ACM 95a/100a Problem Set 4 Solutions
Prepared by: Zhiyi Li
October 26, 2007
Total: 92 points
Include grading section: 2 points
Problem 1 (10 points)
Consider the function
f
(
z
) =
z
1
/
2
=
√
re
iθ/
2
,

π/
2
< θ
≤
3
π/
2
and the contour C corresponding to the positively oriented boundary of the half disk 0
≤
r
≤
1, 0
≤
θ
≤
π
in
the upper half plane. Show by separate parametric evaluation of the semicircle and the two radii constituting
the boundary that
Z
C
f
(
z
)
dz
= 0
.
Does the CauchyGoursat theorem apply here?
Solution 1
Let
C
≡
C
1
+
C
2
+
C
3
, where
C
1
≡
'
z
=
re
iθ
:
r
= 0 to
r
= 1
, θ
= 0
“
(1)
C
2
≡
'
z
=
re
iθ
:
r
= 1
, θ
= 0 to
θ
=
π
“
(2)
C
3
≡
'
z
=
re
iθ
:
r
= 1 to
r
= 0
, θ
=
π
“
(3)
For
C
2
,
dz
=
ire
iθ
=
ie
iθ
. For
C
1
and
C
3
,
dz
=
e
iθ
, (
θ
= 0 or
π
). So,
Z
C
f
(
z
)
dz
=
Z
C
1
f
(
z
)
dz
+
Z
C
2
f
(
z
)
dz
+
Z
C
3
f
(
z
)
dz
=
Z
1
0
√
re
i
0
2
e
i
0
dr
+
Z
π
0
√
1
e
i
θ
2
ie
iθ
dθ
+
Z
0
1
√
re
i
π
2
e
iπ
dr
=
2
3
r
3
2
fl
fl
fl
fl
1
r
=0
+
2
3
e
i
3
θ
2
fl
fl
fl
fl
π
θ
=0
+
2
3
r
3
2
e
i
3
π
2
fl
fl
fl
fl
0
r
=1
=
2
3

0
¶
+

i
2
3

2
3
¶
+
0


i
2
3
¶¶
=
0
(4)
Though this integral is zero,
CauchyGoursat doesn’t apply
. The integrand has a singularity (a branch
point) on
C
and therefore isn’t analytic on
C
.
Problem 2 (10 points)
Evaluate the following contour integral using antiderivatives and justify your approach.
Z
C
z
i
dz
=
1 +
e

π
2
(1

i
)
1
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with
z
i
=
e
i
Log
z
,

π <
Arg
z
≤
π.
C joins
z
1
=

1 and
z
2
= 1, lying above the real axis except at the end points. (Hint: redefine
z
i
so that it
remains unchanged above the real axis and is defined continuously on the real axis.)
Solution 2
We justify using antiderivatives two ways:
1. By showing that the conditions for using the Fundamental Theorem for Contours hold.
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 Fall '06
 NilesA.Pierce
 Line integral, Methods of contour integration, dz, real axis

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