Weird question because I feel like I sort of "accidentally" did something right, and I would rather be able to do it on purpose.
So I have $f(z)={1 \over z^3-z^4}$ which I need to expand around the centre $z_0=0$ with radius 1, excluding the origin. To briefly touch upon what I did: $$ f(z)={1 \over z^3(1-z)}= {1\over z}+{1\over z^2}+{1\over z^3}+{1\over 1-z}$$ The first three addendums are already centred around $0$, and the last one is in the form of a geometric series, so we get $$f(z)={1 \over z^3(1-z)}= {1\over z^3}+{1\over z^2}+{1\over z}+\sum_{n=0}^\infty z^n=\sum_{n=-3}^\infty z^n$$
What I'm wondering is: where did the radius and the exclusion of the origin come into place? What should I have done different, if for example I wanted $R>1$? I think I remember that the radius can be defined as the shortest distance greater than 0 between the centre of expansion and a singularity, which in this case is clearly 1 (from the 4th addendum). Does that mean that $R=1$ is the only radius of expansion that this function allows?
Thanks!
The equality$$\sum_{n=0}^\infty z^n=\frac1{1-z}$$holds if and only if $\lvert z\rvert<1$. That's where you used the given condition.
If someone had asked you to do the same thing still with center at $0$ but this time with radius $2$, then that person should be fired, because then the problem would have no solution.