Undergrad Showing that a function is analytic

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The discussion focuses on demonstrating that the function defined by the sum of zeros of the entire function family ##P_k(z)## is analytic in ##k##. The proposed method utilizes the residue theorem to express the sum as an integral, which clearly shows its analyticity through Morera's theorem. Participants confirm that the integral representation is a standard approach for such problems. Clarification is sought on the summation notation used in the original query. Overall, the conversation emphasizes the effectiveness of the residue theorem in establishing the analyticity of the function.
OscarAlexCunning
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Say we have ##P_k(z)## a family of entire functions, and they depend analytically on ##k## in ##\Delta##. Assume ##P_k(z)## is nonzero on ##S^1## for all ##k##. How do I see that for each ##t \ge 0##, we have that$$\sum_{|z| < 1, P_k(z) = 0} z^t$$is an analytic function of ##k##? Here, the zeros of ##P_k(z)## are regarded with multiplicity.

Now, I can show this myself, but I am wondering if you guys have any alternative solutions.

My way of doing this. Via the residue theorem, for each ##k## in ##\Delta## we have$$\sum_{|z| < 1, P_k(z) = 0} z^t = {1\over{2\pi i}} \int_{\partial \Delta} {{P_k'(z)z^t}\over{P_k(z)}}dz.$$This integral representation makes it evident that our function of interest is analytic in ##k## by, for instance, the theorem of Morera.
 
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I don't understand your summation notation. Can you explain it?
 
Your way is the standard method for this type of problem. No other easy method comes to (my) mind.
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

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