(2 intermediate revisions by the same user not shown)
Line 1: Line 1:
 +
[[Category:MA425Fall2009]]
 +
 
==Homework 1==
 
==Homework 1==
  
Line 6: Line 8:
  
 
<math>f'(z_0)=\lim_{z\to z_0}\frac{f(z)-f(z_0)}{z-z_0}</math>
 
<math>f'(z_0)=\lim_{z\to z_0}\frac{f(z)-f(z_0)}{z-z_0}</math>
 +
 +
Here's a hint on I.8.3 --[[User:Bell|Steve Bell]]
 +
 +
It is straightforward to show that
 +
 +
<math>(z,w)\mapsto z+w</math>
 +
 +
is a continuous mapping
 +
 +
<math>(\mathbb C\times \mathbb C)\to\mathbb C</math>
 +
 +
because
 +
 +
<math>|(z+w)-(z_0+w_0)|\le|z-z_0|+|w-w_0|</math>
 +
 +
and to make this last quantity less than epsilon, it suffices to take
 +
 +
<math>|z-z_0|<\epsilon/2</math>
 +
 +
and
 +
 +
<math>|w-w_0|<\epsilon/2.</math>
 +
 +
To handle complex multiplication, you will need to use a standard trick:
 +
 +
<math>zw-z_0w_0 = zw-zw_0+zw_0-z_0w_0=z(w-w_0)+w_0(z-z_0)</math>.

Latest revision as of 06:43, 23 September 2009


Homework 1

HWK 1 problems

This is where members of the class could exchange ideas about the homework. Here is an example of a math formula that is easy to input:

$ f'(z_0)=\lim_{z\to z_0}\frac{f(z)-f(z_0)}{z-z_0} $

Here's a hint on I.8.3 --Steve Bell

It is straightforward to show that

$ (z,w)\mapsto z+w $

is a continuous mapping

$ (\mathbb C\times \mathbb C)\to\mathbb C $

because

$ |(z+w)-(z_0+w_0)|\le|z-z_0|+|w-w_0| $

and to make this last quantity less than epsilon, it suffices to take

$ |z-z_0|<\epsilon/2 $

and

$ |w-w_0|<\epsilon/2. $

To handle complex multiplication, you will need to use a standard trick:

$ zw-z_0w_0 = zw-zw_0+zw_0-z_0w_0=z(w-w_0)+w_0(z-z_0) $.

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood