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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.

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    $\begingroup$ If $|z - 5i| < 1$, the series you've written converges to $\frac{1}{1 - (z-5i)}$, not $\frac{1}{1-z}$. $\endgroup$ – Michael Albanese Aug 17 '14 at 20:08
  • $\begingroup$ 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 $\frac{1}{1-z}$ $\endgroup$ – 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

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

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  • $\begingroup$ Alternatively, before bringing the constant into the sum, you have a geometric series in $\frac{z-5i}{1-5i}$. $\endgroup$ – Michael Albanese Aug 17 '14 at 20:33
  • $\begingroup$ @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. $\endgroup$ – Adam Hughes Aug 17 '14 at 20:34
  • $\begingroup$ 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 $z=1$? $\endgroup$ – etothepitimesi Aug 17 '14 at 20:36
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    $\begingroup$ @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. $\endgroup$ – Daniel Fischer Aug 17 '14 at 20:37
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    $\begingroup$ @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. $\endgroup$ – Adam Hughes Aug 18 '14 at 0:09
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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 $$ |1-5i|=\sqrt{26} $$

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  • $\begingroup$ 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. $\endgroup$ – mark leeds Aug 17 '14 at 23:56
  • $\begingroup$ 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. $\endgroup$ – mark leeds Aug 18 '14 at 0:20
  • $\begingroup$ The function being expanded is $\dfrac1{1-z}$. We use the series $\dfrac1{1-z}=1+z+z^2+z^3+\dots$ to get $$\frac1{1-5i}\frac1{1-\color{#C00000}{\frac{z-5i}{1-5i}}}=\frac1{1-5i}\sum_{k=0}^\infty\left(\color{#C00000}{\frac{z-5i}{1-5i}}\right)^k$$ $\endgroup$ – robjohn Aug 18 '14 at 3:33
  • $\begingroup$ 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. $\endgroup$ – mark leeds Aug 18 '14 at 19:24
  • $\begingroup$ 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. $\endgroup$ – Adam Hughes Aug 18 '14 at 19:39

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