Mathematica 9 is now available
Services & Resources / Wolfram Forums
-----
 /
MathGroup Archive
1999
*January
*February
*March
*April
*May
*June
*July
*August
*September
*October
*November
*December
*Archive Index
*Ask about this page
*Print this page
*Give us feedback
*Sign up for the Wolfram Insider

MathGroup Archive 1999

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: Some problems with complex functions like Sqrt[z]

  • To: mathgroup at smc.vnet.net
  • Subject: [mg18207] Re: [mg18168] Some problems with complex functions like Sqrt[z]
  • From: Daniel Lichtblau <danl at wolfram.com>
  • Date: Mon, 21 Jun 1999 22:50:46 -0400
  • References: <199906200354.XAA27497@smc.vnet.net.>
  • Sender: owner-wri-mathgroup at wolfram.com

Robert Prus wrote:
> 
> Hi,
> 
> For any complex (holomorphic) function f[z] I should obtain 0 as the result
> of the following calculations:
> 
> Mathematica 3.0 for Silicon Graphics
> Copyright 1988-97 Wolfram Research, Inc.
>  -- Motif graphics initialized --
> 
> In[1]:= f[z_]:=z^3
> 
> In[2]:= re=Re[f[a+I b]]//ComplexExpand
> 
>          3        2
> Out[2]= a  - 3 a b
> 
> In[3]:= im=Im[f[a+I b]]//ComplexExpand
> 
>            2      3
> Out[3]= 3 a  b - b
> 
> In[4]:= D[{re,im},a].D[{re,im},b]//Simplify
> 
> Out[4]= 0
> 
> But if I choose more complicated function, like Sqrt[z], I obtain:
> 
> In[5]:= f[z_]:=Sqrt[z]
> 
> In[6]:= re=Re[f[a+I b]]//ComplexExpand
> 
>                                Arg[a + I b]
> Out[6]= Sqrt[Abs[a + I b]] Cos[------------]
>                                     2
> 
> In[7]:= im=Im[f[a+I b]]//ComplexExpand
> 
>                                Arg[a + I b]
> Out[7]= Sqrt[Abs[a + I b]] Sin[------------]
>                                     2
> 
> In[8]:= D[{re,im},a].D[{re,im},b]//Simplify
> 
>         I               2               2              2
>         - (Abs'[a + I b]  + Abs[a + I b]  Arg'[a + I b] )
>         4
> Out[8]= -------------------------------------------------
>                           Abs[a + I b]
> 
> In the following one can use the substitutions:
> 
> Abs'[x_+I y_] -> x/Sqrt[x^2+y^2]
> 
> Arg'[x_+I y_] -> -y/(x^2+y^2)
> 
> (one can check them using:
> 
> z=Abs'[x+I y]-x/Sqrt[x^2+y^2] (* or z=Arg'[x+I y]+y/(x^2+y^2) *)
> Plot3D[Chop[z],{x,-10,10},{y,-10,10}]
> Plot3D[Chop[z],{x,Random[],Random[]},{y,Random[],Random[]}]
> )
> 
> As the final result I obtain:
> 
> In[10]:= %8/.{Abs'[x_+I y_] -> x/Sqrt[x^2+y^2], Arg'[x_+I y_] -> -y/(x^2+y^2)}/.Abs[x_+I y_] -> Sqrt[x^2+y^2]//Together
> 
>                I
>                -
>                4
> Out[10]= -------------
>                2    2
>          Sqrt[a  + b ]
> 
> Under Mathematica 2.0 as the final result I have:
> 
> In[9]:= %8/.{Abs'[x_+I y_] -> x/Sqrt[x^2+y^2], Arg'[x_+I y_] -> -y/(x^2+y^2)}/.Abs[x_+I y_] -> Sqrt[x^2+y^2]//Together
> 
>                     b
> Out[9]= -------------------------
>                           2    2
>         4 (a - I b) Sqrt[a  + b ]
> 
> But it should be equal to zero !!!
> 
> Any comments?
> 
> Robert Prus, robert at fuw.edu.pl
> Institute of Theoretical Physics, Warsaw University
> Hoza 69, 00-681 Warsaw, Poland

I guess the problem lies in assuming that Abs'[...] and Arg'[...] make
sense. In your Plot3D version 3 of Mathematica does a numerical
differentiation and gets the "correct" result (version 4 just leaves
Abs'[...] unevaluated). But really you want to assume the function is
real analytic in order for the manipulations you did to make sense, and
this Abs is not. You need to replace Abs'[...] by a/Sqrt[a^2+b^2] in one
place, and by b/(I*Sqrt[a^2+b^2]) in another. Similar considerations
hold for Arg'[...].

How do I know this? You can rewrite your re and im using
ComplexExpand[..., TargetFunctions->{Re,Im}] (in order to avoid the Abs
and Arg) and then you get expressions like

In[62]:= re2 = ComplexExpand[re, TargetFunctions->{Re,Im}]
           2    2 1/4     ArcTan[a, b]
Out[62]= (a  + b )    Cos[------------]
                               2

In[63]:= im2 = ComplexExpand[im, TargetFunctions->{Re,Im}]
           2    2 1/4     ArcTan[a, b]
Out[63]= (a  + b )    Sin[------------]
                               2

When you take derivatives with respect to a or b and compare to D[re,a]
resp. D[im,b] you will see that the formal derivative Abs'[...] gets
replaced by functions that have different values in the complex plane
(specifically, the two I gave above). Same story for the other pair of
derivatives.

Note that it is now simple to verify the Cauchy-Riemann equations
directly. For example:

In[64]:= D[re2,a]-D[im2,b]
Out[64]= 0

In[66]:= D[re2,b]+D[im2,a]
Out[66]= 0


Daniel Lichtblau
Wolfram Research


  • Prev by Date: Re: problem with Part function
  • Next by Date: Re: problem with Part function
  • Previous by thread: Some problems with complex functions like Sqrt[z]
  • Next by thread: Re: Some problems with complex functions like Sqrt[z]