5. Show that if f(2) is analytic for |2| < 1+ €, then for any z = reio with r < 1, eit z dt = 0 1– žeit 1 (a) 27 eit -dt = f(z) 1 (b) 27 eit
5. Show that if f(2) is analytic for |2| < 1+ €, then for any z = reio with r < 1, eit z dt = 0 1– žeit 1 (a) 27 eit -dt = f(z) 1 (b) 27 eit
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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[Complex Variables] How do you solve this, thanks
![Trigonometric Integral via Cauchy's Formula
• Basic Idea: Switch trigonometric function into rational function, and then use
Cauchy's Formula.
• Procedure: Given an integral of the type
1
a cos 0 + b sin 0 +c°
-d0, a, b, с € C
1. Use the change of variables
sin 0 = (: -),
dz
where z = et0
cos 0
(z+
OP
iz
and plug in the integral, making it into a curve integral
1
;dz, y(t) = e"t, t e [0,1]
pz² + qz +l'
2. Perform the factorisation pz² + qz +l = p(z – 21)(z – 22) (you can do this
thanks to the fundamental theorem of algebra).
3. Identify the position of 21, 2 with respect to the unit circle (the closed contour
along which your curve integral is evaluated).
4. Now there are three cases:
1
pz2 + qz +1
(a). if [21|,|22| > 1 then the function f(2)
is analytic and
the integral vanishes.
1
(note that g(2)
(b). if w.l.o.g. [21| > 1 > |22], then take g(z)
as a function of z, and 21 is a constant parameter). g is analytic in the
unit circle, and apply Cauchy's formula for g(z) at z = 2, i.e.
p(z – 21)
g(2)
z – 22
(c). if [21| < 1,|22| < 1, then the cases is slightly more complicated, and we
g(22) =
don't discuss this in detail here.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb1b8153a-0999-4fb8-8f29-17a3988513a2%2Feec2a3ab-9216-4e6c-a3cc-fbb24885773c%2Fqbpfwzf_processed.png&w=3840&q=75)
Transcribed Image Text:Trigonometric Integral via Cauchy's Formula
• Basic Idea: Switch trigonometric function into rational function, and then use
Cauchy's Formula.
• Procedure: Given an integral of the type
1
a cos 0 + b sin 0 +c°
-d0, a, b, с € C
1. Use the change of variables
sin 0 = (: -),
dz
where z = et0
cos 0
(z+
OP
iz
and plug in the integral, making it into a curve integral
1
;dz, y(t) = e"t, t e [0,1]
pz² + qz +l'
2. Perform the factorisation pz² + qz +l = p(z – 21)(z – 22) (you can do this
thanks to the fundamental theorem of algebra).
3. Identify the position of 21, 2 with respect to the unit circle (the closed contour
along which your curve integral is evaluated).
4. Now there are three cases:
1
pz2 + qz +1
(a). if [21|,|22| > 1 then the function f(2)
is analytic and
the integral vanishes.
1
(note that g(2)
(b). if w.l.o.g. [21| > 1 > |22], then take g(z)
as a function of z, and 21 is a constant parameter). g is analytic in the
unit circle, and apply Cauchy's formula for g(z) at z = 2, i.e.
p(z – 21)
g(2)
z – 22
(c). if [21| < 1,|22| < 1, then the cases is slightly more complicated, and we
g(22) =
don't discuss this in detail here.
Transcribed Image Text:5.
Show that if f(2) is analytic for |2| < 1+ €, then for any z = reio
with r < 1,
(a)
eit z
dt = 0
1– žeit
27
eit
-dt = f(2)
(b)
f(e“)-
eit
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