Question

How do you divide `(-2-5i)/(3i)`?

Answer 1

Solution is `-5/3+(2/3)i`

Explanation:

To solve `(−2−5i)/(3i)`

multiply numerator and denominator by `i` (note `i^2=-1`)

and we get

`(i*(−2−5i))/(i*(3i)` or `(-2i-5i^2)/(3i^2)`

or `(-2i-5(-1))/(3*-1)` i.e. `(5-2i)/-3`, which is equivalent to

`-5/3+(2/3)i`

Answer 2

`2/3 i - 5/3`

Explanation:

The denominator of the fraction is required to be real. To achieve this multiply the numerator and denominator by 3i.

`(-2 - 5i )/(3i) xx( 3i)/(3i )`

`= (-6i - 15i^2)/(9i^2 )`

[Note: `i^2 = (sqrt(-1))^2 = -1`]

`= (-6i + 15)/-9 = 2/3 i - 5/3`