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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!

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  • $\begingroup$ I think $f(z)=\sin\frac1z$ works, doesn't it? $\endgroup$
    – saulspatz
    Commented Sep 13, 2020 at 17:28
  • $\begingroup$ How?Can you explain a little more? $\endgroup$ Commented 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
    Commented Sep 13, 2020 at 18:03
  • $\begingroup$ I wonder that statement given by second author is not correct! $\endgroup$ Commented Sep 13, 2020 at 18:08

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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.

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  • $\begingroup$ Nice.One more question,what would we say about statement (in post) given by second author (S Ponnusamy's)?is the statement not correct?? $\endgroup$ Commented 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
    Commented 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 $\endgroup$ Commented Sep 13, 2020 at 18:30
  • $\begingroup$ And BTW how you used the fact of periodicity here? $\endgroup$ Commented 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
    Commented Sep 13, 2020 at 18:36

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