Question

Please help!! If (2 + 3i)^2 + (2 − 3i)^2 = a + bi, a = (blank) and b = (blank).

Answer 1

(2+3i)^2 = (2+3i)*(2+3i)
(2+3i)^2 = 2*(2+3i)+3i*(2+3i)
(2+3i)^2 = 2*(2)+2*(3i)+3i*(2)+3i*(3i)
(2+3i)^2 = 4+6i+6i+9i^2
(2+3i)^2 = 4+6i+6i+9(-1)
(2+3i)^2 = 4+6i+6i-9
(2+3i)^2 = -5+12i
-----------------------------------------
(2-3i)^2 = (2-3i)*(2-3i)
(2-3i)^2 = 2*(2-3i)-3i*(2-3i)
(2-3i)^2 = 2*(2)+2*(-3i)-3i*(2)-3i*(-3i)
(2-3i)^2 = 4-6i-6i+9i^2
(2-3i)^2 = 4-6i-6i+9(-1)
(2-3i)^2 = 4-6i-6i-9
(2-3i)^2 = -5-12i
-----------------------------------------
(2+3i)^2 + (2-3i)^2 = (2+3i)^2 + (2-3i)^2
(2+3i)^2 + (2-3i)^2 = -5+12i + (-5-12i)
(2+3i)^2 + (2-3i)^2 = -5+12i - 5-12i
(2+3i)^2 + (2-3i)^2 = -10+0i

The expression is now in a+bi form where a = -10 and b = 0