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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?

agaminon
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    $z$ is entire and its derivative is $1$. On the other hand, $\overline z$ is somewhat the archetype of a real-analytic, non-holomorphic function $\Bbb R^2\to\Bbb R^2$. – Sassatelli Giulio May 27 '22 at 17:33

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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.

Alex Nolte
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  • Using this notion of complex differentiation I suppose it makes more intuitive sense, but I've never hears of it being described this way. Where can I read more about it? – agaminon May 28 '22 at 14:37
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    It's hiding in the normal definition! This is what it means for the difference quotient to have limit a complex number. I'm not aware of a first course in complex analysis relying heavily on this perspective (it's great for building intuition but less useful for building formal proofs), but I'd expect it to be touched on in sections of good textbooks on conformal mappings. Geometric perspectives on quasiconformal mappings tend to emphasize equivalent perspectives too, I think. – Alex Nolte May 28 '22 at 15:32
  • @agaminon If you mean just the "locally modeled on multiplication by a complex number", there are probably several derivations on this site. One is sort of contained in this answer of mine. – Mark S. May 28 '22 at 16:18