| Line 15: | Line 15: | ||
and to make this last quantity less than <math>\epsilon</math>, it suffices to take | and to make this last quantity less than <math>\epsilon</math>, it suffices to take | ||
| − | <math>|z-z_0|<\epsilon/2< | + | <math>|z-z_0|<\epsilon/2</math> and <math>|w-w_0|<\epsilon/2</math>. |
To handle complex multiplication, you will need to use the standard trick: | To handle complex multiplication, you will need to use the 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>. | ||
Revision as of 08:30, 3 September 2009
Homework 2
Here's a hint on I.8.3 --Bell
It is straightforward to show that
$ (z,w)\mapsto z+w $
is a continuous mapping from $ \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 the standard trick:
$ zw-z_0w_0 = zw-zw_0+zw_0-z_0w_0=z(w-w_0)+w_0(z-z_0) $.
