-1
$\begingroup$

Find the radius of convergence of the power series of $\frac{z^2-1}{z^3-1}$ at $z =2$.

Is there any nice way to find the power series representation of such function? I think I need to take a derivative of some known power series but I don't know what power series could help. Can you answer this?

$\endgroup$
1

1 Answer 1

Reset to default
2
$\begingroup$

Your function can be written as: $$f(z)=\frac{(z+1)(z-1)}{(z-1)(z^2+z+1)}$$ It's easy to see that this function has only two (non-removable) singularities, at the roots of $z^2+z+1$. The radius of convergence of the power series expansion around $z=2$ is equal to the distance from $z=2$ to the nearest point of singularity.

$\endgroup$
1
  • $\begingroup$ Thank you for your answer. Just saying those thing is enough? It's intuititve but I wonder if it's accepted as a proof $\endgroup$
    – one potato two potato
    Apr 18, 2021 at 7:43

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .