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

$(-3+5i)^2=-3\times -3-3\times 5i+5i\times -3+5i\times 5i$

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