Evaluate and write in standard form: (-3 + 2i)^2 - 3(3 - i)(-2 + 2i),

(-3 + 2i)^2 -
3(3-i)(-2+2i)

We need to write into the standard form z= a+
bi

Let us open the
brackets.

==> (-3+2i)^2 = (-3)^2 +2*-3*2i +
4i^2

But i^2 = -1

==>
(-3+2i)^2 = 9 - 12i - 4 = 5- 12i

(3-i)(-2+2i) = -6 + 6i +
2i -2i^2 = -6 + 8i +2 = -4 +8i

==> 5-12i - 3(-4+8i)
= 5-12i + 12 - 24i = 17 - 36i

Then the answer is
:

z= 17 -
36i