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HW4s - Key to Complex Analysis Homework 4 October 3 2011...

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Key to Complex Analysis Homework 4 October 3, 2011 Chapter 3 (in Greene & Krantz’s book) 44. Suppose f : D (0 , 1) C is a function, such that f 2 and f 3 are holomorphic. Prove that f is holomorphic. Proof. Suppose f 6≡ 0; otherwise, it is trivial. Observation : We only need to check those points z 0 D (0 , 1) with f ( z 0 ) = 0 whether f 0 ( z 0 ) exists (since if f ( z 0 ) 6 = 0, then f 2 ( z 0 ) 6 = 0, so there is r 0 > 0 such that f 2 has no zeros on D ( z 0 , r 0 ) D (0 , 1). From f · f 2 = f 3 on D ( z 0 , r 0 ), ¯ z f · f 2 + f ¯ z f 2 = ¯ z f 3 , use the fact that f 2 , f 3 are holomorphic, we get, on D ( z 0 , r 0 ), ¯ z f · f 2 = 0 . Since f 2 is never zero on D ( z 0 , r 0 ), ¯ z f 0 on D ( z 0 , r 0 ). This implies that f is holomorphic on D ( z 0 , r 0 ). Now let z 0 D (0 , 1) with f ( z 0 ) = 0. Then f 2 ( z 0 ) = 0 and f 3 ( z 0 ) = 0. Since f 2 and f 3 are holomorphic, we can write f 2 = ( z - z 0 ) n g 1 ( z ) , f 3 = ( z - z 0 ) m g 2 ( z ) where g 1 , g 2 are holomorphic on D (0 , 1), and g 1 ( z 0 ) 6 = 0 , g 2 ( z 0 ) 6 = 0, thus g 1 , g 2 have no zeros on D ( z 0 , r ) D (0 , 1) with r > 0. Next, we claim that m > n . Indeed, assume the claim is false, and from ( z - z 0 ) m g 2 ( z ) = f 3 = f 2 · f = ( z - z 0 ) n g 1 ( z ) · f , so g 2 ( z 0 ) = ( z - z 0 ) n - m g 1 ( z 0 ) f ( z 0 ). From f 2 ( z 0 ) = 0 we get f ( z 0 ) = 0. Hence g 2 ( z 0 ) = ( z - z 0 ) n - m g z ( z 0 ) f ( z 0 ) = 0 contradicts with the fact that g 2 ( z 0 ) 6 = 0. Hence m > n . Now for z D ( z 0 , r ) D (0 , 1) with z 6 = z 0 , write f = f 3 f 2 = ( z - z 0 ) m - n g 2 g 1 , 1

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then, noticing that g 1 has no zeros on D ( z 0 , r ) lim z z 0 f ( z ) - f ( z 0 ) z - z 0 = lim z z 0 ( z - z 0 ) m - n - 1 g 2 g 1 = ( z 0 - z 0 ) m - n - 1 g 2 ( z 0 ) g 1 ( z 0 ) exists. For any point z D ( z 0 , r ) other than z 0 , since f ( z ) = ( z - z 0 ) m - n g 2 g 1 and g 1 , g 2 have no zeors on D ( z 0 , r ), it is easy to see that f 0 ( z ) exists. Therefore, f is holomorphic on D ( z 0 , r ). Combining all above, we proved that f is holomorphic on D (0 , 1).
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