Prove that an isolated sigularity of cannot be a pole of .
I know the key is to prove that and have no common pole, but how??
Hello! Here is an idea.
It is easy to see that this is true if is an essential or removable singularity of .
In the case of an -th order pole, then I claim that is an essential singularity of . Clearly it is not a regular point, so it suffices to show that does not tend to infinity as . To do this, it suffices to show that every neighbourhood of contains points such that , for some fixed .
Hi Ynj!
I know this is a really late answer but I had forgotten about this problem and I remembered it today and solved it. So here it is.
Let have a pole at . I claim that has an essential singularity at .
The problem is solved if we can show that there exists a sequence such that is bounded as ; this is because if were a pole of then we would have as . This is equivalent to finding a sequence such that remains bounded, since . Since we know that necessarily blows up to infinity as as , it is the imaginary part which must become very big.
Consider . This is clearly holomorphic at and . Let be a sequence of bounded open neighbourhoods of whose intersection is . Then is a sequence of neighbourhoods of ; in particular, each of the is bounded, and thus is a sequence of neighbourhoods of infinity. So there are points of the imaginary axis in each of the ; pick such that . Then the sequence converges to and . Thus for and we are done.