1Analytic functions
IB Complex Methods
1.6 Conformal maps
Sometimes, we might be asked to solve a problem on some complicated subspace
U ⊆ C
. For example, we might need to solve Laplace’s equation subject to
some boundary conditions. In such cases, it is often convenient to transform
our space
U
into some nicer space
V
, such as the open disk. To do so, we will
need a complex function
f
that sends
U
to
V
. For this function to preserve our
properties such that the solution on
V
can be transferred back to a solution on
U
, we would of course want
f
to be differentiable. Moreover, we would like it to
have non-vanishing derivative, so that it is at least locally invertible.
Definition
(Conformal map)
.
A conformal map
f
:
U → V
, where
U, V
are
open subsets of C, is one which is analytic with non-zero derivative.
In reality, we would often want the map to be a bijection. We sometimes call
these conformal equivalences.
Unfortunately, after many hundred years, we still haven’t managed to agree
on what being conformal means. An alternative definition is that a conformal
map is one that preserves the angle (in both magnitude and orientation) between
intersecting curves.
We shall show that our definition implies this is true; the converse is also
true, but the proof is omitted. So the two definitions are equivalent.
Proposition.
A conformal map preserves the angles between intersecting curves.
Proof.
Suppose
z
1
(
t
) is a curve in
C
, parameterised by
t ∈ R
, which passes
through a point
z
0
when
t
=
t
1
. Suppose that its tangent there,
z
0
1
(
t
1
), has a
well-defined direction, i.e. is non-zero, and the curve makes an angle
φ
=
arg z
0
1
(
t
1
)
to the x-axis at z
0
.
Consider the image of the curve,
Z
1
(
t
) =
f
(
z
1
(
t
)). Its tangent direction at
t = t
1
is
Z
0
1
(t
1
) = z
0
1
(t
1
)f
0
(z
1
(t
1
)) = z
0
1
(t
0
)f
0
(z
0
),
and therefore makes an angle with the x-axis of
arg(Z
0
1
(t
1
)) = arg(z
0
1
(t
1
)f
0
(z
0
)) = φ + arg f
0
(z
0
),
noting that arg f
0
(z
0
) exists since f is conformal, and hence f
0
(z
0
) 6= 0.
In other words, the tangent direction has been rotated by
arg f
0
(
z
0
), and this
is independent of the curve we started with.
Now if
z
2
(
t
) is another curve passing through
z
0
. Then its tangent direction
will also be rotated by arg f
0
(z
0
). The result then follows.
Often, the easiest way to find the image set of a conformal map acting on a
set
U
is first to find the image of its boundary,
∂U
, which will form the boundary
∂V
of
V
; but, since this does not reveal which side of
∂V V
lies on, take a point
of your choice within U , whose image will lie within V .
Example.
(i)
The map
z 7→ az
+
b
, where
a, b ∈ C
and
a 6
= 0, is a conformal map. It
rotates by
arg a
, enlarges by
|a|
, and translates by
b
. This is conformal
everywhere.
(ii) The map f(z) = z
2
is a conformal map from
U =
n
z : 0 < |z| < 1, 0 < arg z <
π
2
o
to
V = {w : 0 < |w| < 1, 0 < arg w < π}.
1
U
f
1
V
Note that the right angles between the boundary curves at
z
= 1 and
i
are
preserved, because
f
is conformal there; but the right angle at
z
= 0 is not
preserved because
f
is not conformal there (
f
0
(0) = 0). Fortunately, this
does not matter, because U is an open set and does not contain 0.
(iii) How could we conformally map the left-hand half-plane
U = {z : Re z < 0}
to a wedge
V =
n
w : −
π
4
< arg w ≤
π
4
o
.
U
V
We need to halve the angle. We saw that
z 7→ z
2
doubles then angle, so we
might try
z
1
2
, for which we need to choose a branch. The branch cut must
not lie in
U
, since
z
1
2
is not analytic on the branch cut. In particular, the
principal branch does not work.
So we choose a cut along the negative imaginary axis, and the function is
defined by
re
iθ
7→
√
re
iθ/2
, where
θ ∈
−
π
2
,
3π
2
. This produces the wedge
{z
0
:
π
4
< arg z
0
<
3π
4
}
. This isn’t exactly the wedge we want. So we need
to rotate it through −
π
2
. So the final map is
f(z) = −iz
1
2
.
(iv) e
z
takes rectangles conformally to sectors of annuli:
U
iy
1
iy
2
x
1
x
2
e
x
1
e
x
2
y
1
y
2
V
With an appropriate choice of branch, log z does the reverse.
(v)
M¨obius maps (which are conformal equivalence except at the point that is
sent to
∞
) are very useful in taking circles, or parts of them to straight
lines, or vice versa.
Consider
f
(
z
) =
z−1
z+1
acting on the unit disk
U
=
{z
:
|z| <
1
}
. The
boundary of
U
is a circle. The three points
−
1
, i
and +1 lie on this circle,
and are mapped to ∞, i and 0 respectively.
Since M¨obius maps take circlines to circlines, the image of
∂U
is the
imaginary axis. Since
f
(0) =
−
1, we see that the image of
U
is the
left-hand half plane.
U
V
We can derive this alternatively by noting
w =
z − 1
z + 1
⇔ z = −
w + 1
w − 1
.
So
|z| < 1 ⇔ |w + 1| < |w − 1|,
i.e.
w
is closer to
−
1 than it is to +1, which describes precisely the left-hand
half plane.
In fact, this particular map
f
(
z
) =
z−1
z+1
can be deployed more generally on
quadrants, because it permutes 8 divisions on the complex plane as follows:
23
67
14
58
The map sends 1
7→
2
7→
3
7→
4
7→
1 and 5
7→
6
7→
7
7→
8
7→
5. In
particular, this agrees with what we had above — it sends the complete
circle to the left hand half plane.
(vi)
Consider the map
f
(
z
) =
1
z
. This is just another M¨obius map! Hence
everything we know about M¨obius maps apply to this. In particular, it is
useful for acting on vertical and horizontal lines. Details are left for the
first example sheet.
In practice, complicated conformal maps are usually built up from individual
building blocks, each a simple conformal map. The required map is the com-
position of these. For this to work, we have to note that the composition of
conformal maps is conformal, by the chain rule.
Example.
Suppose we want to map the upper half-disc
|z| <
1,
Im z >
0 to the
full disc
|z| <
1. We might want to just do
z 7→ z
2
. However, this does not work,
since the image does not include the non-negative real axis, say
z
=
1
2
. Instead,
we need to do something more complicated. We will do this in several steps:
(i) We apply f
1
(z) =
z−1
z+1
to take the half-disc to the second quadrant.
(ii)
We now recall that
f
1
also takes the right-hand half plane to the disc. So we
square and rotate to get the right-hand half plane. We apply
f
2
(
z
) =
iz
2
.
(iii) We apply f
3
(z) = f
1
(z) again to obtain the disc.
Then the desired conformal map is
f
3
◦ f
2
◦ f
1
, you can, theoretically, expand
this out and get an explicit expression, but that would be a waste of time.
z 7→
z−1
z+1
z 7→ iz
2
z 7→
z−1
z+1