Showing that a function is analytic

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SUMMARY

The discussion centers on demonstrating that the function $$\sum_{|z| < 1, P_k(z) = 0} z^t$$ is analytic in the parameter ##k## for a family of entire functions ##P_k(z)## that are nonzero on the unit circle ##S^1##. The integral representation $$\sum_{|z| < 1, P_k(z) = 0} z^t = {1\over{2\pi i}} \int_{\partial \Delta} {{P_k'(z)z^t}\over{P_k(z)}}dz$$ is utilized, leveraging the residue theorem and Morera's theorem to establish the analyticity in ##k##. Participants agree that this method is standard for such problems and express a lack of alternative approaches.

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  • Understanding of entire functions and their properties
  • Familiarity with the residue theorem in complex analysis
  • Knowledge of Morera's theorem regarding analyticity
  • Basic comprehension of complex variable summation notation
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Mathematicians, particularly those specializing in complex analysis, as well as students and researchers interested in the properties of analytic functions and entire functions.

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.
 

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