Did you mean
 
(1 + z²)z or 1+z²z?

If you meant (1 + z²)z, then let z = x + yi where x and y are real
(1 + z²)z
[1 + (x+yi)²](x+yi)
[1 + x²+2xyi+y²i²](x+yi)
Change i² to -1
[1 + x²+2xyi+y²(-1)](x+yi)
(1 + x²+2xyi-y²)(x+yi)
Separate real terms from imaginary terms
[(1+x²-y²)+2xyi](x+yi)
USE FOIL:
(1+x²-y²)x+(1+x²-y²)yi+2x²yi+2xy²i²
x+x³-xy²+yi+x²yi-y³i+2x²yi+2xy²(-1)
Change i² to -1
x+x³-xy²+yi+x²yi-y³i+2x²yi-2xy²
Separate real terms from imaginary terms
(x+x³-xy²-2xy²)+(yi+x²yi-y³i+2x²yi)
Combine like terms:
(x+x³-3xy²)+(yi+3x²yi-y³i)
Factor out i:
(x+x³-3xy²)+(y+3x²y-y³)i
Factor out common factors:
x(1+x²-3y²)+y(1+3x²-y²)i
So the real part is x(1+x²-3y²)

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If you meant 1+z²z
 
1+z²z
1+z³
(1+z)(1-z+z²)
(1+x+yi)[1-(x+yi)+(x+yi)²]

Go through that just as I did above and get:

x³-3xy²+1+(3x²y-y³)i

and the real part is x³-3xy²+1

Edwin