Entire function where at each $a \in \Bbb C$, at least one coefficient of the Taylor series at $a$ is real I was reading about this problem yesterday:
Suppose that for each $a\in \mathbb{C}$ at least one coefficient of the Taylor's series $f$ about $a$ is zero. Show that $f$ is a polynomial.
I was wondering, if instead of

Let $f:\mathbb{C}\rightarrow\mathbb{C}$ be a holomorphic function. Suppose that for each $a\in \mathbb{C}$ at least one coefficient of the Taylor's series $f$ about $a$ is zero. Show that $f$ is  a polynomial.

we ask

Let $f:\mathbb{C}\rightarrow\mathbb{C}$ be a holomorphic function. Suppose that for each $a\in \mathbb{C}$ at least one coefficient of the Taylor's series $f$ about $a$ is real. Then, is $f$ still a polynomial?


It seems that I have fall into a logic mistake in approaching it:
For $f(z)=\sum_{n=0}^\infty c_n(z-a)^n$. Still, the coefficient for the series at $n$ is $c_n=\frac{f^{(n)}(a)}{n!}=r$, for $n$ is the index and $r$ is the real coefficient. Does this means the coefficient $c_{n+1}$ should be $c_{n+1}$ because $f^{(n)}(a)=n! r$ seems like a constant? And then everything back to the "zero" statement.
Am I right or wrong? If wrong, is there any other approaches to the new question?
 A: Your approach does not work. For each $a \in \Bbb C$ there is an index $n$ and a real number $r$ such that $\frac{f^{(n)}(a)}{n!} = r$, but that real number depends on $a$.
But one can mimic this proof and define the sets
$$
A_{n}=\left\{z\in\mathbb{C}:f^{(n)}(z) \in \Bbb R\right\} \, .
$$
All $A_n$ are closed, and their union is equal to $\Bbb C$. By the Baire category theorem, one of the sets must have a non-empty interior, i.e. there is an open disk $U$ and an index $k$ such that $U \subset A_k$.
Then use the open mapping theorem to conclude that $f^{(k)}$ is constant on $U$ and therefore constant in $\Bbb C$. It follows that $f$ is a polynomial.

Alternatively one can proceed as suggested by Conrad: With the

Lemma: Let $g: \Bbb C \to \Bbb C$ be a non-constant holomorphic function. Then the set $A = \{ z \in \Bbb C : g(z) \in \Bbb R\}$ has Lebesgue measure zero.

we can argue as follows: If $f$ is not a polynomial then all derivatives $f^{(k)}$ are non-constant, so that all sets $A_n$ have the Lebesgue measure zero. Then the countable union $\bigcup_n A_n$ has Lebesgue measure zero as well. That is a contradiction to the assumption that $\bigcup_n A_n = \Bbb C$.
Remark: The Baire category theorem is not used in this argument.
Proof of the Lemma: First assume that $g(z_0) = a \in \Bbb R$. Then
$$
 g(z) = a + (h(z))^n
$$
in a neighborhood $V$ of $z=z_0$, where $n$ is the multiplicity of $g$ at $z=z_0$, and $h$ is an injective mapping from $V$ to a disk $B_r(a)$. It follows that
$$
 \{ z \in V : g(z) \in \Bbb R \}
$$
is the union of $n$ analytic curves, and therefore has Lebesgue measure zero.
We have thus shown: Every $z\in A$ has a neighborhood $V_z$ such that $V_z \cap A$ has Lebesgue measure zero.
Now one can conclude that all compact sets $A \cap \overline{B_R(0)}$ have Lebesgue measure zero (because they can be covered with finitely many sets $V_z$).
Finally conclude that $A$ has Lebesgue measure zero.
