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Solutions to the sample exam 2 problems
1. (a) Here, the path
1
is parametrized by
z
= (2 +
i
)
t
, with
0
t
1
:
So
z
= (2
±
i
)
t
and
dz
= (2 +
i
)
dt:
Thus
Z
1
zdz
=
Z
1
0
(2
±
i
)
t
(2 +
i
)
dt
=
5
2
:
(b) Here, the curve
2
is the reverse of the curve parametrized by
z
=
1 + 2
e
it
;
0
t
±
for which
z
= 1 + 2
e
it
and
dz
= 2
ie
it
dt:
Thus
Z
2
zdz
=
±
Z
±
0
1 + 2
e
it
±
(2
ie
it
)
dt
=
4
±
4
i±
(c) Let
²
join
0
to
i
and
³
join
i
to
2 +
i:
Then
²
is parametrized by
z
=
iy;
0
y
1
:
The curve
³
is given by
z
=
x
+
i
,
0
x
2
:
Thus
Z
²
+
³
zdz
=
Z
²
zdz
+
Z
³
zdz
=
Z
1
0
(
±
iy
)
idy
+
Z
2
0
(
x
±
i
)
dx
=
5
2
±
2
i:
2. In this problem, the contour integrals are path independent since
e
2
z
has
a global antiderivative,
F
(
z
) =
1
2
e
2
z
:
Thus
(a)
R
1
e
2
z
dz
=
F
(2 +
i
)
±
F
(0) =
1
2
e
4+2
i
±
1
2
e
0
=
1
2
e
4+2
i
±
1
±
(b)
R
2
e
2
z
dz
=
F
(3)
±
f
(
±
1) =
1
2
(
e
6
±
e
2
)
(c) same as 2(a).
3. The sets in (b) and (c) are simply connected. The set in (a) fails to be
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 Spring '11
 Staff

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