# 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? Commented Sep 13, 2020 at 17:28
• How?Can you explain a little more? Commented 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. Commented Sep 13, 2020 at 18:03
• I wonder that statement given by second author is not correct! Commented 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. Commented 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 $n\in \mathbb{Z}$. Commented Sep 13, 2020 at 18:36