Course of complex analysis

38 Pages·1997·0.3 MB·English

by  Arnold D.N.

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
1Theselecturenoteswerepreparedfortheinstructor’spersonaluseinteachingahalf-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).
UnderthisdefinitionR2 becomesafield,denotedC. Notethat(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 ). The latter
1 1 2 2 1 2 1 2 1 2
formula just encapsulates the formula for the sine and cosine of a sum, and gives
argz z = argz +argz . Inparticular, ircisθ = rcis(θ+π/2), somultiplicationby
1 2 1 2
i is 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 corre-
spondence
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
n=0
∞ zn conver(cid:80)ges (to 1/(1 − z)) if |z| < 1. It clearly diverges, in fact its terms
n=0
(cid:80)become unbounded, if |z| > 1.
Weierstrass M-Test. Let M ,M ,... be positive numbers with M < ∞ and
0 1 n
suppose that f : X → C are functions on some set X satisfying s(cid:80)up |f (x)| ≤
n x∈X n
∞
M . Then f (x) is absolutely and uniformly convergent.
n 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
n=0 n
all |z − a| < R and it converges absolute(cid:80)ly and uniformly 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
n
hence the series is 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 =
r
{|z −a| ≤ r}. Choose r˜∈ (r,R). Then, 1/r˜> limsup|a |1/n, so |a |1/n < 1/r˜ for
n n
all n sufficiently 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,
0 1 n n+1
then 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| <
n n+1 n n+1 n
|a zn+1|. Thus the series a zn has terms of increasing magnitude, and so
n+1 n
cannot be convergent. Thus |z(cid:80)| ≥ R. This shows that lim|a /a | ≥ R.
n n+1
Similarly, suppose that z < lim|a /a |. Then for all n sufficiently large
n n+1
|a | > |a z| and |a zn| > |a zn+1|. Thus the series has terms of decreas-
n n+1 n n+1
ing magnitude, and so, by the previous theorem, |z| ≤ R. This shows that
lim|a /a | ≤ R. (cid:3)
n n+1
Remark. On the circle of convergence, many different behaviors are possible. zn
divergesforall|z| = 1. zn/ndivergesforz = 1, elseconverges, butnotabsol(cid:80)utely
(this follows from the fa(cid:80)ct that the partial sums of zn are bounded for z (cid:54)= 1 and
1/n ↓ 0). zn/n2 converges absolutely on |z| ≤ 1.(cid:80)Sierpinski gave a (complicated)
(cid:80)
example of a 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
n=0 n
its formal derivative (cid:80)∞ na (z−a)n−1 = ∞ (n+1)a (z−a)n. Now clearly
n=1 n n=0 n+1
a (z −a)n h(cid:80)as the same radius of c(cid:80)onvergence as ∞ a (z −a)n since
n=0 n+1 n=0 n
(cid:80) (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 =
n+1
limsup|a |1/n = 1/R. Now lim(n + 1)1/n = 1 as is easily seen by taking logs.
n
Moreover, it is easy to see that if limsupb = b and limc = c > 0, then
n n
limsupb c = bc. Thus limsup|(n+1)a |1/n = 1/R. This shows that the formal
n n n+1
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 (as-
sumes 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 continous at f(a), and g(f(z)) = z for all
(cid:48) (cid:48)
z in a neighborhood of a. Then if g (f(a)) exists and is non-zero, then f (a) exists
(cid:48)
and equals 1/g (f(a)).
5
Definition. Let f be a complex-valued function defined on an open set G in C.
(cid:48)
Then f 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.
We now show that a power series is differentiable at every point in its disk
of convergence 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
n n
radius of convergence is R and that |z | < R. We mu(cid:80)st show that for any (cid:15) > 0,
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 sufficie(cid:12)ntly close to z , whe(cid:12)re g(z) = ∞ na zn−1. Let
0 n=1 n
s (z) = N a zn, R (z) = ∞ a zn. Then (cid:80)
N 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 (z)−(cid:12)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)
(cid:12) (cid:12)
(cid:48)
Now s (z ) is just a partial sum for g(z ), so for N sufficiently large (and all z),
N 0 0
T ≤ (cid:15)/3. Also,
2
∞
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)
(cid:12) (cid:12)
Since a nrn−1 is convergent, 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
3
the differentiability of the polynomial s shows that T ≤ (cid:15)/3 for all z sufficiently
N 1
close to z .
0
We thus know that if f(z) = a zn, then, within the disk of convergence,
n
f(cid:48)(z) = na zn−1, and by indu(cid:80)ction, f(cid:48)(cid:48)(z) = n(n − 1)a zn−2, etc. Thus
n n
(cid:80) (cid:48) (cid:48)(cid:48) (cid:48)(cid:48)(cid:48) (cid:80)
a = f(0), a = f (0), a = f (0)/2, a = f (0)/3!, etc. This shows that any
0 1 2 3
convergent power series is the sum 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.
Note that different branches of the logarithm assign values to logz that differ
by addition 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.
Define a path in the complex plane as a continuous map γ : [a,b] → C. If
γ(cid:48)(t) ∈ C exists 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)
(cid:48)
curve and the horizontal. (Note: if γ (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 under f, e.g., of
ρ(t) = f(γ(t)). Then ρ(cid:48)(a) = f(cid:48)(z)γ(cid:48)(a), ρ˜(cid:48)(a˜) = f(cid:48)(z)γ˜(cid:48)(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
(cid:48)
of a point z such a map behaves to first order like multiplication by f (z ). Since
0 0
(cid:48)
f (z ) = rexpiθ, near z f simply behave like rotation by θ followed by dilation.
0 0
By contrast, a 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 contin-
uous 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
h→0
first 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, for a
proof. We do need at least continuity, 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. The “only
1 2
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
(x,y) 0 0
let τ be 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)
It is essential that this quantity doesn’t depend on the choice of path, or, equiva-
lently, that
(cid:90) f ·τ ds = 0
γ
9
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
a b
complex constant. To each MT we associate the nonsingular matrix (cid:18) (cid:19) of its
c d
coefficients, which is 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).
IfS andT areMTs, thensoisS◦T, itscoefficientmatrixbeingtheproductofthe
coefficient matrices of S and T. Any MTs admits an inverse, namely the transform
with the inverse coefficient matrix. With the understanding that S(∞) = a/c and
S(−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).
Sincethegeometricactionoftranslation, dilation, androtationarequiteevident,
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.
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 .
1
10
Substituting 1/z for z and 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
2
1/z ∈ C ⇐⇒ z ∈ C , showing that inversion maps circles in the sphere onto
1 2
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
1 2 3
will map to the region 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
2 3 4
MT 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
i
infer 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
1 2 3 4
image of z under the transformation taking z , z , z to 1, 0, and ∞, respectively,
1 2 3 4
so
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
This is simply the line integral of f along γ. It can be defined analogously to the
Riemann integral as the limit of sums of the form (f ◦γ)(τ )[γ(t )−γ(t )], so
k k k−1
is the Riemann–Stieltjes integral of f ◦ γ with res(cid:80)pect to γ. Using this definition
it can be extended to rectifiable paths, i.e., ones for which γ is only of bounded
variation.
If φ is a piecewise smooth increasing function from [c,d] onto [a,b] and we let
ρ = γ ◦φ (so ρ is the same path as γ but with a different parametrization), then
f(z)dz = f(z)dz.
(cid:82)ρ (cid:82)γ