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Math 412 Group Work #13 Spring 2009
SOLUTIONS
Problem 1. Evaluate
C
x dz from − 4 to 4 along the upper half of the circle
|z| = 4, oriented clockwise.
SOLUTION: If we parametrize C via z(t) = (4 cos t) − i(4 sin t), where −π ≤ t ≤ 0,
then z
′ (t) = (−4 sin t) − i(4 cos t) and dz = [(−4 sin t) − i(4 cos t)]dt. Hence,
C
x dz =
−π
(4 cos t)[(−4 sin t) − i(4 cos t)]dt
−π
[
−16 sin t cos t − i(16 cos
2 t)
]
dt
∫ (^) −π
0
[
16 sin t cos t + i(16 cos
2 t)
]
dt
= 16i
π
0
cos
2 t dt
= 16i ·
π
= 8πi.
Problem 2. Let C be the unit circle in the complex plane, oriented coun-
terclockwise.
(a): Show that ∣ ∣ ∣ ∣
C
dz
12 + 5z
2 π
SOLUTION: We wish to apply Theorem 6.3. Since the circumference of the unit
circle is 2π, the length of the contour C is L = 2π. The maximum value of f (z) =
1
12 + 5z
occurs when |12 + 5z| is as small as possible (i.e. when z = −1). Thus,
M =
. Thus, by Theorem 6.3, we conclude that
C
dz
12 + 5z
≤ M L =
· 2 π =
2 π
(b): By considering the right and left halves of C, show that
∣ ∣ ∣ ∣
C
dz
12 + 5z
20 π
SOLUTION: Let C 1 denote the left-half of C and let C 2 denote the right-half of C,
both oriented counterclockwise. Then C = C 1 + C 2. We have
∣ ∣ ∣ ∣
C
dz
12 + 5z
C 1 +C 2
dz
12 + 5z
C 1
dz
12 + 5z
C 2
dz
12 + 5z
C 1
dz
12 + 5z
C 2
dz
12 + 5z
Now for C 1 , the length of half of the unit circle is L 1 = π. The maximum value
of f (z) on C 1 occurs when |12 + 5z| is as small as possible (i.e. z = −1). Thus,
M 1 =
. Thus, by Theorem 6.3, we conclude that
C 1
dz
12 + 5z
≤ M 1 L 1 =
· π =
π
Next, for C 2 , we have L 2 = π, and the maximum value of f (z) on C 2 occurs when
|12 + 5z| is as small as possible (i.e. when z = ±i, which are the points on the circle
closest to −12). Since | 12 ± 5 i| = 13, we have M 2 =
. Thus, by Theorem 6.3, we
conclude that (^) ∣ ∣ ∣ ∣
C 2
dz
12 + 5z
≤ M 2 L 2 =
· π =
π
Hence, ∣ ∣ ∣ ∣
C
dz
12 + 5z
π
π
20 π
