Indentify the plane defined by $Re(z\overline{z})=\frac{lm(\overline{z})}{2}$ I tried:
$$Re(z\overline{z})=\frac{lm(\overline{z})}{2} \Leftrightarrow \\
Re((x+yi)(x-yi))=\frac{lm(x-yi)}{2} \Leftrightarrow \\
Re(x^2+y^2)= \frac{lm(x-yi)}{2} \Leftrightarrow \\
x^2 = \frac{-y}{2} \Leftrightarrow \\
-2x^2 = y$$
But my book states:

Circunference of center $(0,-\frac{1}{4})$ and radius $\frac{1}{4}$.

What went wrong?

Let $z=x+iy$ then $z\bar{z} = (x^2 + y^2)  + 0i$ . 
So $\Re(z\bar{z}) = x^2 + y^2$. Similarly $\bar{z} = x-iy$ so $\Im(\bar{z})=-y$. Hence the required plane is $$x^2 + y^2 = -\frac{y}{2} \iff x^2 + \left(y + \frac{1}{4}\right)^2 = \left(\frac{1}{4}\right)^2$$ which is a circle centered at $(0, -1/4)$ and radius $1/4$. 

Recall that whilst $\Re(z) = x$, it need not be the case that $\Re(z\bar{z}) = x^2$. In fact, if you let $z\bar{z} = u+iv$  you see that $u = x^2 + y^2$ and $v=0$. Then clearly $\Re(z\bar{z}) = u = x^2 + y^2$.
