Complex analysis

39 Pages·1997·0.301 MB·English

by  Douglas Arnold

COMPLEX ANALYSIS1
Douglas N. Arnold2
References:
John B. Conway, Functions of One Complex Variable, Springer-Verlag, 1978.
Lars V. Ahlfors, Complex Analysis, McGraw-Hill, 1966.
Raghavan Narasimhan, Complex Analysis in One Variable, Birkha¨user, 1985.
CONTENTS
I. The Complex Number System.............................................2
II. Elementary Properties and Examples of Analytic Fns................3
Differentiability and analyticity ........................................... 4
The Logarithm............................................................6
Conformality..............................................................6
Cauchy–Riemann Equations...............................................7
M¨obius transformations...................................................9
III. Complex Integration and Applications to Analytic Fns. ............ 11
Local results and consequences...........................................12
Homotopy of paths and Cauchy’s Theorem...............................14
Winding numbers and Cauchy’s Integral Formula.........................15
Zero counting; Open Mapping Theorem..................................17
Morera’s Theorem and Goursat’s Theorem ............................... 18
IV. Singularities of Analytic Functions....................................19
Laurent series............................................................20
Residue integrals.........................................................23
V. Further results on analytic functions.................................26
The theorems of Weierstrass, Hurwitz, and Montel ....................... 26
Schwarz’s Lemma........................................................28
The Riemann Mapping Theorem.........................................29
Complements on Conformal Mapping .................................... 31
VI. Harmonic Functions......................................................32
The Poisson kernel.......................................................33
Subharmonic functions and the solution of the Dirichlet Problem ......... 36
The Schwarz Reflection Principle.........................................39
1These lecture notes were prepared for the instructor’s personal use in teaching a half-semester course
on complex analysis at the beginning graduate level at Penn State, in Spring 1997. They are certainly not
meant to replace a good text on the subject, such as those listed on this page.
2Department of Mathematics, Penn State University, University Park, PA 16802.
Email: [email protected]. Web: http://www.math.psu.edu/dna/.
1
2
I. The Complex Number System
R is a field. For n > 1, Rn is a vectorspace over R, so is an additive group, but doesn’t
have a multiplication on it. We can endow R2 with a multiplication by
(a,b)(c,d) = (ac−bd,bc+ad).
Under this definition R2 becomes a field, denoted C. Note that (a/(a2+b2),−b/(a2+b2))
is the multiplicative inverse of (a,b). (Remark: it is not possible to endow Rn with a field
structure for n > 2.) We denote (0,1) by i and identify x ∈ R with (x,0), so R ⊂ C. Thus
(a,b) = a+bi, a,b ∈ R. Note that i2 = −1. C is generated by adjoining i to R and closing
under addition and multiplication. It is remarkable that the addition of i lets us not only
solve the equation x2 +1 = 0, but every polynomial equation.
For a and b real and z = a + bi we define Rez = a, Imz = b, z¯ = a − bi, and
|z| = (a2 +b2)1/2. Then
Rez = (z +z¯)/2, Imz = (z −z¯)/(2i),
1 z¯
|z|2 = zz¯, = ,
z |z|2
z ±w = z¯±w¯, zw = z¯w¯,
z/w = z¯/w¯, |z +w| ≤ |z|+|w|.
The map θ (cid:55)→ (cosθ,sinθ) defines a 2π-periodic map of the real line onto the unit
circle in R2. In complex notation this map is θ (cid:55)→ cisθ := cosθ + isinθ. Every nonzero
complex number can be written as rcisθ where r > 0 is uniquely determined and θ ∈ R
is uniquely determined modulo 2π. The number 0 is equal to rcisθ where r = 0 and θ
is arbitrary. The relation z = rcisθ determines the relations z (cid:55)→ r which is simply the
function r = |z| and z (cid:55)→ θ. The latter is denoted θ = argθ. Note that for z (cid:54)= 0, argθ is
determined modulo 2π (while arg0 is arbitrary). We can normalize arg by insisting that
argz ∈ (−π,π]. Note that if z = rcisθ and z = rcisθ then z z = r r cis(θ + θ ).
1 1 2 2 1 2 1 2 1 2
The latter formula just encapsulates the formula for the sine and cosine of a sum, and gives
argz z = argz +argz . In particular, ircisθ = rcis(θ +π/2), so multiplication by i is
1 2 1 2
just the operation of rotation by π/2 in the complex plane. Multiplication by an arbitrary
complex number rcisθ is just rotation by argθ followed by (or preceded by) dilation by a
factor r. Further, zn = rncis(nθ). Every nonzero z ∈ C admits n distinct nth roots: the
√
nth roots of rcisθ are n rcis[(θ+2πk)/n], k = 0,1,... ,n.
Lines and circles in the plane. Circles given by |z − a| = r where a ∈ C is the center
and r > 0 is the radius. If 0 (cid:54)= b ∈ C then the line through the origin in the direction
b is the set of all points of the form tb, t ∈ R, or all z with Im(z/b) = 0. If t ∈ R
and c > 0 then (t + ci)b = tb + cib represents a point in the half plane to the left of
b determined by the line tb, i.e., {z : Im(z/b) > 0} is the equation of that half-plane.
Similarly, {z : Im[(z −a)/b] > 0} is the translation of that half-plane by a, i.e., the half-
plane determined by the line through a parallel to b and in the direction to the left of
b.
3
Stereographic projection determines a one-to-one correspondence between the unit
sphere in R3 minus the north-pole, S, and the complex plane via the correspondence
x +ix
z ↔ 1 2,
1−x
3
2Rez 2Imz |z|2 −1
x = , x = , x = .
1 1+|z|2 2 1+|z|2 3 |z|+1
If we define C = C ∪ {∞}, then we have a one-to-one correspondence between S and
∞
C . This allows us to define a metric on C , which is given by
∞ ∞
2|z −z | 2
d(z ,z ) = 1 2 , d(z,∞) = .
1 2 (1+|z |2)(1+|z |2) 1+|z|2
1 2
(cid:112) (cid:112)
II. Elementary Properties and Examples of Analytic Functions
For z (cid:54)= 1, N zn = (1 − zN+1)/(1 − z). Therefore the geometric series ∞ zn
n=0 n=0
converges (to 1(cid:80)/(1−z)) if |z| < 1. It clearly diverges, in fact its terms become unb(cid:80)ounded,
if |z| > 1.
Weierstrass M-Test. Let M ,M ,... be positive numbers with M < ∞ and suppose
0 1 n
that f : X → C are functions on some set X satisfying sup (cid:80)|f (x)| ≤ M . Then
n x∈X n n
∞
f (x) is absolutely and uniformly convergent.
n=0 n
(cid:80)
Theorem. Let a ,a ,··· ∈ C be given and define the number R by
0 1
1
= limsup|a |1/n.
n
R
Then (1) for any a ∈ C the power series ∞ a (z − a)n converges absolutely for all
n=0 n
|z−a| < R and it converges absolutely and u(cid:80)niformly on the disk |z−a| ≤ r for all r < R.
(2) The sequence a (z − a)n is unbounded for all |z − a| > R (and hence the series is
n
certainly divergent).
Thus we see that the set of points where a power series converges consists of a disk
|z−a| < R and possibly a subset of its boundary. R is called the radius of convergence of
its series. The case R = ∞ is allowed.
Proof of theorem. For any r < R we show absolute uniform convergence on D = {|z−a| ≤
r
r}. Choose r˜ ∈ (r,R). Then, 1/r˜ > limsup|a |1/n, so |a |1/n < 1/r˜ for all n sufficiently
n n
large. For such n, |a | < 1/r˜n and so
n
sup |a (z −a)n| < (r/r˜)n.
n
z∈D
r
Since (r/r˜)n < ∞ we get the absolute uniform convergence on D .
r
(cid:80)
If |z − a| = r > R, take r˜ ∈ (R,r). Then there exist n arbitrarily large such that
|a |1/n ≥ 1/r˜. Then, |a (z −a)n| ≥ (r/r˜)n, which can be arbitrarily large. (cid:3)
n n
4
Theorem. If a ,a ,... ∈ C and lim|a /a | exists as a finite number or infinity, then
0 1 n n+1
this limit is the radius of convergence R of a (z −a)n.
n
(cid:80)
Proof. Without loss of generality we can suppose that a = 0. Suppose that |z| >
lim|a /a |. Then for all n sufficiently large |a | < |a z| and |a zn| < |a zn+1|.
n n+1 n n+1 n n+1
Thus the series a zn has terms of increasing magnitude, and so cannot be convergent.
n
Thus |z| ≥ R. T(cid:80)his shows that lim|a /a | ≥ R.
n n+1
Similarly, supposethatz < lim|a /a |. Thenforallnsufficientlylarge|a | > |a z|
n n+1 n n+1
and |a zn| > |a zn+1|. Thus the series has terms of decreasing magnitude, and so, by
n n+1
the previous theorem, |z| ≤ R. This shows that lim|a /a | ≤ R. (cid:3)
n n+1
Remark. Onthecircleofconvergence, manydifferentbehaviorsarepossible. zn diverges
for all |z| = 1. zn/n diverges for z = 1, else converges, but not absolutely(cid:80)(this follows
from the fact th(cid:80)at the partial sums of zn are bounded for z (cid:54)= 1 and 1/n ↓ 0). zn/n2
converges absolutely on |z| ≤ 1. Sier(cid:80)pinski gave a (complicated) example of a(cid:80)function
which diverges at every point of the unit circle except z = 1.
As an application, we see that the series
∞
zn
(cid:88) n!
n=0
converges absolutely for all z ∈ C and that the convergence is uniform on all bounded sets.
The sum is, by definition, expz.
Now suppose that ∞ a (z−a)n has radius of convergence R, and consider its formal
n=0 n
derivative ∞ na ((cid:80)z−a)n−1 = ∞ (n+1)a (z−a)n. Nowclearly a (z−a)n
n=1 n n=0 n+1 n=0 n+1
has the sam(cid:80)e radius of convergenc(cid:80)e as ∞ a (z −a)n since (cid:80)
n=0 n
(cid:80)
N N+1
(z −a) a (z −a)n = a (z −a)−a ,
(cid:88) n+1 (cid:88) n 0
n=0 n=0
and so the partial sums on the left and right either both diverge for a given z or both
converge. This shows (in a roundabout way) that limsup|a |1/n = limsup|a |1/n =
n+1 n
1/R. Nowlim(n+1)1/n = 1asiseasilyseenbytakinglogs. Moreover,itiseasytoseethatif
limsupb = b and limc = c > 0, then limsupb c = bc. Thus limsup|(n+1)a |1/n =
n n n n n+1
1/R. This shows that the formal derivative of a power series has the same radius of
convergence as the original power series.
Differentiability and analyticity. Definition of differentiability at a point (assumes
function is defined in a neighborhood of the point).
Most of the consequences of differentiability are quite different in the real and complex
case, but the simplest algebraic rules are the same, with the same proofs. First of all,
differentiability at a point implies continuity there. If f and g are both differentiable at a
point a, then so are f ±g, f ·g, and, if g(a) (cid:54)= 0, f/g, and the usual sum, product, and
quotient rules hold. If f is differentiable at a and g is differentiable at f(a), then g ◦ f
is differentiable at a and the chain rule holds. Suppose that f is continuous at a, g is
(cid:48)
continous at f(a), and g(f(z)) = z for all z in a neighborhood of a. Then if g (f(a)) exists
(cid:48) (cid:48)
and is non-zero, then f (a) exists and equals 1/g (f(a)).
5
Definition. Let f be a complex-valued function defined on an open set G in C. Then f
(cid:48)
is said to be analytic on G if f exists and is continuous at every point of G.
Remark. We shall prove later that if f is differentiable at every point of an open set in C
it is automatically analytic; in fact, it is automatically infinitely differentiable. This is of
course vastly different from the real case.
If Q is an arbitrary non-empty subset of C we say f is analytic on Q if it is defined and
analytic on an open set containing Q.
Wenowshowthatapowerseriesisdifferentiableateverypointinitsdiskofconvergence
and that its derivative is given by the formal derivative obtained by differentiating term-
by-term. Since we know that that power series has the same radius of convergence, it
follows that a power series is analytic and infinitely differentiable in its convergence disk.
For simplicity, and without loss of generality we consider a power series centered at zero:
f(z) = a zn. Suppose that the radius of convergence is R and that |z | < R. We must
n n 0
(cid:80)
show that for any (cid:15) > 0, the inequality
f(z)−f(z )
(cid:12) 0 −g(z )(cid:12) ≤ (cid:15)
(cid:12) z −z 0 (cid:12)
(cid:12) 0 (cid:12)
is satisfied for all z sufficiently(cid:12)close to z , where g((cid:12)z) = ∞ na zn−1. Let s (z) =
0 n=1 n N
N a zn, R (z) = ∞ a zn. Then (cid:80)
n=0 n N n=N+1 n
(cid:80) (cid:80)
f(z)−f(z ) s (z)−s (z )
(cid:12) 0 −g(z )(cid:12) ≤ (cid:12) N N 0 −s(cid:48) (z )(cid:12)
(cid:12) z −z 0 (cid:12) (cid:12) z −z N 0 (cid:12)
(cid:12) 0 (cid:12) (cid:12) 0 (cid:12)
(cid:12) (cid:12) (cid:12) R(cid:12) (z)−R (z )
+|s(cid:48) (z )−g(z )|+(cid:12) N N 0 (cid:12) =: T +T +T .
N 0 0 (cid:12) z −z (cid:12) 1 2 3
(cid:12) 0 (cid:12)
Now s(cid:48) (z ) is just a partial sum for g(z ), so for N (cid:12)sufficiently large (a(cid:12)nd all z), T ≤ (cid:15)/3.
N 0 0 2
Also,
∞
R (z)−R (z ) zn −zn
N n 0 = a 0 .
z −z (cid:88) n z −z
0 0
n=N+1
Now |z | < r < R for some r, and if we restrict to |z| < r, we have
0
zn −zn
(cid:12)a 0 (cid:12) = |a ||zn−1 +zn−2z +···+zn−1| ≤ a nrn−1.
(cid:12) n z −z (cid:12) n 0 0 n
(cid:12) 0 (cid:12)
Since a nrn(cid:12)−1 is conve(cid:12)rgent, we have for N sufficiently large and all |z| < r then
n
(cid:80)
T < (cid:15)/3. Now fix a value of N which is sufficiently large by both criteria. Then the
3
differentiability of the polynomial s shows that T ≤ (cid:15)/3 for all z sufficiently close to z .
N 1 0
We thus know that if f(z) = a zn, then, within the disk of convergence, f(cid:48)(z) =
n
na zn−1, andbyinduction, f(cid:48)(cid:48)((cid:80)z) = n(n−1)a zn−2, etc. Thusa = f(0), a = f(cid:48)(0),
n n 0 1
(cid:80) (cid:48)(cid:48) (cid:48)(cid:48)(cid:48) (cid:80)
a = f (0)/2, a = f (0)/3!, etc. This shows that any convergent power series is the sum
2 3
of its Taylor series in the disk of convergence:
fn(a)
f(z) = (z −a)n.
(cid:88) n!
(cid:48)
In particular, exp = exp.
6
Lemma. A function with vanishing derivative on a region (connected open set) is constant.
Indeed, it is constant on disks, since we can restrict to segments and use the real result.
Then we can use connectedness to see that the set of points where it takes a given value
is open and closed.
Use this to show that exp(z)exp(a−z) ≡ expa. Define cosz and sinz, get cos2+sin2 ≡
1, expz = cosz +isinz, cisθ = exp(iθ), expz = exp(Rez)cis(Imz).
The Logarithm. If x (cid:54)= 0, the most general solution of expz = w is z = log|w|+iargw+
2πin, n ∈ Z. There is no solution to expz = 0.
Definition. If G is an open set and f : G → C is a continuous function satisfying
exp(f(z)) = z, then f is called a branch of the logarithm on G.
Examples: C\R (principal branch), C\R , a spiral strip.
− +
By the formula for the derivative of an inverse, a branch of the logarithm is analytic
with derivative 1/z.
A branch of the logarithm gives a branch of zb := exp(blogz) (understood to be the
principal branch if not otherwise noted). Note that eb = expb.
Notethatdifferentbranchesofthelogarithmassignvaluestologz thatdifferbyaddition
of 2πin. Consequently different branches of zb differ by factors of exp(2πinb). If b is an
integer, all values agree. If b is a rational number with denominator d there are d values.
Conformality. Consider the angle between two line segments with a common vertex in
the complex plane. We wish to consider how this angle is transformed under a complex
map. The image will be two smooth curves meeting at a common point, so we have to
define the angle between two such curves.
Defineapathinthecomplexplaneasacontinuousmapγ : [a,b] → C. Ifγ(cid:48)(t) ∈ Cexists
at some point and is not zero, then it is a vector tangent to the curve at γ(t), and hence
its argument is the angle between the (tangent to the) curve and the horizontal. (Note:
if γ(cid:48)(t) = 0, the curve may not have a tangent; e.g., γ(t) = t2 − it3.) Let γ : [a,b] → C
and γ˜ : [a˜,˜b] → C be two smooth curves, such that γ(a) = γ˜(a˜) = z, and γ(cid:48)(a),γ˜(cid:48)(a˜) (cid:54)= 0.
Then argγ˜(cid:48)(a˜)−argγ(cid:48)(a) measures the angle between the curves. Now consider the images
(cid:48) (cid:48) (cid:48) (cid:48) (cid:48) (cid:48)
under f, e.g., of ρ(t) = f(γ(t)). Then ρ (a) = f (z)γ (a), ρ˜(a˜) = f (z)γ˜ (a˜). Therefore
argρ˜(cid:48)(a˜)−argρ(cid:48)(a) = argγ˜(cid:48)(a˜)+argf(cid:48)(z)−argγ(cid:48)(a)−argf(cid:48)(z), i.e., the angle is invariant
in both magnitude and sign. This says that an analytic mapping is conformal, whenever
its derivative is not zero. (Not true if the derivative is zero. E.g., z2 doubles angles at the
origin.)
The conformality of an analytic map arises from the fact that in a neighborhood of a
(cid:48) (cid:48)
point z such a map behaves to first order like multiplication by f (z ). Since f (z ) =
0 0 0
rexpiθ, near z f simply behave like rotation by θ followed by dilation. By contrast, a
0
smooth map from R2 → R2 can behave like an arbitrary linear operator in a neighborhood
of a point.
At this point, show the computer graphics square1.mps and square2.mps.
7
Cauchy–Riemann Equations. Let f : Ω → C with Ω ⊂ C open. Abuse notation
(cid:48)
slightly by writing f(x,y) as an alternative for f(x + iy). If f (z) exists for some z =
x+iy ∈ Ω, then
f(z +h)−f(z) f(x+h,y)−f(x,y) ∂f
(cid:48)
f (z) = lim = lim = ,
h∈R h h∈R h ∂x
h→0 h→0
and
f(z +ih)−f(z) f(x,y +h)−f(x,y) ∂f
f(cid:48)(z) = lim = lim −i = −i .
h∈R ih h∈R h ∂y
h→0 h→0
Thus complex-differentiability of f at z implies not only that the partial derivatives of f
exist there, but also that they satisfy the Cauchy–Riemann equation
∂f ∂f
= −i .
∂x ∂y
If f = u+iv, then this equation is equivalent to the system
∂u ∂v ∂v ∂u
= , = − .
∂x ∂y ∂x ∂y
Another convenient notation is to introduce
∂f 1 ∂f ∂f ∂f 1 ∂f ∂f
:= (cid:18) −i (cid:19), := (cid:18) +i (cid:19).
∂z 2 ∂x ∂y ∂z¯ 2 ∂x ∂y
(These are motivated by the equations x = (z +z¯)/2, y = (z −z¯)/(2i), which, if z and z¯
were independent variables, would give ∂x/∂z = 1/2, ∂y/∂z = −i/2, etc.)
In terms of these, the Cauchy–Riemann equations are exactly equivalent to
∂f
= 0,
∂z¯
which is also equivalent to
∂f ∂f
= .
∂z ∂x
Thus, if f is analytic on Ω, then the partial derivaties of f exist and are continuous on
Ω, and the Cauchy-Riemann equations are satisfied there. The converse is true as well:
Theorem. If the partial derivatives of f exist and are continuous on Ω, and the Cauchy-
Riemann equations are satisfied there, then f is analytic on Ω.
Proof. Let z = x+iy ∈ Ω. We must show that f(cid:48)(z) exists. Let r be small enough that
the disk of radius r around z belongs to Ω and choose h = s + it with 0 < |s + it| < r.
Then, by the mean value theorem,
f(z +h)−f(z) f(x+s,y +t)−f(x,y)
=
h s+it
f(x+s,y +t)−f(x,y +t) s f(x,y +t)−f(x,y) t
= +
s s+it t s+it
∂f s ∂f t
∗ ∗
= (x+s ,y +t) + (x,y +t ) ,
∂x s+it ∂y s+it
8
where |s∗| < s and |t∗| < t. Note that
∂f ∂f
(x+s∗,y +t)− (x,y) → 0
∂x ∂x
as h → 0, and similarly for the second partial derivative. Moreover |s/(s + it)| stays
bounded (by 1). Thus
f(z +h)−f(z) ∂f s ∂f t
= (x,y) + (x,y) +R(h),
h ∂x s+it ∂y s+it
where lim R(h) = 0. Now if the Cauchy–Riemann equations hold, then the the first
h→0
two terms on the right hand side sum to ∂f/∂x(x,y), which is independent of h, so the
limit exists and is equal to ∂f/∂x(x,y). (cid:3)
Remark. The theorem can be weakened to say that if f is continuous on Ω and the partial
derivatives exist and satisfy the Cauchy–Riemann equations there (without assuming that
the partial derivatives are continuous), then the complex derivative of f exists on Ω (which
is equivalent to f being analytic on Ω. This is the Looman–Menchoff Theorem. See
Narasimhan,Complex Analysis in One Variable,foraproof. Wedoneedatleastcontinuity,
since otherwise we could take f to be the characteristic function of the coordinate axes.
Note that
∂ ∂ ∂ ∂ 1
= = ∆.
∂z ∂z¯ ∂z¯∂z 4
This shows that any analytic function is harmonic (equivalently, its real and imaginary
parts are harmonic). It also shows that the conjugate of an analytic function, while not
analytic, is harmonic.
If u(x,y) = Ref(x+iy), then u is harmonic. Conversely, if u is harmonic on a region
Ω, does there exist a harmonic conjugate, i.e., a function v on Ω such that f = u+iv is
analytic? Clearly, if there exists v it is determined up to addition of a real constant (or
f is determined up to addition of an imaginary constant). The book gives an elementary
construction of the harmonic conjugate in a disk and in the whole plane. We sketch the
idea of a more general proof. We require that the domain satisfy the condition that if γ
is any piecewise simple smooth closed curve in Ω, then γ is the boundary of a subset G of
Ω. In other words, Ω has no holes (is simply-connected). Our proof follows directly from
the following result, which holds on simply-connected domains: let f : Ω → R2 be a C1
vectorfield. Then there exists v : Ω → R such that f = ∇v if and only if ∂f /∂y = ∂f /∂x.
1 2
The “only if” part is obvious. To prove the “if” part, fix a point (x ,y ) in Ω and for any
0 0
(x,y) ∈ Ω, let γ be a piecewise smooth path in Ω from (x ,y ) to (x,y), and let τ be
(x,y) 0 0
the unit tangent to the path pointing from (x ,y ) towards (x,y). Define
0 0
v(x,y) = (cid:90) f ·τ ds.
γ
(x,y)
9
It is essential that this quantity doesn’t depend on the choice of path, or, equivalently,
that
(cid:90) f ·τ ds = 0
γ
for all piecewise smooth simple closed paths γ. But
(cid:90) f ·τ ds = (cid:90) (f2,−f1)·nds = (cid:90)(cid:90) ∇·(f2,−f1)dxdy = 0,
γ γ
by the divergence theorem. It is easy to check that ∇v = f.
For a non-simply connected region, there may exist no harmonic conugate. E.g., if
u = log|z| on C\{0}, then u+iargz is analytic on C \{z ≤ 0}, but cannot be extended
to an analytic function on C\{0}.
M¨obius transformations.
Definition. Let a, b, c, d be complex numbers with ad (cid:54)= bc. Then the mapping
az +b
S(z) = , z ∈ C, z (cid:54)= −d/c,
cz +d
is a Mo¨bius transformation.
Remarks. 1) Note that if ad = bc the same expression would yield a constant. 2) The
coefficents aren’t unique, since we can multiply them all by any nonzero complex constant.
a b
To each MT we associate the nonsingular matrix (cid:18) (cid:19) of its coefficients, which is
c d
determined up to a non-zero multiple. 3) The linear (but non-constant) polynomials are
MTs, namely the ones for which c = 0. 4) If z (cid:54)= −d/c then S(z) ∈ C and S(cid:48)(z) =
(ad − bc)/(cz + d)2 (cid:54)= 0. 5) We can define S(−d/c) = ∞ and S(∞) = a/c, so S can be
viewed of as a map from C into itself (which is, as we will now see, 1-1 and onto).
∞
If S and T are MTs, then so is S ◦ T, its coefficient matrix being the product of the
coefficient matrices of S and T. Any MTs admits an inverse, namely the transform with
the inverse coefficientmatrix. Withthe understandingthatS(∞) = a/candS(−d/c) = ∞
(or, if c = 0, then S(∞) = ∞, the MT maps C 1-1 onto itself. This is evident from the
∞
existence of its inverse S−1(z) = (dz−b)/(−cz+a). Moreover, the composition of M¨obius
transformations is again one, so they form a group under composition. If S is a non-linear
MT we can write it as
az +b b−ad
= a+ .
z +d z +d
This shows that any MT can be written as a composition of translations, rotations and
dilations (multiplication by a complex number), and inversion (reciprocals).
Since the geometric action of translation, dilation, and rotation are quite evident, let’s
consider inversion. It takes the unit disk D to C \D¯. We now show that inversion maps
∞
lines and circles to other lines and circles.
10
Consider the equation |z −p| = k|z −q| where k > 0, p,q ∈ C. This is
(x−a)2 +(y −b)2 = k2[(x−c)2 +(y −d)2]
which is clearly the equation of a circle if k (cid:54)= 1 and the equation of a line if k = 1. In either
case we have a circle in the Riemann sphere, which we call C . Substituting 1/z for z and
1
doing some simple manipulations gives the equation |z −1/p| = (k|q|/|p|)|z −1/q|, which
is another such circle, C . Thus we have 1/z ∈ C ⇐⇒ z ∈ C , showing that inversion
2 1 2
maps circles in the sphere onto other circles in the sphere. Since the same property is
evident for translation, dilation, and rotation, we conclude that this property holds for all
MT.
Some further analysis, which will be omitted, establishes two further properties. 1) A
MT is orientiation preserving in the sense that, if we traverse a circle in the order of three
distinct points on it, z , z , z , the region to the left of the circle will map to the region
1 2 3
to the left of the image circle, with respect to the image orientation. 2) A MT preserves
the property of two points being symmetric with respect to a circle, i.e., lying on the same
ray from the center, and such that the geometric mean of their distances from the center
equals the radius.
Let z , z , z be distinct points in C. Then it is easy to see that there is a unique MT
2 3 4
taking these points to 1, 0, and ∞, respectively, namely
z −z z −z
3 2 4
Sz = .
z −z z −z
4 2 3
We can also handle, as a special case, the possibility that one of the z is ∞. We infer
i
from this and the invertibility of the MTs that given two ordered triples of distinct points
in C , there is a unique MT taking the first onto the second.
∞
If we have any two disks or halfplanes, or one of each, we can map one to the other by
a MT, and can arrange that the image of any three points on the boundary of one go to
three points on the boundary of the other.
The book makes heavy use of the the notation (z ,z ,z ,z ) (cross ratio) for the image
1 2 3 4
of z under the transformation taking z , z , z to 1, 0, and ∞, respectively, so
1 2 3 4
z −z z −z
1 3 2 4
(z ,z ,z ,z ) = .
1 2 3 4 z −z z −z
1 4 2 3
From the definition, (Tz ,Tz ,Tz ,Tz ) = (z ,z ,z ,z ) for any MT T.
1 2 3 4 1 2 3 4
Example: z (cid:55)→ (1−z)/(1+z) takes the right half plane onto the unit disk.
III. Complex Integration and Applications to Analytic Functions
If γ : [a,b] → C is piecewise smooth, and f is a continuous complex-valued function
defined on the image of γ, define
b
(cid:48)
(cid:90) f(z)dz = (cid:90) f(γ(t))γ (t)dt.
γ a