Taylor series problem

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February 23rd 2011, 07:17 AM
skyking

Taylor series problem

I need some help with the following

Let $f(z)=\sum_{n=0}^{\infty}a_nz^n$ analytic for $|z|<R$ and $|f(z)|\leq M$ for all $|z|<R$ .
Show that for every $|z|<\frac {R|a_0|}{M+|a_0|}$ , $f(z)\neq 0$ .

I got stuck by trying the following:

By Cacuchy's inequality we get that for $r<R$ , $\forall n\in N, |a_n|\leq \frac{M}{r^n}$ . I need some help on how to take this further in order to show what's required.
February 23rd 2011, 01:00 PM
Opalg

Just a suggestion. Let $g(z) = f(z)-a_0$ . Then $|g(z)|\leqslant M+|a_0|$ whenever $|z|<R$ , and $g(0)=0$ . I seem to remember that if an analytic function vanishes at the origin and is bounded in a given disc, then there are results limiting how large it can be in a smaller disc (both discs centred at the origin). If you can find a disc within which $|g(z)|<|a_0|$ then clearly $f(z)$ cannot vanish inside that disc.
February 23rd 2011, 04:33 PM
skyking

Quote:

Originally Posted by Opalg

Just a suggestion. Let $g(z) = f(z)-a_0$ . Then $|g(z)|\leqslant M+|a_0|$ whenever $|z|<R$ , and $g(0)=0$ . I seem to remember that if an analytic function vanishes at the origin and is bounded in a given disc, then there are results limiting how large it can be in a smaller disc (both discs centred at the origin). If you can find a disc within which $|g(z)|<|a_0|$ then clearly $f(z)$ cannot vanish inside that disc.

OK, i've been scratching my head for hours and digging through the books, but can't find a result similar to what you mentioned. Can you please eleborate a bit.

Thanks,

SK
February 23rd 2011, 04:43 PM
TheEmptySet

Quote:

Originally Posted by skyking

OK, i've been scratching my head for hours and digging through the books, but can't find a result similar to what you mentioned. Can you please eleborate a bit.

Thanks,

SK

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