# How do I show that if $f$ is entire and $\{\lvert f(z)\rvert < M\}$ is connected for all $M$, then $f$ is a power function?

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| 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 $$B_r=\{z:|z|. I was thinking of connecting this to connectedness of $$\{z: \left|f(z)\right|.

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.

• 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. Mar 2 '16 at 18:51
• If it was a polynomial, then think about the roots of the polynomial. What happens when $M$ is small and close to $0$? Mar 2 '16 at 18:56
• @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...
– user312648
Mar 2 '16 at 19:02
• Do you know the Casorati-Weierstraß theorem? Mar 2 '16 at 19:51
• Note that instead of $\lt$ you can use $\le$ by taking the closure. Then for $M=0,$ the set of zeros must be connected. Aug 19 '20 at 23:35

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 $$\{|z| Thus all zeros of $$f$$ lie in $$\{|z| It follows that $$f$$ has only one zero, namely the one at $$0.$$ Hence $$g(z)$$ never vanishes.

Again, $$\{|f(z)| < m\}$$ lies in $$\{|z| Thus if $$|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.

Let $$D_M=\{z:|f(z)| open connected set; since $$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 $$|w|>r, |f(z)| there is a path in $$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!

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 $f(z)=cz^n$.

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 $f(z)=cz^n$.