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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
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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
= 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 =^
z − 3 +^
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
0
= z 1 z 2 − z^21 + z^22 − 2 z 1 z 2 + z^21 2
(z 22 − z^21 ).
2