Originally Posted by

**CSM** So I had to calculate the radius of convergence of

$\displaystyle \frac{(z-i)}{z^3-z}$ around the point $\displaystyle z=2i$

This is how I tried it. I'd love to hear what is wrong with my attempt:

First I rewrote it

$\displaystyle \frac{(z-i)}{z^3-z}=\frac{(z-i)}{z(z-i)(z+i)}=\frac{1}{z(z+i)}$

Then with partial fractioning:

$\displaystyle \frac{a}{z}+\frac{b}{z+1}\Rightarrow a(z+i)+bz=z-i\Rightarrow a=-1,b=2$

So we can write are original function as :

$\displaystyle \frac{-1}{z}+\frac{2}{z+1}=\frac{1}{2i-(z-2i)}-\frac{2}{-i-(z-2i)}$

And use the geometric series formula:

$\displaystyle =2i\sum(\frac{z-2i}{2i})^n+i\sum(\frac{z-2i}{i})^n$

And finally apply the ratio test to the coefficients:

$\displaystyle a_n=(2i)^{n-1}, a_m=i^{n-1}$

$\displaystyle |\frac{a_n}{a_{n+1}}|=|\frac{(2i)^{n-1}}{(2i)^n}|=|2i|=2$

$\displaystyle |\frac{a_m}{a_{m+1}}|=|\frac{(i)^{n-1}}{(i)^n}|=|i|=1$

So I guess R would be one.

Please, not only advise me how to find the radius of convergence. Please tell me if I made any mistakes above (particularly in method.)