MATH 185 SOLUTIONS OF HW VII
(1) (41.5)(30pts)
By triangle inequality,
Logz
z
2
≤

ln
R

+

iθ

R
2
<
ln
R
+
π
R
2
since
z
= ln
R
+
iθ
where
z
=
Re
iθ
and

π < θ < π
. (N.B.

ln
R

= ln
R
since
R >
1.) Hence,
C
R
Logz
z
2
dz
≤
C
R
Logz
z
2
dz <
2
πR
ln
R
+
π
R
2
= 2
π
ln
R
+
π
R
Moreover,
lim
R
→∞
2
π
(ln
R
+
π
)
R
=
lim
R
→∞
2
π
(1
/R
)
1
= 0
by L’Hospital’s rule, we can conclude that
lim
R
→∞
C
R
Logz
z
2
dz
= 0
NOTICE. If you finish to solve the problem, you need to check WHAT
YOU HAVE PROVEN. Even though you write the same sentence appear
ing the problem, it is important because it can clarify your assertion.
(2) (43. 5)(30pts)
First we may replace the branch

π < θ < π
by

π/
2
< θ <
3
π/
2 since

1 is not defined in
Logz
and change of the branch cut does not affect to
the result in this situation.
So, we can use the antiderivative
z
1+
i
1+
i
of
z
i
.
Hence,
1

1
z
i
dz
=
z
1+
i
1 +
i
1

1
=
1

i
2
(
(1)
i

(

1)
i
)
=
1

i
2
(1 +
e

π
)
(3) (46. 4)
(a)(20pts) Let
I
1
, I
2
, I
3
, I
4
be the line segment from

a
to
a
, from
a
to
a
+
bi
,
from
a
+
bi
to

a
+
bi
, from

a
+
bi
to

a
, respectively. Since
f
(
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 Fall '07
 Lim
 Math, Calculus, Trigraph, e−x dx, lim e−

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