Arsthbt

From the document we know that

Arg[z] gives the gives the argument of the complex number z.

Then how about Arg'[z]? This seems to be meaningless, but Mathematica returns something if z is a non-exact number, for example

Arg'[1. + I]
(* -0.5 *)

So my question is:

1. How is the numeric value of Arg'[z] defined?




2. Why does Arg' behave like this? What's the potential usage of this behavior?

The internal Trace[] Kuba advises shows calculations consistent with the numeric approximation of the partial derivative with respect to the real part:

D[ComplexExpand[Arg[x + I y], TargetFunctions -> Re, Im], x] /. 
 x -> 1 /. y -> 1
(* -(1/2) *)

This is what Mathematica does with the derivative of a numeric function with approximate input.

Other examples:

ClearAll[f, g];
f[x_?NumericQ] := Re[x]^2;
g[x_?NumericQ] := Im[x]^2;

f'[1. + I]
g'[1. + I]
(*
 1.999999999999995` 
 -2.7506672371246275`*^-15 
*)

It seems like the wrong way to evalutate Derivative.

The definition of the argument is $arg(z)=textIm(ln(z))$. Its partial derivative with respect to $z$ would then be

$$
fracpartial arg(z)partial z=
fracpartialpartial zfracln(z)-ln(z^*)2i
= -fraci2z.
$$

What you see looks like twice the real part of this expression:

With[z = 2. + 5 I,
 Arg'[z], 2Re[-I/(2z)]]

(* -0.172414, -0.172414 *)

I don't know in which sense this is the "correct" answer. It could be that what is actually calculated is not the partial derivative with respect to $z$, but rather the sum of partial derivatives

$$
fracpartial arg(z)partial z
+fracpartial arg(z)partial z^*
= -fraci2z+fraci2z^*
= -fractextIm(z)
$$

With[z = 2. + 5 I,
 Arg'[z], 2Re[-I/(2z)], -Im[z]/Abs[z]^2]

(* -0.172414, -0.172414, -0.172414 *)

Funny Answer,
Just ignore the method it uses to calculate and use your own method of calculation based on the result..

The reason for using this is to extended divided by two.

Arg'[1./2]