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I am trying to find the set $\Omega$ where the series $$ \sum_{n \geq 1} \frac{1}{n} \left(\frac{2z-i}{2+iz}\right)^n $$ exhibits pointwise convergence.

I have thought of several approaches:

  • Turning (somehow) the series into some power series and then computing the convergence radius using the tools I've been given.
  • Seeing where $\frac{|2z-i|}{|2 + iz|} < 1$ but I haven't managed to have success through the computations.

Any idea of how I should tackle this problem?

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  • $\begingroup$ @hardmath fixed. Thanks for the tip $\endgroup$ – bvaldivielso Jun 14 '16 at 23:46
  • $\begingroup$ In the complex plane the open unit disk $|u|\lt 1$ is bounded by a circle. It is worth knowing that a linear fractional transformation (Moebius map) sends circles & lines to circles & lines, so the preimage of the unit disk under that map will not be very complicated. $\endgroup$ – hardmath Jun 14 '16 at 23:50
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Let us see where $\left\|\frac{2z-i}{2+iz}\right\|<1$ holds. Assuming $z=s+it$ with $s,t\in\mathbb{R}$, the previous constraint is equivalent to:

$$\|2z-i\|^2 = (2s)^2+(2t-1)^2 < (2-t)^2+s^2 = \|2+iz\|^2 $$ or to: $$ s^2+t^2 < 1 $$ i.e. the series is uniformly convergent over any compact subset of the open disk $\|z\|<1$ and not convergent outside that disk. Now you just have to study what happens on the boundary of such disk. It is useful to recall that for any $z$ such that $\|z\|<1$ the identity: $$ \sum_{n\geq 1}\frac{z^n}{n} = -\log(1-z)$$ holds, and: $$\frac{2e^{i\theta}-i}{2+ie^{i\theta}}=e^{-i\theta}\frac{2e^{i\theta}-i}{2e^{-i\theta}+i}=e^{-i\theta}\frac{w}{\bar{w}}=e^{i\varphi}. $$

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