Question

Compute the special products. (-3+5i)2

Answer 1

we are given

`(-3+5i)^2`

we can expand it

`(-3+5i)^2=(-3+5i)(-3+5i)`

now, we can FOIL it

`(-3+5i)^2=-3* -3-3* 5i+5i* -3+5i* 5i`

now, we can simplify it

`(-3+5i)^2=9-15i-15i+25i^2`

we know that

`i^2=-1`

so, we can plug it

`(-3+5i)^2=9-30i+25(-1)`

`(-3+5i)^2=9-30i-25`

`(-3+5i)^2=-16-30i`..............Answer