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<math>u^2+v^2=c.</math> | <math>u^2+v^2=c.</math> | ||
| − | Take the partial derivative of this equation with respect to x to get one equation. Take it with respect to y to get another. Use the Cauchy-Riemann equations to conclude that the gradients of both u and v must be identically zero. | + | Take the partial derivative of this equation with respect to x to get one equation. Take it with respect to y to get another. Use the Cauchy-Riemann equations to conclude that the gradients of both u and v must be identically zero. (Note: The case c=0 is easy. If c is not zero, then it follows that it is not possible for u and v to vanish simultaneously.) |
Revision as of 09:57, 10 September 2009
Homework 2
Here's a hint for II.3.1 (ii) --Bell:
$ \frac{f(z)g(z)-f(z_0)g(z_0)}{z-z_0}=\frac{f(z)g(z)-f(z)g(z_0)+f(z)g(z_0)-f(z_0)g(z_0)}{z-z_0}= $
$ f(z)\frac{g(z)-g(z_0)}{z-z_0}+g(z_0)\frac{f(z)-f(z_0)}{z-z_0}. $
Here is a hint for II.8.1 (c) --Bell:
If the modulus of $ f=u+iv $ is constant, then
$ u^2+v^2=c. $
Take the partial derivative of this equation with respect to x to get one equation. Take it with respect to y to get another. Use the Cauchy-Riemann equations to conclude that the gradients of both u and v must be identically zero. (Note: The case c=0 is easy. If c is not zero, then it follows that it is not possible for u and v to vanish simultaneously.)
