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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] | ||
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+ | Here's a hint on I.8.3 --[[User:Bell|Bell]] | ||
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+ | 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 | ||
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+ | <math>|(z+w)-(z_0+w_0)|\le|z-z_0|+|w-w_0|</math> | ||
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+ | and to make this last quantity less than <math>\epsilon</math>, it suffices to take | ||
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+ | <math>|z-z_0|<\epsilon/2<\math> and <math>|w-w_0|<\epsilon/2</math>. | ||
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+ | To handle complex multiplication, you will need to use the standard trick: | ||
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+ | <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 07:26, 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 $ 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) $.