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Unformatted text preview: ASSIGNMENT 6 SOLUTIONS MAT 572 A FALL 2007 Problem 1 ( cf. [1, Exercise IV.2 #2]) . Prove that if G C is open and : I G is a rectifiable curve, and : ( I ) G C is continuous and g : G C is defined by g ( z ) = Z ( w, z ) d w, then g is continuous. Also prove that if /z exists and is continuous on ( I ) G , then g is analytic on G , with (1.1) g ( z ) = Z z d w. Proof. For continuity, fix z G and > 0 such that K = B ( z ) G . Then is uniformly continuous on the compact set ( I ) K , so given > 0 we can choose > 0 such that < , and such that | ( w, z )- ( w , z ) | < /V ( ) for all ( w, z ) and ( w , z ) in ( I ) K with d (( w, z ) , ( w , z )) < . In particular, for any z G with | z- z | < , Proposition III.1.17 gives | g ( z )- g ( z ) | = Z ( w, z ) d w- Z ( w, z ) d w Z | ( w, z )- ( w, z ) || d w | V ( ) Z | d w | = , so g is continuous at z . For analyticity, we only need to establish (1.1), since then continuity of g follows from the above. Again, fix z G and > 0 such that K = B ( z ) G . Also fix > 0, and write 2 for z . Then 2 is uniformly continuous on...
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