Question

How do you simplify `(1+i)^2/(-3+2i)^2`?

Answer 1

`(2i)/(5 -12i)`

Explanation:

`(1 + i)^2/(-3 + 2i)^2 = (1 + 2i + i^2)/(9 - 12i +4i^2)`

`i^2 = -1`, therefore

`(1 + 2i + i^2)/(9 - 12i +4i^2) = (1 + 2i +(-1))/(9 - 12i +4(-1))`

`= (1 + 2i -1)/(9 - 12i - 4) = (2i)/(5 -12i)`

Answer 2

The answer is `color(red)((2i)/(5-12i))`.

Explanation:

According to problem(ATP),
`((1+i)^2)/((-3+2i)^2)`
=`(1-1+2i)/(4(i)^2-12i+9)`[as `i^2=-1`]
=`(2i)/(5-12i)`

Answer 3

`(1+i)^2/(-3+2i)^2 = -24/169+10/169i`

Explanation:

Given:

`(1+i)^2/(-3+2i)^2`

Let us first simplify:

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

and then square it...

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

`color(white)((1+i)/(-3+2i)) = (-3-2i-3i-2i^2)/(9-4i^2)`

`color(white)((1+i)/(-3+2i)) = (-3-2i-3i+2)/(9+4)`

`color(white)((1+i)/(-3+2i)) = (-1-5i)/13`

So:

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

`color(white)((1+i)^2/(-3+2i)^2) = ((-1-5i)/13)^2`

`color(white)((1+i)^2/(-3+2i)^2) = (-24+10i)/169`

`color(white)((1+i)^2/(-3+2i)^2) = -24/169+10/169i`