Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

If a is not in the image of f, and f is entire, then we know that there is a sequence in a domain whose value of function converges to a. Now the problem I faced is that even further we can choose a continuous curve p whose parameter is t in[0,1) such that p(t) goes to inf as t goes to 1, while f(p(t)) converges to a.

That's much stronger than image of entire function is dense in the plane .. Actually I faced two difficulties.

  1. How can we find a sequence whose value of functions converges to a? With its modulus goes to infinity?

  2. How can we connect these sequence to make a continuous curve whose value of function is not much varying between two points of consecutive points?

share|cite|improve this question
up vote 1 down vote accepted

The answer to your first question is easy. Since you know that the image of $f$ is dense in $\mathbb{C}$, you can find a sequence $(z_n)$ with $f(z_n) \to a$. If $(z_n)$ does not diverge to $\infty$, there exists a subsequence converging to some point $z^* \in \mathbb{C}$. Continuity then implies $f(z^*) = a$, contradicting the assumption that $a$ is not in the image of $f$.

The second question is in fact quite a bit harder, the result is a special case of Iversen's theorem. Just picking any such sequence it might be impossible to connect even a subsequence into a curve that works. If $a\ne \infty$, you can replace $f$ by $\tilde{f}(z)=\frac{1}{f(z)-a}$, which is again an entire function, and any curve on which $\tilde{f}$ tends to $\infty$ will be a curve on which $f$ tends to $a$. So you can assume that $a=\infty$. One standard proof of this result is via the Gross star theorem, which says that any local inverse near $w_0 = f(z_0)$ of an entire function can be continued along almost any ray emanating from $w_0$. If $R$ is any such ray, and $\gamma$ is the image of $R$ under the continuation of this local inverse, then $f \to \infty$ along $\gamma$, which in turn implies that $\gamma$ has to tend to $\infty$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.