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Math 412 Week #12 Practice Quiz Solutions for Complex Analysis - Prof. Scott Annin, Quizzes of Mathematics

The solutions to problem 1 and problem 2 from a practice quiz in math 412, a university-level course on complex analysis. The solutions involve evaluating contour integrals using various theorems and formulas, and proving cauchy's theorem directly.

Typology: Quizzes

Pre 2010

Uploaded on 08/16/2009

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April 20-24, 2009 Week #12 Practice Quiz Name:
Math 412
SOLUTIONS
Problem 1. (4 points each) Evaluate each contour integral below:
(a):
ZC
cos z
2dz,
where Cis the line segment from z= 0 to z=π+ 2i.
SOLUTION: Since fis entire, we can apply Theorem 6.9: We have
ZC
cos z
2dz = 2 sin z
2
π+2i
0
= 2 sin π
2+i= 2 cosh 1,
where we have used identity (5-33) on p. 178 of the text. 2
(b):
ZC
cos z
z(z2+ 8)dz,
where Cis the square with vertices (±2,±2), with counterclockwise orientation.
SOLUTION: The function f(z) = cos z
z2+ 8 is analytic in and on C, so we apply Cauchy’s Integral
Formula with z0= 0 to get
2πif (0) ·1
8=
4.
2
(c):
ZC
4ez
z22z3dx,
where Cis a circle of radius 5centered at the origin and with clockwise orientation.
SOLUTION: We write 1
z22z3=A
z3+B
z+ 1.
Cross-multiplying, we find that A(z+ 1) + B(z3) = 1, so that A+B= 0 and A3B= 1. We
quickly find that
A=1
4and B=1
4.
Thus, we have
ZC
4ez
z22z3dz =ZC
ez
z3dz ZC
ez
z+ 1dz =ZC
ez
z3dz +ZC
ez
z+ 1dz,
pf2

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Download Math 412 Week #12 Practice Quiz Solutions for Complex Analysis - Prof. Scott Annin and more Quizzes Mathematics in PDF only on Docsity!

April 20-24, 2009 Week #12 Practice Quiz Name:

Math 412

SOLUTIONS

Problem 1. (4 points each) Evaluate each contour integral below:

(a): (^) ∫

C

cos

( (^) z 2

dz,

where C is the line segment from z = 0 to z = π + 2i.

SOLUTION: Since f is entire, we can apply Theorem 6.9: We have

C

cos

( (^) z 2

dz = 2 sin

( (^) z 2

π+2i

0

= 2 sin

( (^) π 2

  • i

= 2 cosh 1,

where we have used identity (5-33) on p. 178 of the text. 2

(b): (^) ∫

C

cos z z(z^2 + 8) dz,

where C is the square with vertices (± 2 , ±2), with counterclockwise orientation.

SOLUTION: The function f (z) = cos^ z z^2 + 8 is analytic in and on C, so we apply Cauchy’s Integral

Formula with z 0 = 0 to get

2 πif (0) ·

iπ 4

(c): (^) ∫

C

4 ez z^2 − 2 z − 3 dx,

where C is a circle of radius 5 centered at the origin and with clockwise orientation.

SOLUTION: We write 1 z^2 − 2 z − 3 =^

A

z − 3 +^

B

z + 1.

Cross-multiplying, we find that A(z + 1) + B(z − 3) = 1, so that A + B = 0 and A − 3 B = 1. We quickly find that

A =

and B = −

Thus, we have ∫

C

4 ez z^2 − 2 z − 3 dz =

C

ez z − 3 dz −

C

ez z + 1 dz = −

−C

ez z − 3 dz +

−C

ez z + 1 dz,

where we have used the last step to correct the orientation of our curve. Now using Cauchy’s Integral Formula with f (z) = ez^ , we get

−(2πi)e^3 + 2πie−^1 = (2πi)

e − e^3

Problem 2. (4 points) Give a precise statement of Cauchy’s Theorem.

SOLUTION: Theorem 6.5 on page 217. 2

Problem 3. (4 points) Prove directly (i.e. by parametrization) that if z 1 , z 2 ∈ C, then ∫

C

z dz =

2 (z

(^22) − z (^21) ),

where C is the line segment from z 1 to z 2.

SOLUTION: We parametrize C via z(t) = z 1 + t(z 2 − z 1 ), where 0 ≤ t ≤ 1. Then z′(t) = z 2 − z 1 and dz = (z 2 − z 1 )dt. Hence, ∫

C

z dz =

0

[

z 1 + t(z 2 − z 1 )

]

(z 2 − z 1 )dt

[

(z 1 z 2 − z 12 )t + t^2 (z 2 − z 1 )^2 2

] 1

0

= z 1 z 2 − z^21 + z^22 − 2 z 1 z 2 + z^21 2

(z 22 − z^21 ).

2