If the principal part converges for, say, $|z|>1$, and the analytic part (the positive powers in ($z-z_0$)) converges for $|z|<2$, then does the Laurent series, as a whole, converge in the annulus $1<|z|<2$?

So, this would be like taking the overlapping region of convergence of the principal part and the analytic part of the Laurent series.

Is this thinking correct?

  • 1
    $\begingroup$ Yes, that is correct. $\endgroup$
    – Martin R
    Commented Dec 15, 2015 at 18:57

1 Answer 1


Yes, that is correct.

A Laurent series $$\sum_{k=-\infty}^{\infty} a_k (z-z_0)^k$$ will converge on a (possibly empty) annulus, $r < |z-z_0| < R$. The simplest way to see this is to split the series $$ \sum_{k=-\infty}^{\infty} a_k (z-z_0)^k = \sum_{k=-\infty}^{-1} a_k (z-z_0)^k + \sum_{k=0}^{\infty} a_k (z-z_0)^k, $$ and by definition the doubly-infinite series converges iff both these split series do. The second one is a power series, and will have a radius of convergence $R$. The first can be viewed as a power series in $w = \frac{1}{z-z_0}$ and will converge on $|w| < \rho$ for some $\rho$, i.e. for $|z-z_0| > r = \frac1\rho$.


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