# How do you simplify #(8 - 6i)/(3i)#?

##### 2 Answers

# -2 -8/3 i #

#### Explanation:

require the denominator to be real. To achieve this multiply numerator and denominator by 3i

# 3i xx 3i = 9i^2# [ remember

# i^2 = (sqrt-1)^2 =-1 # ]

# rArr (8-6i)/(3i )= (3i(8-6i))/(3i .3i) =( 24i - 18i^2)/(9i^2)#

# =( 24i + 18)/-9 = 18/-9 +( 24i)/-9 = -2 -8/3 i #

This one is reasonably simple: multiply top and bottom by

#### Explanation:

Since there is a single complex number - and only the imaginary part of it - on the bottom row, I decided to multiply top and bottom of the fraction by a number that will yield a bottom line of 1. That number is

Written in more traditional form, this is