I am doing a multiple choice test for complex analysis, and I am stuck a bit at the following one.

Let $f$ be holomorphic with an essential singularity at $0$. Then for every $z_0\in \mathbb{C}$ the Laurent expansion around $z_0$ has a nonvanishing principal part.

My idea is that it is wrong, but I didn't found a counterexample. My idea is taking a function with an essential singularity like $e^{-x^{-2}}$, adding a term with no singularity like $\frac{1}{1-z}$ and choosing $z$ so that the Laurent series around $z$ has a smaller radius of convergence than the distance to the essential singularity. Does it work?


1 Answer 1


If $f$ is holomorphic on a neighbourhood of $z_0$, the Laurent expansion of $f$ around $z_0$ will be the same as the Taylor expansion around $z_0$. (I.e. the principal part vill vanish.)

(Assuming you mean the Laurent series that converges on some $0 < |z-z_0| < r$. There are other possible interpretations.)

  • $\begingroup$ So the Statement is false and my counterexpamle works ? $\endgroup$ Feb 26, 2013 at 13:10
  • $\begingroup$ @DominicMichaelis The statement is false. Take any $f$ with an essential singularity at $0$ and see what will happen, e.g. $f(z) = \exp(1/z)$. $\endgroup$
    – mrf
    Feb 26, 2013 at 13:12

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