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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. Image

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:

  1. Is the radius of convergence really the distance to the nearest singularity?
  2. 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?

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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 $-99-20i$.

enter image description here

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  • $\begingroup$ I had little knowledge about the branches so I had to google, and I think I got what you are talking about.But I don’t understand how the example works. I guess the example function is the example on how a power expansion on different points can give different analytic functions, but I don’t see why it’s the case. $\endgroup$ – Kaira Jul 14 '20 at 10:21
  • $\begingroup$ Thank you for your edit, but I have a question about the example in the edit. so, at $z=-99-20i$, we are considering the principal branch of $\sqrt{z}$. but at $z=-99+20i$, we are considering the other branch. Isn't the other branch a completely different function? It feels to me that the Taylor expansion at $z=-99+20i$ is not about the function $f(z)$, but about the completely different function. $\endgroup$ – Kaira Jul 14 '20 at 13:53
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    $\begingroup$ There are two possible square roots of any nonzero number. The branch cut is chosen to be the negative real axis (but the Taylor series doesn't know that, it just depends on the function and its derivatives at the expansion point). If we use $\sqrt{\ }$ to denote the principal branch of square root, the analytic function corresponding to the series around $z=a$ is $f(z) = 1/(\sqrt{z}+1-10i)$ when $z$ is on the same side of the branch cut as $a$, but $1/(-\sqrt{z}+1-10i)$ when $z$ is on the other side of the branch cut. $\endgroup$ – Robert Israel Jul 14 '20 at 14:18
  • $\begingroup$ Ooh, I think I got it. Let me explain to see if I got it. Let me use this figure from the pdf. The orange branch is the principal branch here, and the red line is the branch cut(negative real line). So, assume that we are going to do a power series expansion of $f(z)$ at $z=a$ and a is on the second quadrant. Power series expansion at z=a is to find an analytic function $g(z)$ that coincides with $f(z)$ at small enough disk around z=a,... $\endgroup$ – Kaira Jul 14 '20 at 15:39
  • $\begingroup$ ...and the domain of the analytic function $g(z)$ should be as large as possible. If $z$ is above the branch cut we can just take $g(z)=f(z)$, but if z is below the branch cut we cant take $g(z)=f(z)$, because $f(z)$ is discontinuous at the branch cut. Instead, we will take the blue branch(from the graph) so that the $\sqrt{z}$ will be continuous. This is the reason the corresponding analytic function is different below the branch cut. So the expansion at $z=-99+20i$ and the expansion at $z=-99-20i$ is indeed a different analytic function, but it is indeed the expansion of $f(z)$. $\endgroup$ – Kaira Jul 14 '20 at 15:39

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