Conformal Mapping
JunFeng Yin
In this chapter, we discuss the property of analytic function from
the viewpoint of geometric.
Remarks:
1. Local properties and global properties.
2. one-to-oneness.
Theorem 1
If f is analytic at z0 and f 0 (z0 ) 6= 0, then there is an open disk D
centered at z0 such that f is one-to-one on D.
Definition 2
The mapping f is said to be conformal at z0 if the angle between
two lines are preserved.
Theorem 3
An analytic function f is conformal at every point z0 for which
f 0 (z0 ) 6= 0.
Definition 4
A function is said to be an open mapping if the image of every
open set in its domain is, it self, open; that is, the function maps
open sets to open sets.
Theorem 5
An analytic function that is nonconstant on domains is an open
mapping.
Theorem 6
(Riemann Mapping Theorem) Let D be any simply connected
domain in the plane other than the entire plane itself. Then there
is a one-to-one analytic function that maps D onto the open unit
disk. Moreover, one can prescribe an arbitrary point of D and a
direction through that point which are to be mapped to the origin
and the direction of the positive real axis, respectively. Under such
restrictions the mapping is unique.
Demos:
1. w = z + 2.
2. w = 2z.
3. w = z + z0 , z0 = 2 + 3i.
4. w = z1 z, z1 = 1 + i.
Demos:
1. w = z + 2.
2. w = 2z.
3. w = z + z0 , z0 = 2 + 3i.
4. w = z1 z, z1 = 1 + i.
Transformation: shift, magnification, rotation.
Example 1
Find the linear transformation that rotates the entire complex
plane through an angle θ about a given point z0 .
Example 1
Find the linear transformation that rotates the entire complex
plane through an angle θ about a given point z0 .
Solution. We know that the mapping
w1 = e iθ z
rotates the plane through the angle θ about the origin.
Since every straight line gets rotated through the angles θ, and z0
is left fixed. Thus
w = w1 + (z0 − e iθ z0 ) = e iθ z + (1 − e iθ )z0 .
Example 2
Find a linear transformation that maps the circle C1 : |z − 1| = 1
onto the circle C2 : |w − 3i/2| = 2.
Example 2
Find a linear transformation that maps the circle C1 : |z − 1| = 1
onto the circle C2 : |w − 3i/2| = 2.
Solution:
w1 = z − 1
w2 = 2w1 = 2z − 2
w
= w2 + 3i/2 = 2z − 23i/2
Inversion transformation
Definition:
1
z
The inversion is a one-to-one mapping of the extended complex
plane onto itself, and it turns the unit circle inside out, and vice
versa.
Note that the point z = ρe iθ is mapped to
w = f (z) =
w=
1
1
= e −iθ
iθ
ρ
ρe
• Inversion maps generalized circle to generalized circle.
Möbius transformation
Definition 7
A Möbius (or Moebius) transformation (sometimes known as a
fractional linear transformation or bilinear transformation) is any
function of the form
w = f (z) =
az + b
cz + d
with the restriction that ad 6= bc (so that w is not a constant
function).
Note that since
f 0 (z) =
ad − bc 2
cz + d
does not vanish, the Möbius transformation f (z) is conformal at
every point except its pole z = −d/c
Note that since
f 0 (z) =
ad − bc 2
cz + d
does not vanish, the Möbius transformation f (z) is conformal at
every point except its pole z = −d/c
It is easy to see that Möbius transformation can be decomposed
into elementary transformations since
az + b
=
cz + d
a
c (cz
+ d) − ad
b − ad
c +ba
c
+
cz + d
c
cz + d
Theorem 8
Let f be any Möbius transformation. Then
1. f can be expressed as the composition of a finite sequence of
translation, magnification, rotations, and inversions.
2. f maps the extended complex plane one-to-one onto itself.
3. f maps the class of circles and lines to itself.
4. f is conformal ta every point except its pole.
Example 3
Find the image of the interior of the circle C : |z − 2| = 2 under
the Möbius transformation
w = f (z) =
z
2z − 8
Example 3
Find the image of the interior of the circle C : |z − 2| = 2 under
the Möbius transformation
w = f (z) =
z
2z − 8
Since f has a pole at z = 4 and this point lies on C , the image has
to be a straight line.
w = f (0) = 0
and w = f (2 + 2i) = −
w = f (2) =
2
1
=−
4−8
2
i
2
Example 4
Find a conformal map of the unit disk |z| < 1 onto the right
half-plane Rew > 0.
Example 4
Find a conformal map of the unit disk |z| < 1 onto the right
half-plane Rew > 0.
Find a transformation maps the circle |z| = 1 onto the imaginary
axis.
Then
z +1
w = f1 (z) =
z −1
maps 1 to ∞ and −1 to 0.
To see which half-plane is the image of the interior of the circle,
we check the point z = 0, i.e., w = −1 in the left half-plane.
Finally, f = −f1 , i.e.,
w = f (z) =
1+z
1−z
Homework
1. p. 392: 3,4,5,7.
Given any Möbius transformation
w = f (z) =
az + b
cz + d
its inverse f −1 (w ) yields
z = f −1 (w ) =
is again a Möbius transformation.
dw − b
−cw + a
Given any Möbius transformation
w = f (z) =
az + b
cz + d
its inverse f −1 (w ) yields
z = f −1 (w ) =
dw − b
−cw + a
is again a Möbius transformation.
Moreover, it is easy to verify the composition of any two Möbius
transformations is also a Möbius transformation.
Since any three distinct noncollinear points uniquely determine a
circle.
Hence, assume that z1 , z2 , z3 on Cz and w1 , w2 , w3 on Cw , the
mapping
f (z1 ) = w1 ,
f (z2 ) = w2 ,
f (z3 ) = w3 .
exist and is unique.
Assume that w1 = 0, w2 = 1, and w3 = ∞, then
T (z) =
abbreviated by (z, z1 , z2 , z3 ).
(z − z1 )(z2 − z3 )
(z − z3 )(z2 − z1 )
It is easy seen that a Möbius transformation map
w1 → 0,
is
S(w ) =
w2 → 1,
w3 → ∞
(w − w1 )(w2 − w3 )
(w − w3 )(w2 − w1 )
It is easy seen that a Möbius transformation map
w1 → 0,
w2 → 1,
is
S(w ) =
w3 → ∞
(w − w1 )(w2 − w3 )
(w − w3 )(w2 − w1 )
Thus
T (z) = S(w )
and the wanted Möbius transformation is
w = f (z) = S −1 T (z)
Example
Find a Möbius transformation that maps 0, 1 and −1 to i, 2 and 4.
Example
Find a Möbius transformation that maps 0, 1 and −1 to i, 2 and 4.
Solution The appropriate cross-ratios are given by
(z, 0, 1, −1) =
and
(w , i, 2, 4) =
(z − 0)(1 − (−1))
2z
=
(z − (−1))(1 − 0)
z +1
(w − i)(2 − 4)
−2(w − i)
=
(w − 4)(2 − i)
(w − 4)(2 − i)
Hence, solving the equation
2z
−2(w − i)
=
(w − 4)(2 − i)
z +1
for w yields the desired transformation
w=
(16 − 6i)z + 2i
(6 − 2i)z + 2
Example
Find a Möbius transformation that maps the region D1 : |z| > 1
onto the region D2 : Rew < 0.
Example
Find a Möbius transformation that maps the region D1 : |z| > 1
onto the region D2 : Rew < 0.
Solution We choose three points on the circle |z| = 1 clockwisely
z1 = 1,
z2 = −i,
z3 = −1,
w1 = 0,
w2 = i,
w3 = ∞,
and on y-axis
The appropriate cross-ratios are given by
(z − 1)(−i − (−1))
(z, 1, −i, −1) =
(z − (−1))(−i − 1)
and
(w − 0)
(w , 0, i, ∞) =
(i − 0)
Hence, solving the equation yields the desired transformation
(z − 1)(1 + i)
(1 − z)
w=
=
(z + 1)(−i − 1)
(1 + z)
Definition 9
Two points z1 and z2 are said to be symmetric with respect to a
circle C if every straight line or circle passing through z1 and z2
intersects C orthogonally.
Theorem 10
(Symmetry principle) Let Cz be a line or circle in the z-plane,
and let w = f (z) be any Möbius transformation. Then two points
z1 and z2 are symmetric with respect to Cz if and only if their
image w1 = f (z1 ), w2 = f (z2 ) are symmetric with respect to the
image of Cz under f .
Homework
1. p. 403-404: 1,3,5.