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Hi: Th next question in John D'Angelo's text is exercise 4.8: where does the series for $\frac{1}{1-z}$ about the point $5i$ converge ?

I understand that the expansion is : $\sum_{n=0}^{\infty} (z - 5i)^{n}$.

Now, for the series to converge, $|z-5i|$ has to be less than 1 because the series is geometric. So is that the answer ? that $|z-5i|$ < 1$. This exercise is after another exercise which was much harder ( required abel's convergence for complex series test ) so I'm thinking that maybe I'm not correct. Thanks.

mark leeds
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    If $|z - 5i| < 1$, the series you've written converges to $\frac{1}{1 - (z-5i)}$, not $\frac{1}{1-z}$. – Michael Albanese Aug 17 '14 at 20:08
  • I assume you should be talking about $\frac{1}{(1-5i)(1-x)}$ where $x=\frac{z-5i}{1-5i}$. This x has radius of convergence |(1-5i)| in terms of (z-5i). This is analytic extension. You should check the analytic extension of $\frac{1}{1-z}$ – user45765 Aug 17 '14 at 20:24

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You want to write

$${1\over 1-z}={1\over 1-5i-(z-5i)}$$

Then this is just

$${1\over 1-5i}\left({1\over 1-{z-5i\over 1-5i}}\right)$$

Which, by geometric series, instantly gives

$${1\over 1-5i}\sum_n {(z-5i)^n\over (1-5i)^n}=\sum_n{(z-5i)^n\over (1-5i)^{n+1}}$$

and moreover we know geometric series have convergence if and only if the common ratio, $r$, has $|r|<1$ i.e. for

$$\left|{z-5i\over 1-5i}\right|<1\iff |z-5i|<|1-5i|=\sqrt{26}$$

Adam Hughes
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  • Alternatively, before bringing the constant into the sum, you have a geometric series in $\frac{z-5i}{1-5i}$. – Michael Albanese Aug 17 '14 at 20:33
  • @MichaelAlbanese I already changed it. I realized after I wrote it, the only reason the two series are equal is because of geometric series anyway, so I may as well stay on topic and just use that reasoning throughout. – Adam Hughes Aug 17 '14 at 20:34
  • is it just a coincidence that the radius of convergence is exactly the distance from $z=5i$ to the nearest (and only) singularity of $\frac{1}{1-z}$, namely $z=1$? – etothepitimesi Aug 17 '14 at 20:36
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    @el.Salvador No, not a coincidence at all. The radius of convergence of a power series is always the distance to the nearest singularity of the function. – Daniel Fischer Aug 17 '14 at 20:37
  • @DanielFischer, I know. The goal of my question was to make the point that, to answer the OP's question, you don't need to go through the hassle of finding the expression of the Taylor series centered at $5i$ :) – etothepitimesi Aug 17 '14 at 20:40
  • Having looked up the table of contents of the referenced text, it doesn't look like the machinery you are calling for is developed by chapter 4. – jxnh Aug 17 '14 at 20:43
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    @el.Salvador the theorem that proves that is far more sophisticated than a simple geometric series. There are often several answers to any given question, each independent one making it unnecessary to go through the "hassle" of the others. – Adam Hughes Aug 18 '14 at 00:09
  • I wrote my comment in the wrong place so in case some missed it, I will write it again. – mark leeds Aug 18 '14 at 00:36
  • Thank you. I read the answers and one question I have is: since the series being expanded is 1/z what does it really mean to expand the series about the point 5i. to me the expansion is equivalent to the expansion about z = 0 since we subtracted and added 5i.thanks. – mark leeds Aug 18 '14 at 00:37
  • It's for a different set, that's all. Remember power series converge on disks, this one converges on a different one that the one centered around $z=0$. – Adam Hughes Aug 18 '14 at 00:41
  • hi adam: last bother. is the one that the answer was done for ( not mine. the right one ), a disk centered at the point in the complex plane (0,5i) ? thanks. – mark leeds Aug 18 '14 at 02:12
  • Yes, the disk for the series I wrote is centered at $z=5i$. – Adam Hughes Aug 18 '14 at 02:14
  • @Adam: If it's not simple, don't worry about it but $\frac{1}{z} = \frac{1}{1-5i -(z-5i)}$ so how can the same series converge on two different disks at the same time ? It might be obvious to most but not to me. thanks. – mark leeds Aug 18 '14 at 03:55
  • Its not the same series, they are different series for the same function. The function is only equal to the series on the disk of convergence. Different desk, naturally a different series. Also you said ${1\over 1-z}$, just for the record. – Adam Hughes Aug 18 '14 at 03:59
  • thanks adam. I had a typo in my question. I meant $\frac{1}{1-z}$. I'm gonna print out your wisdom and the other people's and go over carefully. this stuff definitely makes me play twister with my brain. – mark leeds Aug 18 '14 at 04:27
  • hi adam: not at all. I usually check all of them and meant to this time also. is there some etiquette about only checking one or it okay to check them all if you like them all ? all of youir answers were tremendous. i just still need to ponder the last part about a "different set". thanks again. – mark leeds Aug 18 '14 at 19:16
  • adam: I checked your answer now also but I'm not sure if it's allowed to do it later. if it doesn't stick, then my apologies. – mark leeds Aug 18 '14 at 19:18
  • @mark leeds: you are only allowed to accept one answer per question. It's usually to think the one who made it clearest to you with their explanation. That's why I asked if there was something I could have done better, I assumed that you found the other answer superior in some way, and I was going to try to improve my future answers if that was the case. – Adam Hughes Aug 19 '14 at 02:20
  • adam: thanks for explaining. no. it was my mistake. definitely your answer was great. I think I even get the concept of it converging on a different disk now, thanks to you. – mark leeds Aug 20 '14 at 03:40
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As was noted by Adam Hughes, the series for $\frac1{1-z}$ about the point $5i$ is $$ \begin{align} \frac1{1-z} &=\frac1{(1-5i)-(z-5i)}\\ &=\frac1{1-5i}\frac1{1-\color{#C00000}{\frac{z-5i}{1-5i}}}\\ &=\frac1{1-5i}\sum_{k=0}^\infty\left(\color{#C00000}{\frac{z-5i}{1-5i}}\right)^k\\ &=\sum_{k=0}^\infty\frac{(z-5i)^k}{(1-5i)^{k+1}} \end{align} $$ which converges by the ratio test for $|z-5i|\lt|1-5i|=\sqrt{26}$.

Another indicator of the radius of convergence of the Taylor series is the distance from the center of the expansion to the nearest singularity. Since the only singularity is at $z=1$, the radius of convergence is $$ |1-5i|=\sqrt{26} $$

robjohn
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  • thanks to all. I'm going to print out and read carefully since I clearly did the wrong thing. I will check all of the answers since they are all extremely useful. it's much appreciated. – mark leeds Aug 17 '14 at 23:56
  • I read the answers and one question I have is: since the series being expanded is 1/z what does it really mean to expand the series about the point $5i$. to me the expansion is equivalent to the expansion about z = 0 since we subtracted and added 5i.thanks. – mark leeds Aug 18 '14 at 00:20
  • The function being expanded is $\dfrac1{1-z}$. We use the series $\dfrac1{1-z}=1+z+z^2+z^3+\dots$ to get $$\frac1{1-5i}\frac1{1-\color{#C00000}{\frac{z-5i}{1-5i}}}=\frac1{1-5i}\sum_{k=0}^\infty\left(\color{#C00000}{\frac{z-5i}{1-5i}}\right)^k$$ – robjohn Aug 18 '14 at 03:33
  • my confusion is that $\frac{1}{1-z}$ can be expanded about the point $z = 0$ or $z = 5i$. doing this results in two different series even though the function in both cases is $\frac{1}{1-z}$. Adam explained that the function is on two different "sets" so I'm trying to ponder that. thanks for any insight on my difficulty. – mark leeds Aug 18 '14 at 19:24
  • Remember, that a function is only equal to the power series on the disk that is related to the series. Since we are dealing with a different series, which converges on a different disk we have a different situation. Note that the function is related to each of the individual series, but this series are not really related to one another other than through the function. I think your difficulty comes from the fact that you believe there should be some connection between the two series simply because they happen to deal with the same function. I.e. that they should be the same, but that is not so. – Adam Hughes Aug 18 '14 at 19:39
  • @adam. I will ponder your wisdom some more. and your help and patience is really appreciated. one more bother: is what you said above only true for complex series or series in general. thanks a lot. – mark leeds Aug 18 '14 at 20:20
  • @mark leeds convergence only depends upon the decay rate of the coefficients, so it doesn't matter what field you think about them being in. – Adam Hughes Aug 18 '14 at 20:35
  • @adam: is it correct to say that both series are converging to the function $\frac{1}{1-z}$, but A) the series expanded about $5i$ converges to $\frac{1}{1-z}$ where the center of the disk is (0,5i) and B) the series expanded about $0$ converges to $\frac{1}{1-z}$ where the center of the disk is the point $(0,0i)$. thanks. – mark leeds Aug 18 '14 at 20:50
  • Usually you don't put the $i$ in the ordered pairs notation, but yes. – Adam Hughes Aug 18 '14 at 21:08