0
$\begingroup$

I am studying Picard's great theorem in complex analysis and I saw two different forms of theorem.For instance in convoy the theorem is stated as:

Great Picard's Theorem: If an analytic function $f$ has an essential singularity at a point $w$, then on any punctured neighborhood of $w, f(z)$ takes on all possible complex values, with at most a single exception, infinitely often.

While my text book foundations of complex analysis by S Ponnusamy states the same theorem as follows:

Suppose that in $f$ is analytic in $\Delta(z_0;r)$\{$z_0$}(punctured disk around $z_0$ with radius r) and $z=z_0$ is an essential singularity of $f$.Then $\mathbb{C}$\ $f$ ($\Delta(z_0;r)$ \{$z_0$}) is singleton.

Clearly according to first statement,the image of punctured disk around essential singularity may be either whole complex plane or a complex plane minus one point but according to second statement of the same theorem the the image of punctured disk around essential singularity is complex plane with one exception.If I suppose that statement of first book is more precise(convoy) then there is a possibility of a analytic function such that image of punctured disk around essential singularity is whole complex plane,if such a example exists then please provide it.

Thanks in advance!

$\endgroup$
4
  • $\begingroup$ I think $f(z)=\sin\frac1z$ works, doesn't it? $\endgroup$
    – saulspatz
    Sep 13, 2020 at 17:28
  • $\begingroup$ How?Can you explain a little more? $\endgroup$ Sep 13, 2020 at 17:42
  • $\begingroup$ It seems the discrepancy is: (1) the first one says the whole plane with at most one exception, whereas (2) the second one says the whole plane with exactly one exception. $\endgroup$
    – GEdgar
    Sep 13, 2020 at 18:03
  • $\begingroup$ I wonder that statement given by second author is not correct! $\endgroup$ Sep 13, 2020 at 18:08

1 Answer 1

1
$\begingroup$

One example is given by $$f(z)=\sin\frac1z$$ We must show that given $r>0$ and $w\in\mathbb{C}$ there exists a $\zeta\in\mathbb{C}$ with $|\zeta|<r$ and $\sin\frac1\zeta=w$ or what is the same thing, that there is a $z\in\mathbb{C}$ with $|z|>\frac1r$ and $\sin z =w$.

Solving for $z$, $$\begin{align} &\sin z=w\\ &e^{iz}-e^{-iz}=2iw\\ &e^{2iz}-2iwe^{iz}-1=0\\ &e^{iz}=\frac{2iw\pm\sqrt{-4w^2+4}}2 \end{align}$$ The right-hand side cannot be $0$, so there is a $z$ that satisfies the equation. Then by periodicity of $e^z$, there is a $z$ with arbitrarily large modulus that satisfies the equation.

$\endgroup$
6
  • $\begingroup$ Nice.One more question,what would we say about statement (in post) given by second author (S Ponnusamy's)?is the statement not correct?? $\endgroup$ Sep 13, 2020 at 18:07
  • $\begingroup$ It appears to be false, unless the author is using an unusual meaning of "singleton". I suspect that it's just a mistake. BTW, looking back at my answer, there's no need to split $w=0$ out as a separate case, but of course, it doesn't hurt. $\endgroup$
    – saulspatz
    Sep 13, 2020 at 18:12
  • $\begingroup$ I don't understand why you used the periodicity though the phrase there is a z that satisfies the equation completes the proof $\endgroup$ Sep 13, 2020 at 18:30
  • $\begingroup$ And BTW how you used the fact of periodicity here? $\endgroup$ Sep 13, 2020 at 18:31
  • $\begingroup$ We need to show that there is a $z$ of arbitrarily large modulus that satisfies the equation, but if $z$ satisfies the equation, so does $z+2n\pi$ for any $n\in \mathbb{Z}$. $\endgroup$
    – saulspatz
    Sep 13, 2020 at 18:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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