7. 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 TU Int(T). We can subdivide T into four equal triangles A, by bisecting each of the sides like this: T 42 To arrive at a contradiction, we assume that there is h> 0 such that | f(z)dz| ≥ h. the boundaries A, of the triangles A, can be oriented such that 4 √ 1(z)dz = Σ f(z)dz. (*) asi i=1 From equation (), conclude that there must be a triangle A, such that the inequality Jo, f(z) ≥ holds. Call this triangle T₁. Note that its longest side has length 1/2. Argue that by continuing these steps, there is a sequence of triangles T T₁ T2... such that the longest side of Tn has length 1/2" and |_ f(z)dz > ƏTn h 4n
7. 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 TU Int(T). We can subdivide T into four equal triangles A, by bisecting each of the sides like this: T 42 To arrive at a contradiction, we assume that there is h> 0 such that | f(z)dz| ≥ h. the boundaries A, of the triangles A, can be oriented such that 4 √ 1(z)dz = Σ f(z)dz. (*) asi i=1 From equation (), conclude that there must be a triangle A, such that the inequality Jo, f(z) ≥ holds. Call this triangle T₁. Note that its longest side has length 1/2. Argue that by continuing these steps, there is a sequence of triangles T T₁ T2... such that the longest side of Tn has length 1/2" and |_ f(z)dz > ƏTn h 4n
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
![7. 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
A4
A₂
To arrive at a contradiction, we assume that there is h >0 such that |f₁ f(z)dz| ≥ h.
A; of the triangles A, can be oriented such that
the boundaries
4
[ f(z)dz = Σf f(z)dz.
aDi
i=1
(★)
From equation (★), conclude that there must be a triangle A¿ such that the
inequality Jo, f(z) ≥ holds.
Call this triangle T₁. Note that its longest side has length 1/2.
Argue that by continuing these steps, there is a sequence of triangles TƆ T₁
T₂... such that the longest side of Tn has length 1/2" and
h
So f(z)dz| ≥ M.
4n](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb0d2dbea-0987-4c02-a00c-0b87f219c8f9%2Fa6d60fa1-dc33-4fd5-97ce-b1e88fcaf78a%2F2y3y5ux_processed.png&w=3840&q=75)
Transcribed Image Text:7. 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
A4
A₂
To arrive at a contradiction, we assume that there is h >0 such that |f₁ f(z)dz| ≥ h.
A; of the triangles A, can be oriented such that
the boundaries
4
[ f(z)dz = Σf f(z)dz.
aDi
i=1
(★)
From equation (★), conclude that there must be a triangle A¿ such that the
inequality Jo, f(z) ≥ holds.
Call this triangle T₁. Note that its longest side has length 1/2.
Argue that by continuing these steps, there is a sequence of triangles TƆ T₁
T₂... such that the longest side of Tn has length 1/2" and
h
So f(z)dz| ≥ M.
4n
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