Complex Analytic Functions

Date: 12/08/98 at 10:12:28
From: 10ecgal
Subject: Complex Analytic Functions

Dr. Math,

I am trying to find out if abs(z)*(conjugate z) is analytic and where, 
but I have gotten stuck. I know that since z = x + iy, then 

   abs(z) = sqrt(x^2 + y^2)  and (conjugate z) = x - iy

And:

   abs(z)*(conjugate z) = x * sqrt(x^2 + y^2) - iy * sqrt(x^2 + y^2)

Now I have to use Cauchy's formula and separate into u(x,y) and 
v(x,y). I get u(x,y) = x * sqrt(x^2 + y^2) and 
v(x,y) = -y * sqrt(x^2 + y^2). Now, u_x must equal v_y and u_y must 
equal -v_x to be analytic. I can't solve any further. If you could 
help me I would truly appreciate it.

Thanks.

Date: 12/08/98 at 13:44:31
From: Doctor Pete
Subject: Re: Complex Analytic Functions

Hi,

You're almost there --as you pointed out, all you need to do is compute 
the partial derivatives u_x, v_y, u_y, v_x.  Now, take u_x for example. 
We see that u is of the form:

     u = f(x) g(x)

where f(x) = x, and g(x) = sqrt(x^2+y^2). Here we treat y as a 
constant, as we are taking the partial derivative of u with respect 
to x. The above form is differentiated by the product rule:

     u_x = f'(x) g(x) + f(x) g'(x)
         = 1*sqrt(x^2+y^2) + x*g'(x)

where g'(x) is differentiated by the chain rule, giving:

     g'(x) = (1/2)(x^2+y^2)^(-1/2)(2x)
           = x(x^2+y^2)^(-1/2)

Hence:

     u_x = sqrt(x^2+y^2) + x^2/sqrt(x^2+y^2)
         = (2x^2+y^2)/sqrt(x^2+y^2)

The other derivatives are calculated in a similar way. However, I would 
like to point out to you that it is obvious that u_x is not equal to 
v_y, because:

     u(x,y) = x*sqrt(x^2+y^2)
     v(x,y) = -y*sqrt(x^2+y^2)

and so we see that the function u with respect to x is the negative of 
the function v with respect to y; hence u_x = -v_y. This simple fact 
alone tells you that the Cauchy-Riemann equations are not satisfied 
for general (x,y), and therefore abs(z) * conjugate(z) is not analytic, 
except possibly at (0,0).

- Doctor Pete, The Math Forum
  http://mathforum.org/dr.math/