20
20
Conformal Mappings
EXERCISES 20.1
Complex Functions as Mappings
3.
For
w
=
z
2
,
u
=
x
2
−
y
2
and
v
= 2
xy
. If
xy
= 1,
v
= 2 and so the hyperbola
xy
= 1 is mapped onto the line
v
= 2.
6.
If
θ
=
π/
4, then
v
=
θ
=
π/
4. In addition
u
= log
e
r
will vary from
−∞
to
∞
. The image is therefore the
horizontal line
v
=
π/
4.
9.
For
w
=
e
z
,
u
=
e
x
cos
y
and
v
=
e
x
sin
y
. Therefore if
e
x
cos
y
= 1,
u
= 1. The curve
e
x
cos
y
= 1 is mapped
into the line
u
= 1. Since
v
=
sin
y
cos
y
= tan
y
,
v
varies from
−∞
to
∞
and the image is the line
u
= 1.
12.
For
w
=
1
z
,
u
=
x
x
2
+
y
2
and
v
=
−
y
x
2
+
y
2
. The line
y
= 0 is mapped to the line
v
= 0, and, from
Problem 2, the line
y
= 1 is mapped onto the circle

w
+
1
2
i

=
1
2
. Since
f
(
1
2
i
) =
−
2
i
, the region 0
≤
y
≤
1 is
mapped onto the points in the halfplane
v
≤
0 which are on or outside the circle

w
+
1
2
i

=
1
2
. (The image
does not include the point
w
= 0.)
15.
The mapping
w
=
z
+4
i
is a translation which maps the circle

z

= 1 to a circle of radius
r
= 1 and with center
w
= 4
i
. This circle may be described by

w
−
4
i

= 1.
18.
Since
w
= (1+
i
)
z
=
√
2
e
iπ/
4
z
, the mapping is the composite of a rotation through 45
◦
and a magnification by
α
=
√
2. The image of the first quadrant is therefore the angular wedge
π/
4
≤
Arg
w
≤
3
π/
4.
21.
We first let
z
1
=
z
−
i
to map the region 1
≤
y
≤
4 to the region 0
≤
y
1
≤
3. We then let
w
=
e
−
iπ/
2
z
1
to rotate
this strip through
−
90
◦
. Therefore
w
=
−
i
(
z
−
i
) =
−
iz
−
1 maps 1
≤
y
≤
4 to the strip 0
≤
u
≤
3.
24.
The mapping
w
=
iz
will rotate the strip
−
1
≤
x
≤
1 through 90
◦
so that the strip
−
1
≤
v
≤
1 results.
27.
By Example 1, Section 20
.
1,
z
1
=
e
z
maps the strip 0
≤
y
≤
π
onto the upper halfplane
y
1
≥
0, or
0
≤
Arg
z
1
≤
π
. The power function
w
=
z
3
/
2
1
changes the opening of this wedge by a factor of 3
/
2 so the wedge
0
≤
Arg
w
≤
3
π/
2 results. The composite of these two mappings is
w
= (
e
z
)
3
/
2
=
e
3
z/
2
.
30.
The mapping
z
1
=
−
(
z
−
πi
) lowers
R
by
π
units in the vertical direction and then rotates the resulting region
through 180
◦
. The image region
R
1
is upper halfplane
y
1
≥
0. By Example 1, Section 20
.
1,
w
= Ln
z
1
maps
R
1
onto the strip 0
≤
v
≤
π
. The composite of these two mappings is
w
= Ln(
πi
−
z
).
301
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EXERCISES 20.2
Conformal Mappings
20.2
Conformal Mappings
3.
f
(
z
) = 1 +
e
z
and 1 +
e
z
= 0 for
z
=
±
i
±
2
nπi
. Therefore
f
is conformal except for
z
=
πi
±
2
nπi
.
6.
The function
f
(
z
) =
πi
−
1
2
[Ln(
z
+1)+Ln(
z
−
1)] is analytic except on the branch cut
x
−
1
≤
0 or
x
≤
1, and
f
(
z
) =
−
1
2
1
z
+ 1
+
1
z
−
1
=
−
z
z
2
−
1
is nonzero for
z
= 0,
±
1. Therefore
f
is conformal except for
z
=
x
,
x
≤
1.
9.
f
(
z
) = (sin
z
)
1
/
4
is the composite of
z
1
= sin
z
and
w
=
z
1
/
4
1
. The region
−
π/
2
≤
x
≤
π/
2,
y
≥
0 is mapped to
the upper halfplane
y
2
≥
0 by
z
1
= sin
z
(See Example 2) and the power function
w
=
z
1
/
4
1
maps this upper
halfplane to the angular wedge 0
≤
Arg
w
≤
π/
4. The real interval [
−
π/
2
, π/
2] is first mapped to [
−
1
,
1] and
then to the union of the line segments from
e
i
π
4
to 0 and 0 to 1. See the figures below.
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 TheodoreHolford
 Upper halfplane

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