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How do you find the radius of convergence of a Taylor series for a function centered at point $z_0$ without actually finding the Taylor series?

I know that we can use comparison test, ratio test or root test to find the radius when we find the Taylor series but how do you do this without finding the series?

For example, finding the radius of convergence for the Taylor series of $f(z) = z^i$ centered at $z_0=2$.

Thank you

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    $\begingroup$ For "sufficiently nice" functions, it is the distance to the nearest singularity. $\endgroup$
    – Michael Burr
    Apr 10, 2016 at 20:46
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    $\begingroup$ @MichaelBurr, singularity. $\endgroup$
    – Martín-Blas Pérez Pinilla
    Apr 10, 2016 at 20:47
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    $\begingroup$ @Martín-BlasPérezPinilla Thanks for the catch. $\endgroup$
    – Michael Burr
    Apr 10, 2016 at 20:50
  • $\begingroup$ I'm affraid you need to study first some complex-analysis course on holomorphic (analytic) functions, and prove the Cauchy integral formula, before showing that if $f(z)$ is locally analytic around $z_0$ then the radius of convergence of the power series around $z_0$ is exactly the distance to the nearest singularity from $z_0$. (if someone knows another way to prove it..) $\endgroup$
    – reuns
    Apr 10, 2016 at 21:12

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If all else fails try the distance to the nearest singularity. For $z^i=\exp(i\ln z)$ we have a singularity at $z=0$, hence expect convergence radius $2$.

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    $\begingroup$ what do you mean "try" ? if $f(z)$ is locally analytic around the point then the radius of convergence of the Taylor series at the point is always the distance to the nearest singularity (of the analytic continuation of the function..) $\endgroup$
    – reuns
    Apr 10, 2016 at 21:15

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