else, we use the range of angles method for
√
1

z
2
=
i
√
z

1
√
z
+ 1. If we use
angle range (0
,
2
π
) for
√
z

1 we get a branch cut on [1
,
∞
). Angle range (0
,
2
π
)
Page 1
MATH 301, Homework 5
Do not hand in
for
√
z
+ 1 gives a branch cut on (1
,
∞
).
This function has no branch point at
infinity, and the branch cuts on (1
,
∞
) cancel. Checking the angles we find that
this gives the negative square root on the upper lip, so we reverse the sign (or use
angle range (2
π,
4
π
) for one of the two). In short, the branch is

i
(
z

1)
1
/
2
(
z
+ 1)
1
/
2
z
2
+ 1
,
where the square roots are both using range of angles (0
,
2
π
).
(b) Consider the two contours
C
and Γ in the figure.
C
Γ
If we connect the two contours we can get a simple contour that retraces the
connection between them, and so the sum of the integrals is given by the residue
theorem.
I
Γ
f
(
z
)
dz
+
I
C
f
(
z
)
dz
= 2
πi
[Res(
f
;
i
) + Res(
f
;

i
)]
,
The residues are easy to compute, as the poles are simple:
Res(
f
;
i
) =

i
(
i

1)
1
/
2
(
i
+ 1)
1
/
2
2
i
=

i
√
2
,
Res(
f
;

i
)
=

i
(

i

1)
1
/
2
(

i
+ 1)
1
/
2

2
i
=

i
√
2
.
(c) We compute the integral along the large circle by using the residue at infinity.
The residue at infinity is lim
z
→∞
zf
(
z
), which is

i
, and so the integral around
the large circle is 2
πi
·
(

i
) = 2
π
.
(d) On the right small circle, if

z

1

=
ε
then

f
(
z
)
≤
C
√
ε
, and so the integral on
the circle is at most of order
ε
3
/
2
, and tends to 0 as
ε
→
0. The same applies to
the left small circle.
(e) On the upper lip we get
I
, and on the lower lip the function is the negative of the
value on the upper lip, and the integral is in the opposite direction, so the integral
is also
I
. Summing up in the limit
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 Spring '14
 Staff
 Methods of contour integration, Complex Plane, upper lip, Pole