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Here is a conjecture that I came up with yesterday.

Conjecture: Suppose that $f:\mathbb{C} \to \mathbb{C}$ is entire. Then $f(z) = w$ has $n$ solutions $\forall w \in \mathbb{C} \iff f $ is a polynomial of degree $n$.

The converse follows immediately from the Fundamental Theorem of Algebra, but the forward implication is a lot more difficult. And after thinking about it more, it seems to me that the forward implication is probably not true. Unfortunately, I can't immediately think of a counterexample.

I thought a little about the approach I might take to prove this. Suppose $f$ is as above. Then $f(z)=0$ has $n$ solutions, so we can write $f(z)=p(z)g(z)$ where $p$ is a polynomial of degree $n$ and $g$ is entire with $g\neq 0$. If $g$ is a constant, then we are done. Otherwise, by Liouville and Casaroti-Weierstrass, $g$ is unbounded and gets arbitrarily close to any value in $\mathbb{C}$. We can also write $g=e^h$ for some $h$ entire. (There is also Picard's theorem, but I am reluctant to use this as I don't know the proof.)

The above gets somewhere, but I am not sure it leads to a proof without another insight. If the conjecture is false, then it definitely doesn't lead to a proof.

So, is the conjecture true or false? And if it is true, am I on the right track to prove it? (I would appreciate small hints much more than a full answer.)

legionwhale
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  • What about $f(z)=z^{10}$? Do you count $f(z)=0$ as having 10 solutions? – Randall Apr 15 '22 at 14:35
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    @Randall Yes -- $n$ zeros, counted with multiplicity. – legionwhale Apr 15 '22 at 14:37
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    In case $n=1$ this means: If $f$ is bijective then $f$ is a polynomial of degree $1$. This is considered here (maybe the methods can be extended to $n>1$): https://math.stackexchange.com/questions/29758/entire-one-to-one-functions-are-linear – Gerd Apr 15 '22 at 14:48
  • @Gerd It looks very helpful. Thanks! – legionwhale Apr 15 '22 at 14:53
  • The statement is definitely correct. The proof becomes simple if you can use the Great Picard Theorem: If $f$ is not a polynomial then it has an essential singularity at infinity, and every value with at most one exception is attained infinitely often. – Martin R Apr 15 '22 at 14:58
  • @MartinR The proof of Picard's theorem is (currently) too advanced for me, so I guess I was wondering if there was a more elementary argument. But it is good to know that the conjecture is true -- thanks! – legionwhale Apr 15 '22 at 15:02
  • It does not work for constant polynomials: $\infty$ many solutions for $w=0$. – emacs drives me nuts Apr 15 '22 at 15:46
  • @emacsdrivesmenuts Fair enough. Let $n \in \mathbb{N}$. (Although your choice of $w=0$ is unfortunate, because the zero polynomial is sometimes defined to have degree $\infty$.) – legionwhale Apr 15 '22 at 15:49

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One can easily prove this without Picard as the assumptions imply that (if $f$ nonconstant so $n \ge 1$) then $f$ is a proper entire map since for any compact $K$ then $f^{-1}(K)$ is bounded

(pick $w \in K$ there is a small open neighborhood of it $B_w$ for which $f^{-1}(B_w)$ is a union of at most $n$ open bounded sets - there are exactly $n$ precisely when $w$ is not a critical value of $f$ - cover $K$ with finitely many such $B_w$ etc)

But any proper entire map is a polynomial since then $|f(z)| \to \infty$ as $z \to \infty$

(if there is $|z_n| \to \infty$ st $|f(z_n)|$ bounded, one can pick a subsequence and rename it $z_n$ also, for which $f(z_n) \to w$ and then for a small ball $B_w$ around $w$, one has $f^{-1}(\overline B_w)$ containing elements with unbounded absolute value, hence it is not compact)

Edited: as per comments more detail about why $f$ is proper:

If $w$ is not a critical value of $f$, then by the local form of holomorphic functions, for every $z_k, k=1,..n$ in its preimage, there is a small neighborhood $D_k$ of $z_k$ for which $f$ is injective and has image some open neighborhood $U_k$ of $w$ so one can take $B_w$ any ball in the intersection of the $U_k$ which is open and contains $w$ - and then the preimage of $B_w$ is included in the union of $D_k$ as each contains a root of $f(z)=y$ for every $y \in B_w$, so we account for all $n$ of them; if $w$ is a critical value then for every preimage for which $f'(z_k)=0$ the same applies except that now $f$ is a $m:1$ map from $D_k$ onto $U_k$ where $m$ is the order of the critical point $z_k$ so in other words the order of the zero for $f(z)-f(z_k)$

$K$ is compact and covered by the $B_w$ so by the property of compacity, it is covered by finitely many such.

Conrad
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  • Thanks for the answer. I was unfamiliar with the term "proper", but I looked it up. I do not entirely follow your reasoning for why $f$ is proper. I agree that $f^{-1}(B_w)$ is open, since $f$ is continuous and $|f^{-1}(w)|=n$. I don't see why this implies $f^{-1}(B_w)$ is bounded. I also don't see how $K$ is being covered with finitely many such balls (unless you are using the same radius $r>0$ for each ball). – legionwhale Apr 15 '22 at 15:52
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    added some more explanations – Conrad Apr 15 '22 at 16:09
  • also, note that it is enough to know that the preimage of each point is finite of size $n_w$ that may apriori depend on $w$ as the proof above shows that for each $w$ there is a small ball $w \in B_w$ for which $n_y \ge n_w, y \in B_w$ and then from here (and Hurwitz) it is easy to see that there is a small ball for which $n_y=n_w$ for all $y$ in there so $w \to n_w$ is locally constant and integral valued, hence it is constant – Conrad Apr 15 '22 at 16:28
  • The above is true but requires a slightly more involved argument using the finiteness of $w\to n_w$ – Conrad Apr 15 '22 at 17:10