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==Homework 1== | ==Homework 1== | ||
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<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|Bell]] | + | Here's a hint on I.8.3 --[[User:Bell|Steve Bell]] |
It is straightforward to show that | It is straightforward to show that | ||
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<math>|w-w_0|<\epsilon/2.</math> | <math>|w-w_0|<\epsilon/2.</math> | ||
| − | To handle complex multiplication, you will need to use | + | 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>. | <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 07:43, 23 September 2009
Homework 1
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) $.
