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Consider $f(z) = 0^z$, where $z\in \mathbb{C}$. $f(z)$ is a zero-valued constant function in punctured complex plane $D:= \{\mathbb{C} \setminus \{0\}\}$. Therefore it is holomorphic in $D$. By applying Riemann Removable Singularity, we can extend $f(z) $ to $F(z)$ defined as $F(z) = f(z)$ when $z\ne 0$ and $F(z) =0$ when $z=0$.

The Laurent series of $f(z)$ around $z=0$ is $\sum_{n=-\infty}^{\infty} \frac{a_n}{z}$ where $a_n$ is a contour integral, which is going to be $0$ in this case because $f(z)$ being $0$.

My questions are: (1) As this Laurent series has all the coefficients $0$, do we need $F(z)$? Is t the Laurent series convergent and zero valued on all of the complex plane? (2) I believe there is nothing special about zero valued function and singularity at $z=0$. Similar convergence of Laurent series will be observed for all Riemann Removable Singularities making $F(z)$ definition an overkill. Am I incorrect in asserting this? The residue term will give value to the singularity and all other terms will vanish when we calculate $f(z_0)$ where $z_0$ is the singularity. (Unless the denominators $(z-z_0)$ will cause some problems that I am not accounting for).

Thanks in advance!

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  • $\begingroup$ It is unclear to me why $0^z$ should be constant in $\Bbb C\setminus\{0\}$. $\endgroup$ – Gae. S. Dec 10 '19 at 7:00
  • $\begingroup$ I’m thinking when you raise 0 to any finite Complex number (except for 0 - Where different results are possible based on the function) you get 0. $\endgroup$ – Shree Dec 10 '19 at 15:44
  • $\begingroup$ It is apparent that you are thinking that. I was asking if you opinion is based on something. $\endgroup$ – Gae. S. Dec 10 '19 at 16:27
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(1) The Laurent series is for $f(z)$, so essentially you cannot evaluate the Laurent series at $z=0$, although all the coefficients are $0$. However, you can define a new function using the Laurent series to extend your function to the domain of convergence. Now this function defined by the Laurent series is different from the original function. This extension would be exactly $F(z)$. So you can extend any function using Laurent series: consider the Laurent series near or on the boundary point of the domain, and patch up the function to get the extension.

(2) Yes. Actually that is the definition of the removable singularity: at the removable singularity, if you do the Laurent series expansion, there must be no negative degree term. When you find the Laurent series around the singularity $z_0$, $$ \sum_{k=-\infty}^{\infty} c_n (z-z_0)^n, $$ if the first nonzero coefficient is $c_{-k}$, $k$ is called the order of the singularity. If the order is 1, it is called the simple pole. If the order is smaller than or equal to 0, it is called the removable singularity. If all the negative terms are alive, it is called the essential singularity. So yes, if there is a removable singularity, you can fill the hole using the Laurent series expansion.

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  • $\begingroup$ (1) As all the coefficients are 0, there is no need to patch up. No? $\endgroup$ – Shree Dec 10 '19 at 15:34
  • $\begingroup$ @Shree Using the Laurent series, you are patching up at z=0 to find the extension, since the original function was not defined at that point. $\endgroup$ – J1U Dec 10 '19 at 15:37
  • $\begingroup$ so $F(z) = \sum \cdots$! I get it now.Thanks. $\endgroup$ – Shree Dec 10 '19 at 16:27

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