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When I apply the ratio test, I get:

$\displaystyle\lim_{j \rightarrow \infty}\biggl| \dfrac{2(z-i)^{j+1}(1-i)^{j+1}}{(1-i)^{j+2}(z-i)^{j}}\biggr|$=$\displaystyle\lim_{j \rightarrow \infty} \biggl|\frac{2(z-i)}{(1-i)}\biggr|$. Thus giving me that that the radius of convergence is

$|z-i|<\dfrac{\sqrt{2}}{2}$? I'm not sure if I'm doing the process correctly?

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    $\begingroup$ The $2$ should also appear in the denominator. $\endgroup$ – user647486 Apr 14 at 18:29
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Why not use substitution? With $\;u=\dfrac{z-1}{1-i}$, you obtain the series in $u$: $$1+\frac 2{1-i}\sum_{j=1}^\infty u^j,$$ which converges if and only $|u|<1$, i.e. $\;|z-i|<|1-i|=\sqrt 2$.

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  • $\begingroup$ whenever I'm finding the radius of convergence, do I have to take into account whatever is multiplied with the summation? so, $\dfrac{2}{1-i}$ in this case? $\endgroup$ – K.M Apr 14 at 18:44
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    $\begingroup$ No, unless it is a series. $\endgroup$ – Bernard Apr 14 at 18:45
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I don't know where does that $2$ come from. Your approach is correct (the root test would work too) and the radius of convergence is $\sqrt2\left(=\lvert 1-i\rvert\right)$.

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  • $\begingroup$ I took the $2$ outside the summation and included it with $\dfrac{(z-i)^{j}}{(1-i)^{j}}$ $\endgroup$ – K.M Apr 14 at 18:38
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    $\begingroup$ But if you put it in the numerator, you have to put it in the denominator too. $\endgroup$ – José Carlos Santos Apr 14 at 18:55
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Alternatively:

The series is well known to have unit radius of convergence for the argument $\dfrac{x-i}{1-i}$.

Then

$$\left|\dfrac{x-i}{1-i}\right|<1\iff|x-i|<\sqrt2.$$

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