# What is the image near the essential singularity of z sin(1/z)?

This was part of a homework problem from J.B. Conway's complex analysis text which I was assigned long ago but didn't get. A few years later I was a TA for a course where the problem was assigned. I still didn't know how to solve it, nor did any of my students. If you can give me a simple answer, my gratefulness will far outweigh my embarrassment.

Define $f(z)=z\sin(\frac{1}{z})$ on the punctured plane. What is the image under $f$ of a (small) punctured disk at the origin?

Remarks:

• The corresponding problem for $\sin(\frac{1}{z})$ is straightforward to solve using the definition of $\sin$ in terms of exponentials, the fact that $\exp$ is onto the punctured plane, and periodicity. The answer in this case is the whole plane.
• The corresponding problem for $z\cos(\frac{1}{z})$ is straightforward to solve using Picard's theorem, because the oddness of the function implies that 0 is the only possible excluded point, and it is easy to see that 0 is in the image by taking reciprocals of large zeros of $\cos$. Thus the answer in this case is also the whole plane. (The same argument could be applied to the previous remark, but there no big theorems are needed. Perhaps Picard's theorem isn't needed here either.)
• Another part of the problem I couldn't solve generalizes this, taking $z^n\sin(\frac{1}{z})$ when $n$ is a positive integer. When $n$ is even, there is no problem, because the argument from the previous remark applies. But I don't know how to solve it when $n$ is odd.
• By Picard's theorem, there is at most one point missing. I would be surprised if there is a point missing, but I have no argument to back up this expectation.
• Although an elementary argument would be nice, I do not care whether the argument is "appropriate" for what is covered up to that point in the text. This is pure curiosity.
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Hello, Well I had a picture of w=z sin 1/z to upload and since I'm a newbie I couldn't. I'f anyones's interested I can send it to your email Art –  Arthur Lawrence Jan 11 '11 at 19:18

It can (apparently) be shown that the equation

$\sin z = Az$ has infinitely many roots for any complex number $A$.

Given any $\delta > 0$ there must be such a root with $|z| > \frac{1}{\delta}$, otherwise we can show that $\sin z - Az$ must be identically zero on some bounded disk.

So I suppose this shows that there is no value missed by $z \sin\frac{1}{z}$.

(Am a bit rusty, please pardon if incorrect).

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Great, thanks! I still don't know about $z^n\sin(\frac{1}{z})$ with $z$ odd and bigger than 1. Superficially, it looks like there will be trouble in trying to apply a similar method due to the pole at infinity. –  Jonas Meyer Aug 24 '10 at 0:49
Jonas - the argument follows in similar way in that case. –  George Lowther Aug 24 '10 at 1:07
George - The argument cited relies on the fact that $\frac{\sin(z)}{z}$ is entire (and actually that $\frac{\sin(\sqrt z)}{\sqrt z}$ is entire). I'm still thinking about it, but if you see immediately how to accommodate $n>1$ then I'd be happy if you'd share the idea in an answer. –  Jonas Meyer Aug 24 '10 at 1:19

The image of $f(z)=z^n\sin(\frac1z)$ (for n odd) on any punctured disc about the origin is the whole of $\mathbb{C}$.

The following is (I think) a simplified proof along the lines of that given in Moron's reference. I'll use proof by contradiction, so suppose that there is an $a$ such that $f(z^{-1})=z^{-n}\sin(z)\not=a$ for all large $z$. By symmetry, $f(z^{-1})=g(z^2)$ for an analytic function $g$ with, for $n>1$, a pole at the origin of order $(n-1)/2$. As $g(z)-a$ can have only finitely many zeros, we can decompose it as $$g(z)=a+z^{(1-n)/2}h(z)\prod_{i=1}^m(z-b_i)$$ where $b_i$ are the zeros of $g(z)-a$. Writing out $f(z^{-1})=\frac{1}{2i}z^{-n}(e^{iz}-e^{-iz})$ shows that $f(z^{-1})$ is of order $1$. So $g$ and, hence, $h$ are of order 1/2. Any entire and nonzero function of order less than 1 is constant, so $h$ is constant and we see that $f$ is a rational function, which is false.

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Thanks very much! –  Jonas Meyer Aug 25 '10 at 2:34