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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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 Spring '09

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