= Solution
Write the <Laurent series> on the punctured disc as $f(z)=\sum_{j=-\infty}^{\infty}a_jz^j$. A <removable singularity> means that $f$ extends holomorphically to the origin; exactly all negative-index coefficients vanish. A <pole> of order $k\ge1$ means that $z^kf(z)$ extends with nonzero value at zero; exactly $a_{-k}\ne0$ and every coefficient below $-k$ vanishes. An <essential singularity> is neither of these cases; exactly infinitely many negative-index coefficients are nonzero.
Here is a proof of the required <Casorati-Weierstrass theorem> consequence. If some disc of values around $w$ were missed in a punctured neighbourhood, then $|f(z)-w|\ge\varepsilon>0$ there. The <holomorphic function> $q(z)=1/(f(z)-w)$ would be bounded, and hence have a <removable singularity>. If its extension had $q(0)\ne0$, then $f=w+1/q$ would extend holomorphically. If $q(0)=0$, its zero has a finite positive order, since $q$ is not identically zero; thus $f$ would have a <pole>. Both contradict an <essential singularity>. Therefore every sufficiently small punctured neighbourhood has values arbitrarily close to every $w$. Choosing $0<|z_n|<\min(r/2,1/n)$ with $|f(z_n)-w|<1/n$ proves
$$
\boxed{z_n\to0,\qquad f(z_n)\to w.}
$$
The <open mapping theorem (complex analysis)> states that a nonconstant <holomorphic function> on a domain maps open sets to open sets. If $f$ were injective with an <essential singularity>, choose a small open disc $U$ away from zero and a point $z_0\in U$. Its open image contains a disc about $w=f(z_0)$. The sequence just constructed has $z_n\to0$ outside $U$, but eventually $f(z_n)\in f(U)$. Thus some $y_n\in U$ has $f(y_n)=f(z_n)$ although $y_n\ne z_n$, contradicting injectivity. \b[An injective <holomorphic function> cannot have an isolated <essential singularity>.]
For an injective <entire function> $g$, apply this to $f(z)=g(1/z)$. If its singularity at zero were removable, $g$ would be bounded outside a large disc, hence bounded everywhere and constant by <Liouville's theorem>. This contradicts injectivity. It must therefore be a <pole>, say of order $k$, so $|g(z)|\le C|z|^k$ for large $|z|$. For any Taylor coefficient of index $j>k$, the <Cauchy coefficient formula> gives $|a_j|\le CR^{k-j}$ on arbitrarily large circles; hence $a_j=0$. Thus $g$ is a nonconstant polynomial. Its derivative never vanishes by the allowed injectivity fact. If its degree exceeded one, the <fundamental theorem of algebra> would give a zero of this nonconstant derivative. Therefore <injective entire functions are affine>:
$$
\boxed{g(z)=az+b,\qquad a\ne0.}
$$
Back to article page
