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Let $f$ be a non constant entire function satisfying the following conditions :

  1. $f(0)=0$
  2. for every positive real $M$, the set $\{z: \left|f(z)\right|<M\}$ is connected.

Prove that $f(x)=cz^n$ for some constant $c$ and positive integer $n$.

Let $f(z)=a_nz^n+\cdots+a_1z+a_0$ be function that satisfies the given conditions. As $f(0)=0$ we have $a_0=0$ and $f(z)=a_nz^n+\cdots+a_1z$.

As $f$ is non-constant function, its zeros are isolated. So, there exists an $r>0$ such that $f$ is non-zero on $B_r=\{z:|z|<r\}$. I was thinking of connecting this to connectedness of $\{z: \left|f(z)\right|<M\}$.

I wanted to check what goes wrong in case of $f(z)=z^2+z$. I want to check if the given set is connected for this but failed in doing so.

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    $\begingroup$ First of all, there is no immediate reason the function must be a polynomial. You must also check that, for instance, $\sin(z)$ and $e^z-1$ both fail condition 2. $\endgroup$
    – Arthur
    Mar 2 '16 at 18:51
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    $\begingroup$ If it was a polynomial, then think about the roots of the polynomial. What happens when $M$ is small and close to $0$? $\endgroup$ Mar 2 '16 at 18:56
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    $\begingroup$ @StevenGubkin : As $M$ is close to $0$ then we have finitely many distinct roots in case of polynomials other than $f(z)=az^n$ which is clearly disconnected... $\endgroup$
    – user312648
    Mar 2 '16 at 19:02
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    $\begingroup$ Do you know the Casorati-Weierstraß theorem? $\endgroup$ Mar 2 '16 at 19:51
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    $\begingroup$ Note that instead of $\lt$ you can use $\le$ by taking the closure. Then for $M=0,$ the set of zeros must be connected. $\endgroup$
    – Bumblebee
    Aug 19 '20 at 23:35
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We can write $f(z) = z^kg(z)$ for some $k\in \mathbb N,$ where $g$ is entire and $g(0)\ne 0.$ Choose $r>0$ such that $g\ne 0$ in $\{|z|\le r\}.$ Then

$$m= \min_{|z|=r}r^k|g(z)|>0.$$

Now $0\in \{|f(z)| < m\},$ and this set doesn't intersect $\{|z|=r\}.$ Because $\{|f(z)| < m\}$ is given to be connected, it must lie in $\{|z|<r\}.$ Thus all zeros of $f$ lie in $\{|z|<r\}.$ It follows that $f$ has only one zero, namely the one at $0.$ Hence $g(z)$ never vanishes.

Again, $\{|f(z)| < m\}$ lies in $\{|z|<r\}.$ Thus if $|z|\ge r,$ we must have $|f(z)| \ge m.$ But an entire function that behaves this way cannot have an essential singularity at $\infty.$ Thus $f$ has at most a pole at $\infty,$ which means $f$ is a polynomial. But a polynomial with a $k$th order zero at $0$ and no other zeros, has the form $cz^k.$ That is the desired result.

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Let $D_M=\{z:|f(z)|<M\}$ open connected set; since $f$ is non-constant $D_M$ is not the plane for any $M$; given any Jordan curve $J \subset D_M$, the interior of $J$ is contained in $D_M$ by maximum modulus, hence $D_M$ is simply connected.

Let now $B_{2r}$ a small disc of radius $2r$ around $0$ where $f$ vanishes only at $z=0$ and $2a=\min_{|z|=r}|f(z)| >0$; since then $f(\partial B_r) \cap D_a = \emptyset$ it follows that $D_a \subset B_r$ (as otherwise if there is $|w|>r, |f(z)| <a$ there is a path in $D_a$ joining $w$ and $0 \in B_r$ and that must intersect $\partial B_r$) so in particular $D_a$ bounded and $f$ vanishes only at zero.

It follows that $f$ is a polynomial since $\infty$ is not an essential singularity ($|f(z)| > a, |z|>r$ and since $f$ vanishes only at $0$ we are done!

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Suppose $f$ has Taylor series $$ f(z)=\sum_{n=0}^{\infty}a_nz^n \quad\text{and }\quad g(z) =f(1/z) =\sum_{n=0}^{\infty}\frac{a_n}{z^n} $$ If $f$ is not polynomial, then $\infty$ is an essential singularity of $f$ ($0$ is an essential singularity of $g$). By Casorati-Weierstrass theorem, for $0\in \Bbb{C}$, there is a sequence $z_n'\to0$ such that $\lim_{n\to\infty}g(z_n')=0$, i.e. there is $z_n=1/z_n'\to\infty$ such that $\lim_{n\to\infty}f(z_n)=0$. This means that $f$ has always $z_0$ that $f(z_0)=0$ near $\infty$.

Since zero of analytic function is isolated and $f(0)=0$, by choosing $M$ so that $z_0\in \{z: \left|f(z)\right|<M\}$ $(0\in \{z: \left|f(z)\right|<M\}$ is obvious$)$, we conclude that $\{z: \left|f(z)\right|<M\}$ can not be connected because $z_0$ and $0$ are separated. Thus $f$ must be polynomial.

If $f$ has nonzero root, i.e. $f(z_1)=0,\: z_1\ne 0$, then $\{z: \left|f(z)\right|<M\}$ including $z_1$ can not be connected for $z_1$ and $0$ are separated. So $f(z)=0$ can only has zero root. Thus we conclude that $f(z)=cz^n$.

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We may assume that $$ f(z)=cz^n(1+c_1z+c_2z^2+\cdots)=cz^ng(z), $$ where $c>0$ (for simplicity) and $g(z)$ has no zeros in $|z|\le r_0$ for some positive $r_0$. Furthermore there is a positive $r_1<r_0$ such that \begin{align} |f(z)|>\frac{c{r_1}^n}{2} \quad \text{on}\,\;|z|=r_1,\tag{1} \end{align} since $|g(z)|\to 1$ $(\,z\to 0\,)$.

Let $M=\displaystyle\frac{c{r_1}^n}{2}$ and $A=\{z:\, |f(z)<M\}$.
If $g(z_0)=0$ for some point $z_0\,(\,|z_0|>r_0\,)$, then $A$ contains $0$ and $z_0$, and hence by the connectedness there is a curve $\gamma \subset A$ joining $0$ and $z_0$. But this is impossible by $(1)$.
Therefore $g(z)\ne 0$ and hence we can write $g(z)=e^{-h(z)}$ , $h(z)=\sum_{k=1}^\infty a_kz^k$.

Suppose that there exists $a_k\ne 0$. Let $H(r)=\max_{|z|=r} \operatorname{Re}\, h(z)$. Then as well-known $$ |a_k|r^k\le \max\{ 4H(r), 0\}-2\operatorname{Re}\, h(0)= 4H(r).$$ Hence we have \begin{align} \min_{|z|=r} |g(z)|&=\min_{|z|=r} \left|e^{-h(z)}\right|=e^{-H(r)}\le e^{-|a_k|r^k/4}. \end{align}

Thus for sufficiently large $r$ we have $$ |f(z_0)|\le cr^n\cdot e^{-|a_k|r^k/4}<M$$ at a point $z_0$ on $|z|=r$. Then $A$ contains $z_0$, but it is impossible.
Thus all $a_k=0$ and we have $f(z)=cz^n$.

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