Why does conformal transformation at require the function to be analytic at point z instead of differentiable?
If $f(z)$ is differentiable at point $z_0$ and $f'(z_0)\ne 0$,then we could also get the angle-preserving property and result that Scale factor is constant.Because when we proved the angle-preserving property,we got that the angle is constant because the $\arg f(z_0)$ is the angle.So did Scale factor.
So why does conformal transformation require the function to be analytic at point $z_0$ instead of differentiable?
I'm confused with that.I hope you could help me.Thank you!
The geometric intuition about a conformal transformation is that it locally "looks like" a similarity transformation.
This is not necessarily the case if the function is just complex differentiable (with nonzero derivative) at a single isolated point -- such a function might not even be injective on any neighborhood of that point!
"Conformal" is really only interesting when it's true in an entire open set -- and in that case, when we're talking about the complex plane, "differentiable" and "analytic" are well known to be synonyms anyway. You can probably find texts that use either wording in their definitions. In particular, "conformal" makes sense for $\mathbb R^n$ in general rather than just $\mathbb C$, and texts that want to emphasize this connection are likely to select definitions that generalize more directly than "analytic" or "homeomorphic".