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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!

Silence
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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.

  • That's what I thought. So technically asking for $R=1$ is redundant. How sneaky... thank you! – Silence Jan 27 '20 at 17:22
  • Not quite. If the question was to find the Laurent expansion in the region ${z\in\mathbb C,|,\lvert z\rvert>1}$, that would make sense and it would have a different answer. – José Carlos Santos Jan 27 '20 at 17:32
  • Well yeah, he last addendum would commute to $\sum_{n=1}^\infty z^{-n}$ instead. I expressed myself poorly, I meant that specifying a module isn't necessary, since this function only makes sense considered around $R=1$ (either inside or outside). I.e. I could've worded the question "calculate the Laurent expansion within/outside its radius of convergence". Thank you for correcting me, and please correct me again if I'm wrong. – Silence Jan 27 '20 at 21:46
  • In this case, that is correct. – José Carlos Santos Jan 27 '20 at 21:48
  • Brilliant, thank you again. – Silence Jan 27 '20 at 23:48