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/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2013 The Math Forum