Calculate the value of the following expression:
f(z)=-z3+3z2+z+2f(3+2i)
f(z)=-z3+3z2+z+2f(3+2i)=?
f(3+2i)
=-3+2i3+33+2i2+3+2i+2
=-27+3·32·2i+3·3·2i2+2i3+3·9+2·3·2i+2i2+3+2i+2
=-27+54i+9·4i2+8i3+3·9+12i+4i2+3+2i+2
=-27-54i-36-1-8-i+27+36i+12(-1)+3+2i+2
=+36-12+3+2+(-54+8+36+2)i
=29-8i
f(3+2i)=29-8i
Imaginary number
i=-1
The cube of the binomial
( a + b ) 3 = ( a 3 + 3 a 2 b + 3 a b 2 + b 3 )