Let z be in the complex domain. Then define $\hat{z} = e^{j\theta}z$, where $\theta$ is a constant. If the derivative is taken, by using the fact that rotation won't change the area (area preserving operation), can we say that $d\hat{z} = dz $?
Thank you very much.
Treat this as an extended comment, not really an answer.
I think you are confused here. Let $Z=Cz$ with $C\in \mathbb{C}$. Then $dZ = Cdz$ and this has nothing to do with "area". This is a simple differential (more along the lines of length, than area).
If you are actually interested in area, then $dx dy$ is what you are looking for. Then writing $z=x+iy$, $Z=X+iY$ and $C=a+ib$, you get $$ X+iY=(ax-by)+i(ay+bx) $$ Then using a Jacobian $$ dX dY = |\begin{vmatrix}a & -b\\ b & a\end{vmatrix}| dxdy = (a^2+b^2) dxdy = |C|^2 dxdy $$ Clearly, if $C=e^{i\theta}$, then the area element remains unchanged (as it should be).
Alternatively, if you insist on using $dz$ in relation to area, the area element is $dz d\overline{z}$. From which you can again see that $dZd\overline{Z}=|C|^2 dz d\overline{z}$.