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[http://www.math.purdue.edu/~bell/MA425/hwk2.txt HWK 2 problems] | [http://www.math.purdue.edu/~bell/MA425/hwk2.txt HWK 2 problems] | ||
| − | Here's a hint for II.3.1 (ii) --[[User:Bell|Bell]]: | + | Here's a hint for II.3.1 (ii) --[[User:Bell|Steve Bell]]: |
<math>\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}=</math> | <math>\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}=</math> | ||
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<math>f(z)\frac{g(z)-g(z_0)}{z-z_0}+g(z_0)\frac{f(z)-f(z_0)}{z-z_0}.</math> | <math>f(z)\frac{g(z)-g(z_0)}{z-z_0}+g(z_0)\frac{f(z)-f(z_0)}{z-z_0}.</math> | ||
| − | Here is a hint for II.8.1 (c) --[[User:Bell|Bell]]: | + | Here is a hint for II.8.1 (c) --[[User:Bell|Steve Bell]]: |
If the modulus of <math>f=u+iv</math> is constant, then | If the modulus of <math>f=u+iv</math> is constant, then | ||
Latest revision as of 06:44, 23 September 2009
Homework 2
Here's a hint for II.3.1 (ii) --Steve 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) --Steve 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.)
