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Sorry if this question was asked before... I was searching and I dont found something about that.

Suppose that I have a function on the complexes, that is $f:\Bbb C\to\Bbb C$, and I want to evaluate it (total) derivative. I know that I can write the function as a function in $\Bbb R^2$ of the kind $f\mapsto g:=(\Re(f),\Im(f))$, and after see if the function $g$ is totally differentiable and, when it happen, if the Cauchy-Riemann equations holds to see if the original function $f$ is complex differentiable in this region.

But probably there is some shortcut to this workaround, and this is what Im searching for.

EDIT: because some trolls voted to close this question by the supposed reason that "this is not related to the Wolfram mathematical software" I must declare that I want a way to evaluate the derivative of a complex function in Wolfram Mathematica, of course! :)

By example the following code

f[z_]:= D[z^2,z]

when called after, by example for f[3+I] says that 3+Iis not a correct variable. So I must found other method.

J. M.'s missing motivation
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Masacroso
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  • Are you sure you didn't intend to ask this on math.stackexchange.com? – b3m2a1 Sep 04 '17 at 04:35
  • @b3m2a1 of course NOT. Read my edit please. – Masacroso Sep 04 '17 at 04:41
  • See Assuming. e.g., Assuming[z ~Element~ Complexes, D[z^2, z]]. Also maybe give the intro tutorials a look. f[3+I] insert 3+I anywhere it sees z. So that expanded to D[(3+I)^2, 3+I] which is, of course, an improper usage of D. Note that D[(3 + I)^2, z] works. – b3m2a1 Sep 04 '17 at 04:46
  • If the answers in the linked question do not solve your problem, please edit your question to explain why. – J. M.'s missing motivation Sep 04 '17 at 04:47
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    One method is Clear[f, x, z]; f[z_] = D[x^2, x] /. x -> z and then f /@ {x, z, 3 + I} will evaluate the way we want. But if x and z already have values that cannot be cleared, another method is Clear[f]; f[z_] := Module[{x}, D[x^2, x] /. x -> z], and then f/@{x, z, 3+I} will evaluate the way we want. Note the use of Set in the first method and SetDelayed in the second method. – LouisB Sep 04 '17 at 05:45

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