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==Homework 2== | ==Homework 2== | ||
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Here's a hint for II.3.1 (ii): | Here's a hint for II.3.1 (ii): | ||
| − | <math>\frac{f(z)g(z)-f(z_0)g(z_0)}{z-z_0}=</math> | + | <math>\frac{f(z)g(z)-f(z_0)g(z_0)}{z-z_0}=\frac{f(z)g(z)-f(z)g(z_0)+f(z)g(z_0)-f(z_0)g(z_0)}{z-z_0}=</math> |
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| + | <math>f(z)\frac{g(z)-g(z_0)}{z-z_0}+g(z_0)\frac{f(z)-f(z_0)}{z-z_0}.</math> | ||
Revision as of 07:58, 4 September 2009
Homework 2
Here's a hint for II.3.1 (ii):
$ \frac{f(z)g(z)-f(z_0)g(z_0)}{z-z_0}=\frac{f(z)g(z)-f(z)g(z_0)+f(z)g(z_0)-f(z_0)g(z_0)}{z-z_0}= $
$ f(z)\frac{g(z)-g(z_0)}{z-z_0}+g(z_0)\frac{f(z)-f(z_0)}{z-z_0}. $
