1 · 2018-05-07T17:41:25

Via "razionalization"

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 #(a-b)(a+b) = a^2-b^2#, we have

#\frac{4}{2-3i} = \frac{4}{2-3i}\frac{2+3i}{2+3i} = \frac{4(2+3i)}{(2+3i)(2-3i)}#

The numerator is a simple product:

#4(2+3i) = 8+12i#

At the denominator we have the identity discussed before:

#(2+3i)(2-3i) = 2^2 - (3i)^2 = 4-9i^2 = 4+9=13#

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

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