Question

How do you divide `(3i)/(1+i) + 2/(2+3i)`?

Answer 1

`27/26i+47/26`

Explanation:

`(3i)/(1+i)+2/(2+3i)`
`=(3i(1-i))/((1+i)(1-i))+(2(2-3i))/((2+3i)(2-3i)`
`=(3i-3i^2)/(1-i^2)+(4-6i)/(4-9i^2)`
`=(3i-3(-1))/(1-(-1))+(4-6i)/(4-9(-1))` [as, `i^2=-1` ]
`=(3i+3)/(1+1)+(4-6i)/(4+9)`
`=(3i+3)/2+(4-6i)/13`
`=3/2i+3/2+4/13-6/13i`
`=27/26i+47/26`

Answer 2

`47/26+27/26i`

Explanation:

Get a common denominator.

`=(3i(2+3i))/((1+i)(2+3i))+(2(1+i))/((1+i)(2+3i))`

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

`=(2+8i+9i^2)/(2+5i+3i^2)`

This can be simplified since `i^2=-1`.

`=(2+8i-9)/(2+5i-3)`

`=(-7+8i)/(-1+5i)`

Now, multiply by the complex conjugate of the denominator.

`=(-7+8i)/(-1+5i)((-1-5i)/(-1-5i))`

`=(7+27i-40i^2)/(1-25i^2)`

`=(7+27i+40)/(1+25)`

`=(47+27i)/26`

`=47/26+27/26i`