| Line 5: | Line 5: | ||
Here's a hint on I.8.3 --[[User:Bell|Bell]] | Here's a hint on I.8.3 --[[User:Bell|Bell]] | ||
| − | It is straightforward to show that <math>(z,w)\mapsto z+w</math> is a continuous mapping from <math>\mathbb C\times \mathbb C</math> because | + | It is straightforward to show that |
| + | |||
| + | <math>(z,w)\mapsto z+w</math> | ||
| + | |||
| + | is a continuous mapping from <math>\mathbb C\times \mathbb C</math> to <math>\mathbb C</math> because | ||
<math>|(z+w)-(z_0+w_0)|\le|z-z_0|+|w-w_0|</math> | <math>|(z+w)-(z_0+w_0)|\le|z-z_0|+|w-w_0|</math> | ||
Revision as of 08:28, 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<\math> and <math>|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) $.
