How do you simplify \frac { 8+ 6i } { 3- 2i }8+6i32i?

1 Answer
Dec 2, 2017

(12+34i)/1312+34i13

Explanation:

In order to simplify this, you need to make sure ii isn't in the denominator (or bottom) of the fraction. To do this, multiply both the numerator and denominator by the conjugate of 3-2i32i, which is 3+2i3+2i. When doing this, make sure to use the FOIL method:

1)

((8+6i) * (3+2i))/((3-2i) * (3+2i))(8+6i)(3+2i)(32i)(3+2i)

2)

(24+16i+18i-12)/(9+6i-6i-4i^2)24+16i+18i129+6i6i4i2

Since ii equals sqrt(-1)1, i^2i2 equals -11.

3)

(12+34i)/(9+4)12+34i9+4

4)

(12+34i)/(13)12+34i13