20
Conformal Mappings and Applications
Exercises 20.1
1.
For
w
=
1
z
,
u
=
x
x
2
+
y
2
and
v
=
−
y
x
2
+
y
2
.I
f
y
=
x
,
u
=
1
2
1
x
,
v
=
−
1
2
1
x
, and so
v
=
−
u
. The image is the
line
v
=
−
u
(with the origin (0
,
0) excluded.)
2.
If
y
=1,
u
=
x
x
2
+1
and
v
=
−
1
x
2
. It follows that
u
2
+
v
2
=
1
x
2
=
−
v
and so
u
2
+(
v
+
1
2
)
2
=(
1
2
)
2
. This
is a circle with radius
r
=
1
2
and center at (0
,
−
1
2
)=
−
1
2
i
. The circle can also be described by

w
+
1
2
i

=
1
2
.
3.
w
=
z
2
,
u
=
x
2
−
y
2
and
v
=2
xy
f
xy
=1
,
v
= 2 and so the hyperbola
xy
= 1 is mapped onto the line
v
=2.
4.
If
x
2
−
y
2
=4,
u
= 4 and so the hyperbola
x
2
−
y
2
= 4 is mapped onto the vertical line
u
=4.
5.
w
=Ln
z
,
u
= log
e

z

and
v
= Arg
z
. The semicircle

z

,
y>
0 may also be described by
r
,
0
<θ<π
. Therefore
u
= 0 and 0
<v<π
. The image is therefore the open line segment from
z
=0to
z
=
πi
.
6.
If
θ
=
π/
4, then
v
=
θ
=
4. In addition
u
= log
e
r
will vary from
−∞
to
∞
. The image is therefore the
horizontal line
v
=
4.
7.
w
=
z
1
/
2
re
iθ
)
1
/
2
=
r
1
/
2
e
iθ/
2
and
θ
=
θ
0
,
w
=
√
iθ
0
/
2
. Therefore Arg
w
=
θ
0
/
2 and so the image is
the ray
θ
=
θ
0
/
2.
8.
If
r
= 2 and 0
≤
θ
≤
π
2
,
w
=
√
2
e
iθ/
2
. Therefore

w

=
√
2 and 0
≤
Arg
w
≤
4. This image is a circular arc.
9.
w
=
e
z
,
u
=
e
x
cos
y
and
v
=
e
x
sin
y
. Therefore if
e
x
cos
y
,
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.
10.
If
w
=
z
+
1
z
and
z
=
e
it
,
w
=
e
it
−
e
−
it
= 2cos
t
. Therefore
u
t
and
v
= 0 and so the image is the closed
interval [
−
2
,
2] on the
u
axis.
11.
The ±rst quadrant may be described by
r>
0, 0
<θ<π/
2. If
w
/z
and
z
=
iθ
,
w
=
1
r
e
−
iθ
. Therefore
Arg
w
=
−
θ
and so
−
2
<
Arg
w<
0. The image is therefore the fourth quadrant.
12.
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
≤
1is
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.)
13.
Since
w
=
e
x
+
iy
=
e
x
e
iy
and
4
≤
y
≤
2,
4
≤
Arg
w
≤
2 and

w

=
e
x
. The image is therefore the
angular wedge de±ned by
4
≤
Arg
w
≤
2.
14.
Since
w
=
e
x
+
iy
=
e
x
e
iy
and 0
≤
x
≤
1, 0
≤
y
≤
π
, we have

w

=
e
x
and Arg
w
=
y
. Therefore 1
≤
w
≤
e
and 0
≤
Arg
w
≤
π
. These inequalities de±ne a semiangular region in the
w
plane.
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

821
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View Full DocumentExercises 20.1
16.
If
w
=2
z
−
1 and

z

= 1, then, since
z
=
w
+1
2
,
¯
¯
¯
¯
w
2
¯
¯
¯
¯
=1or

w

= 2. The image is a circle with center
at
w
=
−
1 and with radius
r
=2.
17.
The mapping
w
=
iz
is a rotation through 90
◦
since
i
=
e
iπ/
2
. Therefore the strip 0
≤
y
≤
1 is rotated through
90
◦
and so the strip
−
1
≤
u
≤
0 is the image in the
w
plane.
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 Fall '07
 Delahanty
 Boundary value problem, Line segment, arg, Upper halfplane

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