12

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?

xzczd
  • 65,995
  • 9
  • 163
  • 468
  • This can shed some light: Trace[ Arg'[1. + I], TraceInternal -> True ] – Kuba Apr 25 '19 at 08:04
  • It is impossible to understand what the output from Trace[ Arg'[1. + I], TraceInternal -> True ] mean. May be numerics gone mad or something :) so just change 1.0 to 1 in the example given and then Arg will no longer do what you show. – Nasser Apr 25 '19 at 08:08
  • @Kuba Oh blinding light… – xzczd Apr 25 '19 at 08:09
  • @Nasser and xzczd it looks like it calculates derivative value from set of points f or Arg, but I wasn't paying too much attention. – Kuba Apr 25 '19 at 08:11
  • Just speculating, but I wonder if parts of the numerical code are working in the reals. Compare: Limit[(Arg[1 + I + h] - Arg[1 + I])/h, h -> 0, Direction -> #] & /@ {Reals, Complexes} – Greg Hurst Apr 25 '19 at 12:26
  • 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] – Empire of E Apr 25 '19 at 11:55
  • Er… this answer answers which part of my question? – xzczd Apr 25 '19 at 12:16
  • 1
    Related: https://mathematica.stackexchange.com/questions/29329/symbolic-derivatives-are-being-calculated-numerically – Michael E2 May 01 '19 at 15:27

2 Answers2

9

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 evaluate Derivative.

xzczd
  • 65,995
  • 9
  • 163
  • 468
Michael E2
  • 235,386
  • 17
  • 334
  • 747
5

The definition of the argument is $\arg(z)=\text{Im}(\ln(z))$. Its partial derivative with respect to $z$ would then be

$$ \frac{\partial \arg(z)}{\partial z}= \frac{\partial}{\partial z}\frac{\ln(z)-\ln(z^*)}{2i} = -\frac{i}{2z}. $$

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 partial derivative with respect to the real part of $z$:

$$ \frac{\partial \arg(z)}{\partial\text{Re}(z)} =\frac{\partial \arg(z)}{\partial z}\frac{\partial z}{\partial\text{Re}(z)} +\frac{\partial \arg(z)}{\partial z^*}\frac{\partial z^*}{\partial\text{Re}(z)}\\ =\frac{\partial \arg(z)}{\partial z} +\frac{\partial \arg(z)}{\partial z^*} = -\frac{i}{2z}+\frac{i}{2z^*} = -\frac{\text{Im}(z)}{|z|^2} $$

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

(* {-0.172414, -0.172414, -0.172414} *)
Roman
  • 47,322
  • 2
  • 55
  • 121
  • Er… how is derivative of $\ln(z^*)$ defined here? – xzczd Apr 25 '19 at 08:45
  • Any idea why then With[{z = 1.0 + I}, Arg'[z]] not same as With[{z = 1 + I}, Arg'[z]]? Should not these give same result? – Nasser Apr 25 '19 at 08:47
  • @xzczd $z$ and $z^$ are independent variables in complex analysis, so $\partial z^/\partial z=0$ etc. – Roman Apr 25 '19 at 08:48
  • 1
    @Nasser they do give the same result when you apply N: Arg'[1 + I] // N also gives -0.5. – Roman Apr 25 '19 at 08:50