Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Physicist with a light background in mathematics here. I am studying the conformal group in 2d, and facing the result that a 2d conformal transformation (a smooth transformation on the coordinates that preserves the angles) is nothing else than an holomorhic mapping: $$ z \rightarrow f(z) \\ \bar{z} \rightarrow \bar{f}(\bar{z})$$ with $f$ holomorphic. These transformation form a group and so need to be invertible leading to the conclusion that all holomorphic functions are invertible. This seems really powerful (maybe too much to be true?) and totally new to me, I heard nothing like that in my superficial introductory lectures to complex analysis. My prudence and semi intuition are telling me that this could only be true locally (what would be the radius of definition of the inverse function?).

I tried to look at it under the expansion of the function in its Laurent series around a certain $z_0$: $$f(z) = \sum_{n>0}\alpha_n(z-z_0)^n$$ Looking for the inverse of a certain $f(z_1)$ (in the radius of convergence of the expansion) means finding the z satisfying: $$g(z) = \sum_{n>0}\alpha_n(z-z_0)^n -f(z_1)=0$$. (The root of a polynome on $\cal{C}$) Although I have no idea how to prove the existence and uniqueness of the said solution.

share|improve this question

2 Answers 2

up vote 3 down vote accepted

A holomorphic function $f(z)$ is conformal at $a$ only if it is a local isomorphism i.e. if its derivative does not vanish at $a$ . For example $e^z$ is conformal at every $a\in \mathbb C$.
Let me insist that the function needn't be invertible (i.e. injective) in order to be conformal, as witnessed by the example above of the exponential.

The holomorphic function $sq(z)=z^2$ however is not conformal at $a=0$ since it doubles the angles there, as witnessed by the formula $sq(\epsilon e^{i\theta})=\epsilon^2e^{2i\theta}$.
The tell-tale sign of this deficiency is that the derivative $sq'(z)=2\cdot z$ vanishes at zero: $sq'(0)=2\cdot 0=0$.
At any $a\neq0$ however the function $z^2$ is conformal since $sq'(a)=2\cdot a\neq0$.

share|improve this answer

The constant function $z \to a$ for some $a \in \mathbb{C}$ is holomorphic on the whole complex plane, yet obviously not invertible. Polynomials are also holomorphic on the whole complex plane, yet they cannot be invertible if they have more than one zero (since that means more than one value is mapped to zero). If they have degree $\geq 2$, they're in general also not invertible locally, because then their derivative will have zeros too.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.