Solutions to HW 8
Karol Koziol
April 5, 2011
There are several convenient properties of Möbius
A note on Möbius transformations:
transformations that we will use throughout. Firstly, Möbius transformations are all invertible and
holomorphic. This means they are one-to-one and onto, and therefore conformal where they are de-
az+b
0
cz+d , then f (z) 6= 0 for all z ). Since the maps are
holomorphic, they are open mappings, which means that the interior of a disc must be mapped to
ned (one can also quickly check that if
all
all
of the interior or
a half-plane (if
f
f (z) =
of the exterior of a circle (if
f
maps the given circle to another), or all of
maps the given circle to a line). Any Möbius transformation can be obtained by a
nite number of translations, dilations, rotations and inversions. In addition, given any two triples of
3 points
w2 ,
and
(z1 , z2 , z3 ) and (w1 , w2 , w3 ), there
z3 to w3 (more on this later).
is a
unique
Möbius transformation sending
z1
to
w1 , z2
to
p. 392
1. We rst use the transformation
z 7→ 3z
to take the circle of radius 1 and center 0 to a circle of
radius 3 with the same center. Notice that on the circle
i
point, which suggests we should send
map is a rotation by
z 7→ z + 5.
π/2,
|w − 5| = 3,
the point 2 is at the leftmost
to the leftmost point of our current circle. Thus our second
which is given by
z 7→ iz .
Finally, we translate to the right by 5, given by
Composing all of these maps gives the map
f (z) = 3iz + 5.
=(z) = y = 0 and =(z) = y = 1 under the map
z−i
. Since this function has a pole along the rst line y = 0, this means that the point 0 will
z
be sent to ∞, and the image will again be a line. The point 1 maps to 1 − i and the point −1 maps to
2.
Let's examine what happens to the lines
z 7→ w =
<(z) = 1. The function has no poles on
=(z) = 1 will be a circle. It remains to determine
1−i
which circle this will be. The function maps i to 0, 1 + i to
2 , and the point ∞ to 1 (regard this as
a limit). These three points are enough to determine the circle, and we see that the image of =(z) = 1
1
1
is the circle |z − | = . (Draw a picture!) Now, we know that the area between the two lines must
2
2
1
1
map to either the right half-plane with <(z) > 1, the interior of the circle |z − | = , or the region
2
2
1 + i.
Therefore, the image of the line
y=0
will be the line
the second line, and therefore the image of the line
between them. Choosing a point between the two lines and nding its image will tell us which. For
example, the image of the point
3. Consider the circle
i
2 is
|z − 2| = 1.
−1,
so the image is the region between the line and the circle.
This is a circle of radius 1 centered at the point
(a) The image of the circle under
z 7→ w = z − 2i
2.
will be a circle of radius 1 centered at
2 − 2i.
The interior of the original circle will map to the interior of the new circle.
(b) The map
z 7→ w = 3iz
is a composition of a dilation and a rotation (taken in arbitrary order).
The dilation will stretch everything by a factor of 3, meaning the image will be a circle of radius
1
3 centered at the point 6, and the rotation will rotate everything counterclockwise by
Therefore, the image of the circle will be a circle of radius 3 centered at
6i.
π/2
radians.
The interior again maps
to the interior.
z−2
z−1 has a pole on the circle, and therefore the image of the circle will be a
straight line. Therefore, we just need to compute the image of two points to gure out which line it
1+i
1
is. The point 3 maps to , and the point 2 + i maps to
2
2 . Therefore, the image of the circle is the
1
line <(z) = . To determine where the interior of the circle goes, note that the image of the point 2
2
1
is 0. Thus, the interior of the circle maps to <(z) < .
2
z−4
(d) The map z 7→ w =
z−3 also has a pole on the circle, this time at the point z = 3. Again the
3+i
3
image of the circle will be a straight line. The point 1 maps to
2 , and the point 2 + i maps to 2 .
3
Therefore, the image of the circle is the line <(z) = . We must again nd where the interior goes.
2
3
The point 2 maps to 2, and therefore the image of the interior will be the half-plane <(z) > .
2
1
(e) The map z 7→ w =
z has no pole on the circle, and therefore the image will be another circle.
(c) The map
z 7→ w =
1 maps to
2−i
1
, and the point 2 + i maps to
. One then veries that all of these points lie on the
3
5
2
1
1
circle |z − | = , and therefore this must be the image of our original circle. The point 2 maps to ,
3
3
2
which lies inside our new circle, and we see that the interior maps to the interior.
We must nd the images of three points to determine this circle uniquely. We have that
1, 3
maps to
4. We want to nd a Möbius transformation that takes the lower half-plane to the disc
|w + 1| < 1.
We know that if we want to transform a line into a circle, we must use some form of inversion.
A
1
won't work, however, since this will take the real axis back to itself. We therefore
z
1
consider the map z 7→
z−i (we can make many other choices here, meaning the map we will get in
the end won't be unique). This will take the real axis to a circle; it maps the point 0 to i, the point
−1+i
1 to 1+i
2 , and the point −1 to
2 . Therefore, we see that the image of the real axis is the circle
i
1
i
given by |z − | = . Note also that the image of −i is
2
2
2 , so that the image of the lower half-plane
rst guess of
z 7→
interior of the disc. If we dilate this by 2, we will get the radius we want, and multiplying by
2i
i will rotate the image by π/2, giving us the circle |z −1| < 1. The whole map is given by z 7→ w = z−i
.
is the
5. We would like rst to nd a map which has a zero at
z = −i
(so that it maps
−i
to
somewhere else on the circle, so that the image of the circle is a line. Let's try the map
This maps has a pole at
z = i,
so the image of the circle will certainly be a line, and it
1+i
1−i = i, so the image of
maps to −1, meaning the interior maps the left half-plane.
so in fact the image is a line passing through the origin. The point
0
−1 to ensure
z+i
z→
7 w = − z−i
.
the circle is the imaginary axis. The point
We must therefore multiply by
whole map is given by
0), and a pole
z+i
z 7→ w = z−i
.
maps −i to 0,
1
maps to
that the image lands in the right half-plane. Thus, the
az+b
cz+d be a Möbius transformation. Assume rst that c = 0. Then our function takes on
the simple form f (z) = αz + β , where α = a/d, β = b/d. Then, if f (z0 ) = αz0 + β = z0 , either z0 = ∞
6. Let
f (z) =
(which we usually disregard), or we must have
Assume now that
c 6= 0.
α = 1, β = 0.
We then have the equation
az0 + b
= z0 ⇔ az0 + b = cz02 + dz0 ⇔ cz02 + (d − a)z0 − b = 0.
cz0 + d
This equation has either one or two solutions in
C,
the quadratic polynomial is 0 or not.
2
depending on whether or not the discriminant of
7. We rst state a uniqueness property of Möbius transformations. Let's assume that we have two
S
and
T
0, 1, ∞
α1 , α2 , α3 . Since Möbius
T −1 is the inverse of T . This map
will again be a Möbius transformation (check!), and will take each αi back to itself. Therefore, it will
have three xed points, which is impossible by the previous problem. Therefore, S must equal T . This
says that if we can write down a solution to this exercise that works, then it is the unique solution.
b
az+b
One can also do these problems by hand: let f (z) =
cz+d , and solve the equations f (0) = d =
a
a+b
α1 , f (1) = c+d = α2 , “f (∞) = limz→∞ f (z) = c = α3 .
(a) One can quickly guess that the map f (z) = iz takes the points 0, 1, ∞ to 0, i, ∞. By our
Möbius transformations
that map
transformations are invertible, we can form the map
to three given points
S ◦ T −1 ,
where
previous claim, this maps is unique.
(b) We go through and solve this directly. We have the three equations:
b
d
= 0
a+b
f (1) =
c+d
= 1
a
f (∞) =
c
= 2.
f (0)
The rst equation implies
gives
c + d = 2c ⇒ d = c.
b = 0,
=
while the third implies
Therefore, our map becomes
a = 2c. Plugging these into the third equation
2cz
2z
f (z) = cz+c
= z+1
.
(c) We set up our system of equations again, being careful about the image of 1:
=
b
d
−i
lim f (z)
=
∞
f (∞)
=
f (0)
z→1
=
a
c
1.
=
f
a = c,
z = 1, which only occurs when c = −d. The
b = −id. Putting all of these together gives
The second equation implies that
must have a pole at
third equation implies that
and the rst that
f (z) =
−dz−id
−dz+d
=
z+i
z−1 .
(d) We again have the following system:
f (0)
lim f (z)
z→1
f (∞)
b
d
= −1
=
= ∞
=
=
3
a
c
1.
Again, the second equation tells us that we must have
us that
a = c, b = −d.
Our map becomes
f (z) =
c = −d, while
z+1
= z−1
.
−dz−d
−dz+d
4
the rst and third equations tell