sol108c5

sol108c5 - (1) Let F : H C be holomorphic and satisfy | F (...

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Unformatted text preview: (1) Let F : H C be holomorphic and satisfy | F ( z ) | 1 and F ( i ) = 0. Prove that | F ( z ) | z- i z + i , for all z H . Here H = { z C | Im ( z ) > } . Solution: Observe that f ( z ) = ( z- i ) / ( z + i ) is a conformal map between H and D (0 , 1) which maps 0 to 0. Thus F ( f- 1 ( z )) is a holomorphic map from the unit disc to itself. By Schwarzs lemma we then see that | F ( f- 1 ( z )) | | z | . Replacing z by f ( z ) this becomes the desired inequality. (2) Let P t ( z ) = n j =0 a j ( t ) z j with a j continuous in t . Show that if P t has a simple zero at z , then there exists , > 0 such that | t- t | < = ! z D ( z , ) : P t ( z ) = 0 . What happens if P t vanishes at z with order k ? Solution: Without loss of generality we can assume z = 0 and t = 0. In particular we find a (0) = 0, while a 1 (0) 6 = 0. Let us now consider | a ( t ) + a 1 (0) z- P t ( z ) | . We can bound this by | a ( t ) + a 1 (0) z- P t ( z ) | | a 1 (0)- a 1 ( t ) || z | + n X k =2 | a k ( t ) | z k . Now, by continuity of a 1 there exists an 1 such that | a 1 (0)- a 1 ( t ) | | a 1 (0) | / 2 if | t | 1 . Moreover we find that by continuity of a k ( k > 1) there exists an M k such that | a k ( t ) | M k if | t | 1 . Moreover if we now set = min k> 1 ( | a 1 (0) | / 2 nM k ) 1 / ( k- 1) we find that for | t | 1 and | z | | a ( t ) + a...
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sol108c5 - (1) Let F : H C be holomorphic and satisfy | F (...

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