This is related to this question. Tell me if this should be in the original question.
Is it possible for a complex function to have both conditions below?
Let $0<r<R$ and $a>0$ be a real number that satisfies $a+r<R$.
Condition 1: Power series expansion at $z=0$ has a radius of convergence $R$.
Condition 2: Power series expansion at $z=a$ has a radius of convergence $r$.
(Edit: It seems like I should swap R and r to make it consistent with the original question, but it might confuse people who have already read the question so I will leave it as it is.)
My attempt:
I think this is impossible. This is because a disk with radius $R$ and center $z=0$ (we call it a disk A) contains a disk with radius $r$ and center $z=a$(we call it a disk B) like an image below.
This is the case $r=1$,$a=2$ and $R=4$. So, the radius of convergence is the distance to the nearest singularity. By condition 2, this means that there is a singularity on $|z-a|=r$. However, by condition 1, the function is analytic in $|z|<R$. This means that the function is also analytic on $|z-a|=r$. This is a contradiction. So it is impossible.
My concern about this argument:
- Is the radius of convergence really the distance to the nearest singularity?
- Does the condition 1 really mean that the function is analytic in $|z|<R$?
I am not sure about those 2 question, so I am not really confident about this argument.
Is this correct? If not, is it possible for a complex function to have both conditions?
Best Answer
More precisely, the radius of convergence is the radius of the largest open disk centred at the expansion point on which there is an analytic function that coincides with your function near the expansion point. If the series around $0$ has radius of convergence $R$, then the corresponding analytic function is analytic in the disk around $0$ of radius $R$, and therefore in the disk around $a$ of radius $R - |a|$.
However there is a loophole: who says that function is the same as the function you are expanding around $a$? You might have chosen a function that has a branch cut that comes between $0$ and $a$ (with branch point outside the big disk). If the branch cut had been chosen differently, you'd have a function analytic in the big disk. But the way you chose it, the function near $z=a$ is on a different branch, and this branch might have a singularity near $z=a$ that the other branch does not. For example, this can occur with a function of the form
$$ f(z) = \frac{1}{\sqrt{z-p} - q} $$ where $p$ is in the first quadrant and you choose the principal branch of the square root.
EDIT: For concreteness, let's take $$ f(z) = \frac{1}{\sqrt{z} + 1 - 10 i} $$ with the principal branch. It has a branch point at $0$. Note that the principal branch of $\sqrt{z}$ has real part $\ge 0$, so the denominator is never $0$. However, other branches may have a pole at $z = (1-10i)^2 = -99-20i$. The Taylor series around $z=-99-20i$ has radius $101$ (the distance to the branch point at $0$). But the Taylor series around $z=-99+20i$ has radius only $40$, because the analytic continuation in a disk around $-99+20i$ will run into a pole at $-99-20i$.