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O Show that the boundaries A; of the triangles A, can be oriented such that √₁1 (2) dz = Σ/₁₁ f(z)dz. i=1 asi (*) From equation (★), conclude that there must be a triangle Aį such that the inequality Jo, f(2) ≥ holds.

O Show that the boundaries A; of the triangles A, can be oriented such that √₁1 (2) dz = Σ/₁₁ f(z)dz. i=1 asi (*) From equation (★), conclude that there must be a triangle Aį such that the inequality Jo, f(2) ≥ holds.

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter8: Applications Of Trigonometry
Section8.3: Vectors
Problem 60E
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Prove Cauchy's theorem for a triangular contour (see the picture below) T by answering the questions below. You do not have justify the steps that are not marked by (a), (b), etc. To be more precise, let T be a triangle whose longest side has length 1, and let f: R → C be function that is holomorphic on a region R containing T U Int(T). We can subdivide T into four equal triangles A; by bisecting each of the sides like this: T A3 A2 To arrive at a contradiction, we assume that there is h> 0 such that |f₁ f(z)dz| ≥ h. (a) Show that the boundaries A; of the triangles A; can be oriented such that 4 [ f(z)dz = £f f(z)dz. Δ i=1 (*) From equation (*), conclude that there must be a triangle A; such that the inequality |√, ƒ (²) ≥ holds.
Transcribed Image Text:Prove Cauchy's theorem for a triangular contour (see the picture below) T by answering the questions below. You do not have justify the steps that are not marked by (a), (b), etc. To be more precise, let T be a triangle whose longest side has length 1, and let f: R → C be function that is holomorphic on a region R containing T U Int(T). We can subdivide T into four equal triangles A; by bisecting each of the sides like this: T A3 A2 To arrive at a contradiction, we assume that there is h> 0 such that |f₁ f(z)dz| ≥ h. (a) Show that the boundaries A; of the triangles A; can be oriented such that 4 [ f(z)dz = £f f(z)dz. Δ i=1 (*) From equation (*), conclude that there must be a triangle A; such that the inequality |√, ƒ (²) ≥ holds.
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