QUESTION: If a function, $f:\mathbb{C}\rightarrow \mathbb{C}$, is entire and it is non constant is it necessarily locally injective? That is if given some $z_0$, does there exists a disk, $\mathbb{D}_i$ centered around $z_0$ such that $f:\mathbb{D}_i\rightarrow f(\mathbb{D}_i)$ is bijective? If it is so, could someone please provide an explanation why it is so. If it is not true, could you please provide a counterexample if you have the time. Thanks.

  • $\begingroup$ This is true if and only if $f'$ is always nonzero: a holomorphic $f$ is locally injective at $z_0$ if and only if $f'(z_0)\neq 0$. The keyword is 'conformal'. $\endgroup$ – user98602 Dec 9 '14 at 23:41

Some are everywhere locally injective like $e^z$, some aren't, like $z^2$ (which is $2$ to $1$ on disks around zero).
The only globally injective entire functions are the affine functions $az+b, a\neq 0$.

  • $\begingroup$ So what if we put an extra condition on our function. Namely, that it maps the real line to the real line, and the upper half plane to the upper half plane. Can we conclude that it is locally bijective? $\endgroup$ – Enigma Dec 9 '14 at 23:50
  • $\begingroup$ Alright. Sorry, I'll make a new post. $\endgroup$ – Enigma Dec 10 '14 at 0:14

In the case where $f$ is analytic at $z_0$ and in addition $f'(z_0)\neq 0$, then it is true that $f$ is conformal and locally $1$-to-$1$ at $z_0$.

If however, $f$ is analytic at $z_0$ but $f'(z_0)=0$, then unless $f$ is constant, in some sufficiently small open set containing $z_0$, $f$ is a $k$-to-$1$ mapping and $f$ magnifies angles at $z_0$ by a factor of $k$, where $k$ is the least positive integer for which $f^{(k)}(z_0)\neq 0$.

While these theorems are more examples of the wonderful world of complex analysis, proving them is not exactly intuitive. However, if their lack of intuition doesn't deter you (and hopefully it doesn't!), I'd recommend reading the chapter on conformal mapping in Complex Analysis by Joseph Bak and Donald J. Newman. It has pretty clear proofs for both of these theorems.

  • $\begingroup$ Alright. Thanks Alex. I'll check that book out. $\endgroup$ – Enigma Dec 10 '14 at 15:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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