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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 $n$..

I do not know where to start...

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 $f(z)=a_nz^n+\cdots+a_1z$..

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..

Please give only hints...

EDIT :

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

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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 $0$? $\endgroup$ – Steven Gubkin 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? $\endgroup$ – Daniel Fischer Mar 2 '16 at 19:51
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    $\begingroup$ $w_0 = 0$ is an excellent choice. The theorem does not say $z^n e^{g(z)} \to 0$ for $z\to\infty$. It only asserts the existence of sequences $(z_k)$ with $z_k\to\infty$ and $z_k^n e^{g(z_k)} \to 0$ if $f$ is transcendental [which means it has an essential singularity at $\infty$]. $\endgroup$ – Daniel Fischer Mar 2 '16 at 20:06
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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 $m= \min_{|z|=r}|g(z)|>0.$ The connected set $\{|f(z)| < m\}$ thus contains $0$ and doesn't intersect ${|z|=r}.$ Therefore $\{|f(z)| < m\}\subset \{|z|<r\}.$ It follows that $f$ has only one zero, namely the one at $0.$ Hence $g(z)$ never vanishes.

We have $|z^kg(z)|\ge m$ for $|z| > r.$ Thus $|1/g(z)| \le |z|^k/m$ for $|z| > r.$ Because $1/g$ is entire, we can then say $|1/g(z)| \le C +|z|^k/m, z\in \mathbb C,$ for some constant $C.$ Thus $1/g$ is an entire function whose modulus is bounded by a polynomial in $|z|.$ As is well known, that implies $1/g$ is a polynomial. Since $1/g$ has no zeros, it must be constant. Therefore $g$ is some constant $\alpha$, and we have $f(z) =\alpha z^k$ as desired.

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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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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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