Lecture18

Lecture18 - with respect to z . Proof: We know from Thm. 10...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: with respect to z . Proof: We know from Thm. 10 that we have a contour integral representation of f at any point z U : f ( z ) = 1 2 i contintegraldisplay C f ( z ) z z dz. (758) Choose a point z U . The derivative of f at z is given by f ( z ) = lim z f ( z + z ) f ( z ) z , (759) if the limit exists. With (758) we have f ( z + z ) f ( z ) z = 1 2 i z contintegraldisplay C f ( z ) z ( z + z ) dz contintegraldisplay C f ( z ) z z dz = 1 2 i z contintegraldisplay C ( z z ) f ( z ) ( z z z ) f ( z ) ( z z z )( z z ) dz = 1 2 i contintegraldisplay C f ( z ) ( z z z )( z z ) dz. (760) Now we compute contintegraldisplay C f ( z ) ( z z z )( z z ) dz contintegraldisplay C f ( z ) ( z z ) 2 dz = contintegraldisplay C zf ( z ) ( z z z )( z z ) 2 dz. (761) We now show that this integral approaches 0 as z 0. f is continuous on C , and therefore bounded: K > 0 : | f ( z ) | K z C . Let d be the miminum distance from z to any point on C : d := min z C | z z | 1 | z z | 2 1 d 2 z C (762) Furthermore, with the triangle inequality, d | z z | | z z z | + | z | , z C. (763) 128 For | z | d/ 2 we have d 2 d | z | | z z z | 1 | z z z | 2 d , z C. (764) Let L := L ( C ) denote the arc length of the curve C . For | z | d/ 2, we use the ML-inequality to obtain vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle contintegraldisplay C zf ( z ) ( z z z )( z z ) 2 dz vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle | z | K 2 d 1 d 2 = 2 K d 3 | z | , z . (765) This proves that f ( z ) exists, and is given by f ( z ) = 1 2 i contintegraldisplay C f ( z ) ( z z ) 2 dz. (766) Therefore f may be represented at z as a contour integral. This is true for any z U . We repeat the previous argument with f replaced by f and (758) replaced by (766) to obtain a contour integral representation of...
View Full Document

Page1 / 6

Lecture18 - with respect to z . Proof: We know from Thm. 10...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online