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.