From Picard's theorem, the image of $\mathbb{C}$ under an analytic function has to the whole plane or $\mathbb{C}$ minus a single point. What about other open sets? This may be too broad, so how about the unit disc?

Let $B$ be the open unit disc. If $f:B\to \mathbb{C}$ is analytic, what can we say about $f(B)$?

We know that $f(B)$ has to be open. Can something further be said?

For instance, $f(B)$ need not be simply connected (consider $exp(\alpha z)$ for a large enough $\alpha$).


2 Answers 2


In fact $f(B)$ can be any open subset of $\Bbb C$. (Edit: Or rather any connected open set; thanks to Robert Israel for noticing that this is what I meant...)

Bonus: If $V$ is any open connected subset of the plane there exists $f$ with non-vanishing derivative such that $f(B)=V$, and in fact if $V\ne\Bbb C$ we can take $f$ to be a covering map.

Say $V\subset\Bbb C$ is open and connected. Assume first that $\Bbb C\setminus V$ contains more than one point. Now the Uniformiization Theorem shows that there is a simply connected Riemann surface $M$ and a holomorphic covering map from $M$ onto $V$. And $M$ is either the Riemann sphere, the disk or the plane. Except $M$ cannot be the Riemann sphere since $V$ is not compact, and the little Picard theorem shows that $V$ cannot be $\Bbb C$, so $M$ is the disk.

That does it except for $V=\Bbb C\setminus \{0\}$ and $V=\Bbb C$. There is a strip $S$ so that $\exp(S)=\Bbb C\setminus\{0\}$ and a map from the disk onto $S$. And there is a map from the disk onto the region $\Re z>-1$; the square of this map maps $B$ onto $\Bbb C$.

We're done except for the bonus part, and we're almost done with that. If $V\ne\Bbb C$ we've given a holomorphic covering map from $B$ onto $V$. If $V=\Bbb C$ there is no such covering map. But there does exist a holomorphic surjection with non-vanishing derivative:

Say $f$ is the entire function with $f(0)=0$ and $f'(z)=e^{\cos(z)}$. Then $f'(z+2\pi)-f'(z)=0$, so there exists $c$ with $f(z+2\pi)=f(z)+c$. Note that $c=f(0)>0$; in particular $c\ne0$.

Since $f(z+2\pi)=f(z)+c$ with $c\ne0$ the Big Picard theorem shows that $f$ takes every complex value infinitely many times. Hence $f(\Bbb C\setminus\{0\})=\Bbb C$. As above there is a covering map $g$ from $B$ onto $\Bbb C\setminus\{0\}$; now $f\circ g$ is a map from $B$ onto $\Bbb C$ with non-vanishing derivative.

  • $\begingroup$ I think you mean any connected open set. $\endgroup$ Commented Dec 17, 2015 at 19:01
  • $\begingroup$ @RobertIsrael Eewps. Yeah, I'm pretty sure that's what I meant, thanks $\endgroup$ Commented Dec 17, 2015 at 19:09

Bloch-Landau theorem states:
There exists an absolute constant $\ell$ such that, for any holomorphic function $f$ defined in the unit disc, with the property $|f'(0)| =1$, Then:
The image of the unit disc contains a ball with radius greater or equal to $\ell$.

Notice that the condition on the derivative is "just" a normalization constraint.

  • $\begingroup$ An since $f'(0)$ can be smaller than $1$, what actually follows from this is that if $f$ is analytic in $B$ then $f(B)$ contains a disk... $\endgroup$ Commented Dec 17, 2015 at 19:12
  • $\begingroup$ @DavidC.Ullrich In the simple case where $f'(0)\ne 0$ then we can define $g(z)=f(z)/f'(0)$, so $g(z)$ containts balls with radius greater than landau-constant, and then , since $f(z)=g(z) \cdot f'(0)$ - $f$ must contains balls with radius factored by $f'(0)$. The other case is not so trivial to me , and i need to invest some more thinking into it. $\endgroup$
    – Ranc
    Commented Dec 17, 2015 at 19:36
  • $\begingroup$ Right. And a ball factored by $f'(0)$ can be any ball at all, so all this shows is that the image contains a ball. Which is not surprising; have you noticed the other answer showing that $f(B)$ can be any connected open set? $\endgroup$ Commented Dec 17, 2015 at 19:38
  • $\begingroup$ @DavidC.Ullrich Yes, but, I was merely suggesting a different approach to this question. The Bloch-Landau THM gives some information regarding any $f$ holomorphic. If you feel there is some duality ,or my answer is not so informative - I won't mind deleting it. $\endgroup$
    – Ranc
    Commented Dec 17, 2015 at 19:55

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