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Consider $\exp(z):=\Sigma_{n=1}^{\infty}\frac{z^n}{n!}.$ We know that the radius of convergence for this series is infinity, and hence it defines a holomorphic map which is not injective. I have following questions:

  1. If we delete from exponential series those terms with positive even powers, the series we are left with, does that defines a holomorphic function which is injective?

  2. What about in general, i.e., let $\{n_\nu:\nu\in\mathbb{N}\setminus\{1\}\}$ be a strictly incresing sequence and delete those terms in exponential series with power indices coming from this set. Now consider the deleted series. When this defines an injective map?

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    $\begingroup$ ad question 1: the resulting function is $\sinh(x)$ $\endgroup$ – Gottfried Helms Jun 22 '13 at 19:26
  • $\begingroup$ Oh yeah thanks for reminding. $\endgroup$ – Abelvikram Jun 22 '13 at 19:34
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    $\begingroup$ Isn't it the case that the only non-trivial automorphisms of $\mathbb{C}$ are only the maps $f(z) = az + b$, so the only way to make the exponential series injective is by deleting all terms at $z^n$ for $n \geq 2$ ? $\endgroup$ – Jakub Konieczny Jun 22 '13 at 19:50
  • $\begingroup$ Careful - $\exp(z):=\sum_{n=0}^{\infty}\frac{z^n}{n!}$ (rather than the sum for $n\geq 1$). $\endgroup$ – Nick Peterson Jun 22 '13 at 19:54
  • $\begingroup$ Feanor, I think you are correct. Surely the automorphisms of complex plane are as of the form you mentioned. $\endgroup$ – Abelvikram Jun 22 '13 at 20:02
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As the limit of $1/(n!^{1/n})$ is 0 (as $n \to \infty$), deleting any subset of terms produces a power series whose radius of convergence is $\infty$.

In the case where the new power series still has infinitely many terms, then $\infty$ is an essential singularity, so the function cannot be injective, by the big Picard theorem.

In the case where the new power series is a polynomial, then clearly it cannot be injective except in the cases when it's linear.

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