Question

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

Answer 1

`-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`

Answer 2

This one is reasonably simple: multiply top and bottom by `-i/3`. Doing this and rearranging leads to the simplified answer `-2-8/3i`.

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 `-i/3`. Multiplying by (`-i/3`)/( `-i/3`) is the same thing as multiplying by 1, which means the simplified number we end up with is the same number we started with.

`(8-6i)/(3i)*(-i/3)/(-i/3) = (-8/3i-2)/1`

Written in more traditional form, this is `-2-8/3i`.