# Deleted Exponential Series and Injectivity (1)?

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?

• ad question 1: the resulting function is $\sinh(x)$ – Gottfried Helms Jun 22 '13 at 19:26
• Oh yeah thanks for reminding. – Abelvikram Jun 22 '13 at 19:34
• 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 $n \geq 2$ ? – Jakub Konieczny Jun 22 '13 at 19:50
• Careful - $\exp(z):=\sum_{n=0}^{\infty}\frac{z^n}{n!}$ (rather than the sum for $n\geq 1$). – Nick Peterson Jun 22 '13 at 19:54
• Feanor, I think you are correct. Surely the automorphisms of complex plane are as of the form you mentioned. – Abelvikram Jun 22 '13 at 20:02

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.