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Let's say I have a complex function like $$f(z)=z+e^{\frac{1}{z-1}}+\sin\left(\frac{1}{z-2}\right)$$ which is holomorphic in $\mathbb{C}\setminus \left \{ 1,2 \right \}$

Then there exist entire functions $$f_{0}(z)=z$$ $$f_{1}(z)=e^{z}$$ $$f_{2}(z)=\sin(z)$$

Such that, the equality $$f(z)=f_{0}(z)+f_{1}\left(\frac{1}{z-1}\right)+f_{2}\left(\frac{1}{z-2}\right)$$ holds for all points except for $z=1$ and $z=2$ which are essential singularities.

My question is, can this be extended in general?

Precisely put, let's say I have a holomorphic function $f(z)$ except for a finite number of essential singularity points $a_{1},\ldots,a_{n}$ in the complex plane. Is it true in general, that there exist $n+1$ entire functions $f_{0}(z),\ldots,f_{n}(z)$ such that: $$f_{i}(z) \text{ is holomorphic in }\mathbb{C}\setminus \left \{ a_{i} \right \} \text{ for }i=1,\ldots,n $$

$$f(z)=f_{0}(z)+f_{1}\left(\frac{1}{z-a_{1}}\right)+\ldots+f_{n}\left(\frac{1}{z-a_{n}}\right)$$

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  • $\begingroup$ You can take a look at weierstrass factorization theorem and Mittag-Lefler's theorem $\endgroup$ Dec 23, 2016 at 8:45
  • $\begingroup$ Mittag-Leffler's theorem is a kind of decomposition of a function in terms of its singularities. Its relation with Weierstrass factorization theorem is obvious. Many call these theorems "cousins". This is a generalization of OP's idea. $\endgroup$ Dec 23, 2016 at 9:07
  • $\begingroup$ @BeslikasThanos I don't see. OP's idea is about functions with finitely many isolated singularties, and is the answer I wrote, while Weierstrass's factorization theorem and Mittag-Lefler's theorem are about functions with infinitely many zeros and poles (otherwise they are trivial). $\endgroup$
    – reuns
    Dec 23, 2016 at 9:09
  • $\begingroup$ Yes but he cqn see these techniques in order to obtain something similar. I didnt post as an answer but to give motivation $\endgroup$ Dec 23, 2016 at 9:12

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  • If $f(z)$ is holomorphic on $0 < |z| < r$ then the Cauchy integral formula in an annulus gives $f(z) =\frac{1}{2i\pi} \int_{|s|= R_1} \frac{f(s)}{s-z}ds-\frac{1}{2i\pi}\int_{|s|= R_2} \frac{f(s)}{s-z}ds$ for $ 0 < R_2 < |z| < R_1 < r$.

  • Expanding $\frac{1}{s-z}$ in a geometric series we get the Laurent series expansion $f(z) = \sum_{n=-\infty}^\infty a_n z^n$ where $a_n =\frac{1}{2i\pi} \int_{|s|= R} f(s) s^{-n-1}ds$ and $0 < R < r$.

  • Let $g(z) = \sum_{n=1}^\infty a_{-n} z^n$. It converges for $|z| > 1/r$ and hence for every $z$ : it is entire.

  • $f(z)-g(1/z)$ is clearly holomorphic at $z=0$

$\implies$ Every holomorphic function on $\mathbb{C} \setminus \{z_1,\ldots,z_K\}$ can be written as $f_0(z)+\sum_{k=1}^K f_k(\frac{1}{z-z_k})$ where $f_k$ are entire

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  • $\begingroup$ Perfect :) Exactly in the lines of what I thought about and I realised what I missed $\endgroup$
    – zokomoko
    Dec 23, 2016 at 9:36

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