chapter 8: Cauchys theorem Flashcards

1
Q

Thm 8.1 Cauchys Theorem

A

Suppose the function f is analytic on a simply
connected region D. Then ∫_γ f = 0 for all contours γ in D.

MUST BE A SIMPLY CONNECTED REGION ALWAYS CHECK

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2
Q

primitive and cauchys theorem

A

if f has a primitive on D, then ∫_γ f = 0. This
hypothesis is very strong and Cauchy’s statement is far superior.

hard to find a primitive so cauchy’s allows us to without finding

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3
Q

EXAMPLE

f(z) = sin^3 (z) /(1+z^3)

A

unction f is analytic on C \ {−1,exp (iπ/3), exp (−iπ/3)}

so its analytic on the disc U = {z ∈ C : |z| < 1}, a SIMPLY CONNECTED REGION

 Let γ be
any contour in U. Then
∫_γ sin^3 (z) /(1+z^3)  dz = 0
by Cauchy’s Theorem, and we have found the value of the integral without doing
any integration
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4
Q

γ is the contour z = Re^it
(0 ≤ t ≤ 2π)

∫_γ dz / z

A

∫_γ (dz / z) = 2πi.

1/z is analytic on C* =C{0} and γis a contour in C* =C{0}.

BUT region C* is not simply connected and cauchy’s theorem couldn’t have been used

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5
Q

CAUCHY_RIEMANN and cauchys thm

A

Suppose
that the function f is analytic in a simply-connected region Ω containing the contour
γ. Use the standard notation z = x + iy and f = u + iv. Then u, v satisfy the
Cauchy-Riemann equations on Ω.

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6
Q

EXAMPLE

∫_γ (dz / (z^2 +4)) where γ is a contour lying in U = {z ∈ C : |z| < 1}

A

(1/ (z^2 +4) is analytic C\ {± 2i} it is analytic on the disc U

this is a simply connected region containing contour γ

by cauchy’s theorem
∫_γ (dz / (z^2 +4))
=0

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7
Q

EXAMPLE

∫_γ Rez .dz where γ is a contour lying in U = {z ∈ C : |z| < 1}

A

Cauchy’s Theorem is irrelevant as Re z is not analytic on any region. The
value of the integral will depend on γ.

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8
Q

EXAMPLE
∫_γ (z^2 +3)/((z+1) e^z)
where γ is a contour lying in U = {z ∈ C : |z| < 1}

A

γ is a contour,

(z^2 +3)/((z+1) e^z) analytic on C \ {−1}
it is analytic on the disc U, which is a
simply-connected region containing
the contour γ

by cauchy’s theorem =0

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9
Q

EXAMPLE
∫_γ (z^2 +3)/((z+1) e^z)
with γ given by z(t) = 2e^it (0 ≤ t ≤ 2π).

A

γ is a contour,

(z^2 +3)/((z+1) e^z) analytic on C \ {−1}

But −1 belongs to
the interior of γ and, therefore, there is no simply connected region containing γ in
which the integrand is analytic

we dont use cauchys but we use…

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10
Q

THM independence of path

A

Let f be analytic on a simply-connected region D and let γ1 and γ2 be any
two paths in D from a to b. Then

∫_γ_1 f(z).dz = ∫_γ_2 f(z).dz

from contour made from γ1 and γ2

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11
Q

EXAMPLE Let γ be ANY path from −i to i which crosses R only between −1 and 1. Evaluate

∫_γ (dz / (1-z^2))

A

Let λ be the straight line segment from −i to i and let
f(z) = 1/(1 − z^2)

Then the function f is analytic C{± 1}

so f is analytic on the cut plane C∗ = C \ {x ∈ R : |x| ≥ 1} which is a simply-connected region

which contains γ and λ

by cauchys theorem

∫_γ    1/(1 − z^2)  .dz
=
∫_λ    1/(1 − z^2)  .dz
=
∫_[-1,1]    i/(1 − y^2)  .dy
=
i[tan-1 (y) ] evaluated at -1 to 1

= iπ/2

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12
Q

EXAMPLE: ML EST

Let α be any path in D = {z ∈ C : |z| < 2}. Find B so that

∫_α (sinh z)/(9 + e^z) .dz | ≤ B

A

Suppose that the path α has initial and final points z0 and z1.

Let λ be the straight line segment from z0 to z1. Then λ ⊂ D.

(diagram disc any path vs straight line)

Sinhz is analytic on C and for all z in D |e^z| = e^Rez less than e^2 less than 9 as e^x s an increasing function on R

hence 9 +e^z is analytic and non-zero on D. This 1/(9 +e^z) is analytic on D and therefore sinhz /(9+e^z) is analytic on the disc D

D is a simply connected region

hence by thm 8.2 indep of path

where
|sinhz /(9+e^z)| ≤ M on λ
and length of λ is
|z_1 -z_0| which is less than 4

M:
for all z in D 
|9+e^z| ≥ | 9| − | e^z| ≥ 9 − e^2
|sinhz| = | 0.5(e^z  + e^-z)| 
≤  0.5(|e^z|  + |e^-z|)
= 0.5(e^Rez + e^Re(-z))
=  0.5(e^x + e^-x)
= cosh x  ≤ cosh 2.

M= cosh2 / (9-e^2)

B=ML = (4cosh2)/(9-e^2)

∫_α (sinh z)/(9 + e^z) .dz | =
| ∫_λ (sinh z)/(9 + e^z) .dz |
≤ ML

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