Question 29568
(4-i)^2+2(3-2i)-4 
=[(4)^2+2X(4)X(-i)+(-i)^2]+[2X3-2X(3i)]-4
[using (a+b)^2 = a^2 +2ab+b^2 where here a = 4 and b= (-i)]
=[16-8i+i^2]+6-4i-4
=(16-8i-1)+6-4i-4
=(16-1+6-4)+(-8i-4i)  (grouping like terms)
=17-12i
Answer: (4-i)^2+2(3-2i)-4 = 17-12i
Note:(4-i)^2 can also be expanded using 
(a-b)^2 = a^2-2ab+b^2  where here a=4,b= i and therefore 
a^2 = 4^2 = 16, -2ab = -2X4x(i) = -8i and b^2 = (i)^2 = i^2 = -1 so that 
(4-i)^2 =4^2-2X4x(i)+ (i)^2  =16-8i-1