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Let $c$ be a non-zero complex number, and consider the power series \begin{equation} S(z)=\frac{z-c}{c}-\frac{(z-c)^2}{2c^2}+\frac{(z-c)^3}{3c^3}-\ldots. \end{equation} By using the Ratio Test, or otherwise, show that the series has radius of convergence $|c|$. By differentiating term by term, show that $S'(z)= \frac{1}{z}$.

I've never done power series in complex analysis, so this is what I've attempted so far:

\begin{equation} S(z)=\frac{z-c}{c}-\frac{(z-c)^2}{2c^2}+\frac{(z-c)^3}{3c^3}-\ldots\\ =\sum_{n=0}^{\infty} \frac{(-1)^n (z-c)^n}{nc}. \end{equation} Let $x_{n}=\frac{(-1)^n}{nc}$. Using the Ratio test, we have: \begin{equation} lim_{n \rightarrow \infty} \lvert \frac{x_{n+1}(z-c)^{n+1}}{x_{n}(z-c)^n} \rvert = \lvert z-c \rvert lim_{n \rightarrow \infty} \lvert - \frac{n}{n+1} \rvert = -\lvert z-c \rvert <1. \end{equation}

I'm pretty sure this is wrong somewhere, but I have no idea how to continue to show that the radius of convergence is $|c|$.

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  • $\begingroup$ It should be $c^n$ on the bottom of the fraction in the sum. $\endgroup$ – Chappers Jan 16 '16 at 17:57
  • $\begingroup$ What about the 2, 3, ... in the denominator? $\endgroup$ – MOA Jan 16 '16 at 18:00
  • $\begingroup$ Sorry, I meant instead of $c$; the whole denominator should be $nc^n$. $\endgroup$ – Chappers Jan 16 '16 at 18:03
  • $\begingroup$ A ok, thanks! But that leaves me with $\frac{-1}{c} \lvert z-c \rvert$. How do I get to $|c|$? $\endgroup$ – MOA Jan 16 '16 at 18:06
  • $\begingroup$ First, you need to take the absolute value (which you missed in your last equality in the question). Then $\{z:\lvert z-c \rvert < c\}$ is precisely the interior of the circle of radius $c$ centred on the complex number $c$. $\endgroup$ – Chappers Jan 16 '16 at 18:09
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Notice (hint):

First of all:

  • When $n$ 'starts' with $0$ than we've got a problem, because we get (dividing by $0$):

$$\frac{(-1)^0(z-c)^0}{0c}$$

  • Use the ratio test, to proof that this series converges, when $|c-z|<1$.
  • So:

$$\sum_{n=1}^{\infty}\frac{(-1)^n(z-c)^n}{cn}=-\frac{\ln(1+z-c)}{c}\space\space\space\space\space\space\text{when}\space|c-z|<1$$

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  • $\begingroup$ This reproduces a mistake in the OP, signalled in comments: note that the denominators are $nc^n$, not $cn$. $\endgroup$ – Did Jan 17 '16 at 10:49

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