chapter 8: Cauchys theorem Flashcards
Thm 8.1 Cauchys Theorem
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
primitive and cauchys theorem
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
EXAMPLE
f(z) = sin^3 (z) /(1+z^3)
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
γ is the contour z = Re^it
(0 ≤ t ≤ 2π)
∫_γ dz / z
∫_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
CAUCHY_RIEMANN and cauchys thm
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.
EXAMPLE
∫_γ (dz / (z^2 +4)) where γ is a contour lying in U = {z ∈ C : |z| < 1}
(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
EXAMPLE
∫_γ Rez .dz where γ is a contour lying in U = {z ∈ C : |z| < 1}
Cauchy’s Theorem is irrelevant as Re z is not analytic on any region. The
value of the integral will depend on γ.
EXAMPLE
∫_γ (z^2 +3)/((z+1) e^z)
where γ is a contour lying in U = {z ∈ C : |z| < 1}
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
EXAMPLE
∫_γ (z^2 +3)/((z+1) e^z)
with γ given by z(t) = 2e^it (0 ≤ t ≤ 2π).
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…
THM independence of path
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
EXAMPLE Let γ be ANY path from −i to i which crosses R only between −1 and 1. Evaluate
∫_γ (dz / (1-z^2))
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
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
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