Is a differentiable complex function with differentiable complex conjugate also constant? I'm asked to prove that if $f=u+iv$ and $f^*=u-iv$ are complex differentiable (and therefore follow the Cauchy-Riemann conditions), that $f$ is therefore constant.
The proof seems simple and I did it simply defining $u^*=u$ and $v^*=-v$, applying the CR conditions which eventually results in both $u$ and $v$ being constant, meaning $f$ has to also be constant.
The problem then is that this result does not seem intuitive in the slightest. Consider $f(z)=z=x+iy$. It's a continuous function with continuous partial derivatives that hold the CR conditions, so according to the Loomen-Menchoff theorem it should be holomorphic. However, it is not constant. So what is going on?
Edit: I just realized my example is wrong because $z*$ does not follow the CR conditions. Either way, the result is still a bit unintuitive. Can someone explain it a bit more deeply?
 A: A function $f$ being complex differentiable at a point $p$ means that $f(z)$ is locally modelled on complex multiplication at $p$. Complex multiplication by $re^{i\theta}$ is described by scaling the plane by $r$ then rotating by $\theta$. In particular, nonzero complex multiplication preserves orientation (i.e. clockwise and counter-clockwise curves are mapped to clockwise and counter-clockwise curves, respectively).
Conjugation interchanges orientation, and so if $f$ is holomorphic at $p$ with nonzero derivative then $\overline{f}$ will map infinitesimal clockwise and counterclockwise curves to curves going in opposite directions, which prevents $\overline{f}$ from being complex differentiable. Thus if both $f$ and $\overline{f}$ are complex differentiable, the only option is that $f$ have derivative $0$ everywhere.
This can be made into a symbolic argument by looking at the form of real matrices of the differential of $f$ and $\overline{f}$ and comparing with what the Cauchy-Riemann equations imply.
