If $f$ and $\overline{f^2}$ are differentiable complex functions, can $f$ have more than one value?
I can't even find any 'fancy' $f$ such that $f$ and $\overline{f^2}$ are differentiable complex functions. On the other hand, if we take $f=const$ or $f=i\cdot const$ so $f$ is differentiable everywhere but this function always has one value.
I was thinking of finding out by Cauchy-Riemann equations like this:
$$f(z)=u(x,y)+i \cdot v(x,y)$$ $$\overline{{f^2(z)}}=u^2(x,y)-v^2(x,y)-2i\cdot u(x,y) \cdot v(x,y)$$ $$\text{Let} \quad a(x,y)=u^2(x,y)-v^2(x,y)\quad \text{and}\quad b(x,y)=-2\cdot u(x,y) \cdot v(x,y)$$ By Cauchy-Riemann equations we have $$\begin{cases} a_x=2uu_x-2vv_x \\ b_y=-2(u_y+v_y) \\ a_y=2uu_y-2vv_y \\ b_x=-2(u_x+v_x) \end{cases}$$
$$\begin{cases} uu_x-vv_x=-(u_y+v_y) \\ u_x+v_x=uu_y-vv_y \end{cases}$$
And here I stoped for a while. Even wolframalpha cannot solve this system. Can anyone help me?
$f$ holomorphic $\implies f^2$ holomorphic $\implies \Re(f^2) =\frac{f^2 +\bar{f^2}}2$ holomorphic. But holomorphic and real implies locally constant. Using C-R equations, $\Im(f^2)$ is locally constant...