What is the radius of convergence of $i + \dfrac{2}{1-i}\displaystyle\sum_{j=1}^{\infty} \biggl(\dfrac{z-i}{1-i}\biggr)^{j}$?

Question

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?

Answer 1

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. $\;|z-i|<|1-i|=\sqrt 2$.

Answer 2

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 $\sqrt2\left(=\lvert 1-i\rvert\right)$.

Answer 3

Alternatively:

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

Then

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