I wish to show that the following expression is a parabola: $$(a\bar{z}+\bar{a}z)^2=2(b\bar{z}+\bar{b}z)+c$$ with $a,b,c$ complex numbers. Whenever I face such an expression, I usually try to use $z=x+yi$ and try to get to the Cartesian form of whatever region it is. However in this case, after expanding the quadratic term:
$$a^2\bar{z}^2+2|a|^2|z|^2+\bar{a}^2z^2=2(b\bar{z}+\bar{b}z)+c$$ I don't see how replacing $z$ by $z=x+yi$ would help me proceed in any way. any hints?
Note that for any complex numbers $u, v$, the sum of "cross-conjugate product pairs" is twice the real part of one cross-conjugate product, i.e.
$$u\bar{v}+v\bar{u}=2\Re(u\bar{v})=2\Re (\bar{u}v)$$
Let $a=p+iq$ and $b=m+in$.
Hence $$a\bar{z}+\bar{a}z=2\Re (\bar{a}z)=2\Re((p-iq)(x+iy))=px+qy\\ b\bar{z}+\bar{b}z=2\Re (\bar{b}z)=2\Re((m-in)(x+iy))=mx+ny$$
Given equation:
$$\begin{align} (a\bar{z}+\bar{a}z)^2&=2(b\bar{z}+\bar{b}z)+c\\ (px+qy)^2&=2(mx+ny)+c\\ (px+qy)^2-2mx-2ny-c&=0\end{align}$$ which is the general equation for a parabola in the form $$(Ax+Cy)^2+Dx+Ey+F=0$$
See also the wiki entry on Conic sections.