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Suppose $w_1,w_2,w_3,...$ are points on the unit circle. Consider the infinite series $$\sum_{k=1}^{\infty} \frac{1}{2^k} \frac{1}{z-w_k}$$ Let $D=\{z \in \mathbb{C}: |z|<1 \}$

A) Show that series converges for each $z$ in $D$.

B) Show that the sum of the series is an analytic function of $z$ for $z$ in $D$.

My approach for A

Since $|w_k|=1$,

$$ \left |\frac{1}{2^k} \frac{1}{z-w_k} \right | \le \frac{1} {1-|z|} $$ which converges by comparison to $\sum |z|^n$ for $|z|<1$

Is this the right approach?

For B, I don't know how to go about it any help will be appreciated thank you.

User69127
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    Hint: If a sequence of "complex" analytic function converges uniformly in a region $S$ of the complex plane, the limit is also analytic in $S$. For a proof, see answers of this question. The key is Morera's theorem. – achille hui May 13 '14 at 05:41
  • @achillehui I never had this theorem in my course. How would we do it by comparison? – User69127 May 13 '14 at 07:00
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    Let $f_s(z) = \sum\limits_{k=1}^\infty\frac{1}{2^k}\frac{1}{(z-w_k)^s}$ for $s \in \mathbb{N}$. $f_1(z)$ is the limit of your series and it is well defined over $D$ by part $A$. By a similar argument, $f_2(z)$ is well defined over $D$ too. If one differentiate $f_1(z)$ formally term by term, one will get $-f_2(z)$. So what you need to do is prove

    $$\lim_{h\to 0} \frac{f_1(z+h)-f_1(z)}{h} + f_2(z) = 0$$

    which isn't very hard for you specific form of series.

     – achille hui May 13 '14 at 07:26

1 Answers1

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Let $A_{k,m}(z)=-2^{-k}w_k^{-m-1}z^m$ for $k\geq1, m\geq0$, and $z\in D$. We have $\vert A_{k,m}(z)\vert=2^{-k}\vert z\vert^m$, so $$\sum_{k,m}\vert A_{k,m}(z)\vert=\frac{1}{1-\vert z\vert}.$$ This absolute convergence, (in fact, it is uniform on every compact subset of $D$), allows us to interchange the order of summation as follows: $$ \sum_{k=1}^\infty\left(\sum_{m=0}^\infty A_{k,m}(z)\right)= \sum_{m=0}^\infty\left(\sum_{k=1}^\infty A_{k,m}(z)\right) $$ for every $z\in D$. This is equivalent to $$ \sum_{k=1}^\infty\frac{1}{2^k}\cdot\frac{1}{z-w_k}= \sum_{m=0}^\infty a_mz^m $$ with $$ a_m=\sum_{k=1}^\infty\frac{-1}{2^kw_k^{m+1}}.$$ This proves that the function : $\displaystyle f(z)\buildrel{\rm def}\over{=}\sum_{k=1}^\infty\frac{1}{2^k}\cdot\frac{1}{z-w_k}$ is analytic in the unit disk $D$ and gives its power series expansion.

Omran Kouba
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  • There is a loose end in this proof. One should justify why the radius of convergence of $\sum_{m=0}^\infty a_m z^m$ is at least $1$. – achille hui May 13 '14 at 18:33
  • @achillehui A power series that converges for $|z|<1$ must have a radius of convergence (RC) greater or equal to $1$. because the fact that $(|a_m|r^m)_m$ is bounded for every $r\in[0,1)$ implies that $RC\geq1$. – Omran Kouba May 13 '14 at 18:38
  • +1 Oh I see, you are using the fact that LHS converges. This is clever. I was thinking a more explicit justification $|a_m| \le 1$. – achille hui May 13 '14 at 18:42