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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]]:
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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 07:44, 23 September 2009


Homework 2

HWK 2 problems

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.)

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