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If I have two functions $f,g$ that are holomorphic around a point $z_0 \in \mathbb{C}$. Assume the Laurent series are known and both $f$ and $g$ have a finite principal part. $$f(z) = \sum_{n=m_f}^{\infty} a_n(z-z_0)^n\\g(z) = \sum_{n=m_g}^{\infty} b_n(z-z_0)^n$$ Here $m_f,m_g \in \mathbb{Z}$ are the orders of $f$ and $g$ (negative if poles). Now I am interested in the Laurent series of the quotient of $f$ and $g$ so I follow: $$\sum_{n=m}^{\infty} c_n(z-z_0)^n=\frac{f(z)}{g(z)} = \frac{\sum_{n=m_f}^{\infty} a_n(z-z_0)^n}{\sum_{n=m_g}^{\infty} b_n(z-z_0)^n}\\ \Rightarrow \left(\sum_{n=m}^{\infty} c_n(z-z_0)^n\right) \cdot \left(\sum_{n=m_g}^{\infty} b_n(z-z_0)^n\right) = \sum_{n=m_f}^{\infty} a_n(z-z_0)^n\\ \Rightarrow \sum_{n=m_f+m_g}^{\infty} \sum_{k=m_f+m_g}^n c_kb_{n-k}(z-z_0)^n\\ \Rightarrow a_n = \sum_{k=m_f+m_g}^n c_kb_{n-k}$$

And then I can successively calculate the $c_n$, starting with $c_{m_f+m_g}$. Is this a valid way to get the Laurent series of the quotient? And under which premises?

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  • $\begingroup$ Check your bounds for $k$. Note that $f/g$ will have a term in $(z - z_0)^{m_f - m_g}$. Also $n-k$ could go down to $m_g$. $\endgroup$ – Robert Israel May 12 '14 at 15:38
  • $\begingroup$ @RobertIsrael Thanks. I think it should be $m_f+m_g$ then, since they can be negative, right? $\endgroup$ – Leif Sabellek May 12 '14 at 15:45
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If you know how to do the Taylor series for $f/g$ where $f$ and $g$ are analytic and $g(z_0) \ne 0$, then you can just note that $f/g = (z - z_0)^{m_f - m_g} F/G$ where $F(z) = (z - z_0)^{-m_f} f(z)$ and $G(z) = (z - z_0)^{-m_g} g(z)$.

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