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Studying a new command of 12.2 FunctionMeromorphic, I try

FunctionMeromorphic[(Re[z] - I*Im[z])/Abs[z]^2, z]

False

FunctionMeromorphic[(Re[z] - I*Im[z])/Abs[z]^2 // Simplify, z]

False

FunctionMeromorphic[((Re[z] - I*Im[z])/
Abs[z]^2 /. {Re[z] -> (z + Conjugate[z])/2, 
Im[z] -> (z - Conjugate[z])/2, 
Abs[z] -> Sqrt[((z + Conjugate[z])/2)^2 + ((z - Conjugate[z])/2)^2]}) //FullSimplify, z]

False

FunctionMeromorphic[(Re[z] - I*Im[z])/Abs[z]^2 // FullSimplify, z]

True

I suspect the latest result is not true. I am right, aren't I?

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    $\begingroup$ The last result is of course correct since the simplification gives 1/z. The function is almost certainly based mostly on expression structure and will not implicitly simplify (or, if it does, might not attain a meromorphic equivalent when one exists). $\endgroup$ – Daniel Lichtblau Apr 7 at 15:15
  • $\begingroup$ @DanielLichtblau: Thak you for your explanation of the wrong behavior of FunctionMeromorphic. Let me repeat my suggestion "FunctionMeromorphic and FunctionAnalytic hardly handle Re[z], Im[z], and Conjugate[z]. As a palliative measure, these commands would be able to return inputs with Re[z], Im[z], and Conjugate[z] unevaluated". Hope I am clear. $\endgroup$ – user64494 Apr 7 at 17:12
  • $\begingroup$ The downvoter: What is wrong with my question? Deep regard. $\endgroup$ – user64494 Apr 7 at 17:24
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    $\begingroup$ They are treating Re, Im et al as they were designed to do. (I'm not the downvoter. If anything, I might vote to close as "requiring expert explanation" since it's a design question.) $\endgroup$ – Daniel Lichtblau Apr 7 at 18:16
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The function (Re[z] - I*Im[z])/Abs[z]^2 is $$ f(z) = {\bar z \over |z|^2} = {\bar z \over z\, \bar z}={1\over z} $$ which is meromorphic (holomorphic over $\mathbb C\backslash \{0\}$). Actually

(Re[z] - I*Im[z])/Abs[z]^2 // FullSimplify

returns 1/z and

FunctionMeromorphic[1/z, z]

returns True.

As a side note one would be tempted to apply FullSimplify to all functions before checking with FunctionMeromorphic, however this causes other errors. For example the function $z\mapsto \bar z/\bar z$ is defined with value $1$ over $\mathbb C\backslash \{0\}$ and undefined at $z=0$ ; however

FunctionMeromorphic[(Re[z] - I*Im[z])/Conjugate[z] // FullSimplify, z]

incorrectly returns True.

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  • $\begingroup$ Thank you for your response. Concerning FunctionMeromorphic[z/z // FullSimplify, z], the same produces FunctionMeromorphic[z/z ,z] since Mathematica automatically simplifies expressions.. In my question FunctionMeromorphic[((Re[z] - I*Im[z])/ Abs[z]^2 /. {Re[z] -> (z + Conjugate[z])/2, Im[z] -> (z - Conjugate[z])/2, Abs[z] -> Sqrt[((z + Conjugate[z])/2)^2 + ((z - Conjugate[z])/2)^2]}) //FullSimplify, z] is presented, where FullSimpliy does not work. $\endgroup$ – user64494 Apr 7 at 14:04
  • $\begingroup$ Your understanding of a meromorphic funcion is not true: Exp[1/z] is analytic for z!=0, but is not meromorphic (see Wiki for the definition). Thank you again for your interest to the topic. $\endgroup$ – user64494 Apr 7 at 14:11
  • $\begingroup$ @user64494 Thanks, I made a few changes and gave another example where FullSimplify is not appropriate. $\endgroup$ – A.G. Apr 7 at 14:38
  • $\begingroup$ Yes, FunctionMeromorphic and FunctionAnalytic hardly deal with Re[z], Im[z], and Conjugate[z]. As a palliative measure, these commands would be able to return inputs with Re[z], Im[z], and Conjugate[z] unevaluated. However, Re[z], Im[z] and Conjugate[z] are used in important applications. $\endgroup$ – user64494 Apr 7 at 15:02

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