Lecture18

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

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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 πi 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 Δ z → f ( z + Δ z ) − f ( z ) Δ z , (759) if the limit exists. With (758) we have f ( z + Δ z ) − f ( z ) Δ z = 1 2 πi Δ z contintegraldisplay C f ( z ) z − ( z + Δ z ) dz − contintegraldisplay C f ( z ) z − z dz = 1 2 πi Δ z contintegraldisplay C ( z − z ) f ( z ) − ( z − z − Δ z ) f ( z ) ( z − z − Δ z )( z − z ) dz = 1 2 πi contintegraldisplay C f ( z ) ( z − z − Δ z )( z − z ) dz. (760) Now we compute contintegraldisplay C f ( z ) ( z − z − Δ z )( z − z ) dz − contintegraldisplay C f ( z ) ( z − z ) 2 dz = contintegraldisplay C Δ zf ( z ) ( z − z − Δ z )( z − z ) 2 dz. (761) We now show that this integral approaches 0 as Δ 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 − Δ z | + | Δ z | , ∀ z ∈ C. (763) 128 For | Δ z | ≤ d/ 2 we have d 2 ≤ d − | Δ z | ≤ | z − z − Δ z | ⇒ 1 | z − z − Δ z | ≤ 2 d , ∀ z ∈ C. (764) Let L := L ( C ) denote the arc length of the curve C . For | Δ z | ≤ d/ 2, we use the ML-inequality to obtain vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle contintegraldisplay C Δ zf ( z ) ( z − z − Δ z )( z − z ) 2 dz vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle ≤ | Δ z | K 2 d 1 d 2 = 2 K d 3 | Δ z | → , Δ z → . (765) This proves that f ′ ( z ) exists, and is given by f ′ ( z ) = 1 2 πi 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...
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Lecture18 - with respect to z Proof We know from Thm 10...

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