Determining the expression of $f(x,y)$ as a function of $z$

Question

I have an entire function $f:(x,y)\mapsto x^2-y^2-3y+(3x+2xy+1)i$ and I should express $f$ as a function of the complex variable $z$, what I did was to replace $x$ by $\frac{z+\bar z}{2}$ and $y$ by $i\frac{\bar z-z}{2}$ which gave me a quite "complicated" expression, however, here's what was done in the correction I have: let $x=\frac{z+1}{2}$ and $y=\frac{z-1}{2i}$ and so $f(z)=z^2+3iz+i$. My question is : aren't $x$ and $y$ real numbers (The real and imaginary parts of z) ? If so how can we substitue $\frac{z+1}{2}$ for $x$ and $\frac{z-1}{2i}$ for $y$ ?

Thank you for your answers !

Answer 1

$f(x, y)=x^2 - y^2 -3y+(3x +2xy+1)i=(x+iy)^2 +3(xi-y)+i$

$3(xi-y)=3i(xi-y)/i=-3(x+iy)/i=3(x+iy)i$

⇒$f(x, y)=x^2 - y^2 -3y+(3x +2xy+1)i=(x+iy)^2 + 3(x+iy)i +i$

$z= x+iy$ ⇒ $f(x, y)=x^2 - y^2 -3y+(3x +2xy+1)i= z^2 +3zi +i$

Where x and y are real numbers.

$\frac{z-1}{2i}$ in not real number, you can not take for y.

Answer 2

Let $x=z-iy$.

Carying out the computation will yield (If I have not done any mistake)

$f(z) = z^2 + i(1+3z)$