# Image of punctured disk around essential singularity

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.

• I think $f(z)=\sin\frac1z$ works, doesn't it? Sep 13, 2020 at 17:28
• How?Can you explain a little more? Sep 13, 2020 at 17:42
• 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. Sep 13, 2020 at 18:03
• I wonder that statement given by second author is not correct! Sep 13, 2020 at 18:08

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 $$|\zeta| and $$\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.
• 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. Sep 13, 2020 at 18:12
• 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 $n\in \mathbb{Z}$. Sep 13, 2020 at 18:36