Complex Analysis T.K. Subrahmonian Moothathu Contents 1 Basic properties of C 2 2 Holomorphic functions, and the branches of logarithm 7 3 Power series 12 4 Mobius maps 17 5 Integration of a continuous function along a path 23 6 Cauchy’s integral formula and power series representation 28 7 Liouville’s theorem and Zeroes theorem 34 8 Some more fundamental theorems about holomorphic functions 38 9 Winding number and Cauchy’s general theory 43 10 Singularities 49 11 Residues 54 12 More on zeroes and poles 56 13 The automorphism group 60 14 Harmonic functions 66 This is just a first course in Complex Analysis; to master some of the deeper theorems in the subject, the student should study an advanced course in Complex Analysis also. We will study 1
Transcript
Complex Analysis
T.K. Subrahmonian Moothathu
Contents
1 Basic properties of C 2
2 Holomorphic functions, and the branches of logarithm 7
3 Power series 12
4 Mobius maps 17
5 Integration of a continuous function along a path 23
6 Cauchy’s integral formula and power series representation 28
7 Liouville’s theorem and Zeroes theorem 34
8 Some more fundamental theorems about holomorphic functions 38
9 Winding number and Cauchy’s general theory 43
10 Singularities 49
11 Residues 54
12 More on zeroes and poles 56
13 The automorphism group 60
14 Harmonic functions 66
This is just a first course in Complex Analysis; to master some of the deeper theorems in the
subject, the student should study an advanced course in Complex Analysis also. We will study
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differentiable functions f : U → C, where U ⊂ C is a connected open set. Such functions are called
holomorphic or analytic functions. The notion of differentiability in Complex Analysis is so strong
that it imposes many rigid properties on the function f . Some such properties are the following.
Assume f : U → C is holomorphic, where U ⊂ C is a connected open set. Then:
(i) f is differentiable infinitely often.
(ii) Locally f is a combination of a stretching/shrinking and a rotation, and locally f has a power
series expansion. Moreover, f preserves angles provided f ′ is non-vanishing in U .
(iii) Any holomorphic g : U → C agreeing with f on an uncountable subset of U must agree with
f on the whole of U .
(iv) If f is non-constant, then f is an open map, and moreover, the maximum of |f | on any
nonempty compact set K ⊂ U is attained only on the boundary of K.
(v) If U has no ‘holes’, then for any two points z1, z2 ∈ U and any two smooth paths α, β in U
from z1 to z2, we have that∫α f =
∫β f .
(vi) If U = C, then f behaves somewhat like a polynomial. For instance, if f : C → C is holomorphic
and non-constant, then f(C) must contain all complex numbers with one possible exception.
Another aspect of the theory is the study of singularities. Let U ⊂ C be a connected open set
a ∈ U and f : U \ {a} → C be holomorphic. Then a is called a singularity of f . In some cases, f
or (z − a)kf (for some k ∈ N) will have a holomorphic extension to U . Otherwise, a is called an
essential singularity of f , and in this case the behavior of f near a will be quite wild: the f -image
of any deleted neighborhood of a will be dense in C. We also remark that many proofs in Complex
Analysis makes use of the following fact:∫|z−a|=r
dz
z − a= 2πi = 0 for a ∈ C and r > 0.
Some books for further reading:
1. D. Alpay, A Complex Analysis Problem Book.
2. J.B. Conway, Functions of One Complex Variable.
3. T.W. Gamelin, Complex Analysis.
4. T. Needham, Visual Complex Analysis.
5. R. Remmert, Theory of Complex Functions.
1 Basic properties of C
Let us recall the preliminary notions. If z = x + iy ∈ C, where x, y ∈ R, then x and y are called
respectively the real part and imaginary part of z, and we write Re(z) = x and Im(z) = y. The
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conjugate of z is z = x− iy and the absolute value of z is |z| =√x2 + y2. If z = 0, we may write
z in polar coordinates as z = reiθ, where r = |z| > 0 and θ ∈ [0, 2π) (or θ ∈ [−π, π)); here θ is
called the argument of z and we write arg(z) = θ; note that x = r cos θ and y = r sin θ. Also recall
how the addition and multiplication of two complex numbers are defined: for a + ib, c + id ∈ C,
we have that (a+ ib) + (c + id) = (a+ c) + i(b + d) and (a+ ib)(c + id) = (ac− bd) + i(ad+ bc).
The geometric meaning of complex multiplication is better understood in polar coordinates: if
z = r1eiθ1 and w = r2e
iθ2 , then zw = r1r2ei(θ1+θ2). There is scaling as well as rotation involved in
complex multiplication. In particular, we have that zn = rneinθ if z = reiθ, and iz is the complex
number obtained by rotating z by an angle π/2 in the anticlockwise direction since i = eiπ/2.
Exercise-1: For z, w ∈ C, we have that:
(i) z = z, |z| = |z|, z + w = z + w, and zw = z w.
(ii) Re(z) = (z + z)/2, Im(z) = (z − z)/2i, and |z|2 = zz.
Since f is real differentiable, it suffices to check Cauchy-Riemann equations. (i) {−i/2, 2/3 + i/6},
(ii) {0}, (iii) {x+ iy : x = 0 or y = 0}.]
Exercise-22: Let U ⊂ C be a connected open set and f ∈ H(U). If either f(U) ⊂ R or f(U) ⊂ iR,
then f is a constant. [Hint : If f = u+ iv, then v ≡ 0 or u ≡ 0. Use Cauchy-Riemann equations.]
Exercise-23: [A geometric insight] Let U ⊂ C be open and f ∈ H(U). Think of f as a map from
U ⊂ R2 to R2, and let Jf (z) be the Jacobian of f at z. Then,
(i) det(Jf (z)) = |f ′(z)|2.
(ii) If U is bounded, and f is bounded and injective, then Area(f(U)) =∫ ∫
U |f ′(z)|2dxdy.
[Hint : (i) Use Exercise-20 after noting that Jf (z) =
ux(z) uy(z)
vx(z) vy(z)
=
ux(z) −vx(z)
vx(z) ux(z)
by
Cauchy-Riemann equations. (ii) Area(f(U)) =∫ ∫
U det(Jf (z))dxdy by Multivariabele Calculus.]
[105] The (complex) exponential function e : C → C is defined as e(x + iy) = ex(cos y + i sin y).
Then, (i) e ∈ H(C), ez = ez, anddez
dz= ez.
(ii) e is a surjective group homomorphism from (C,+) onto (C \ {0}, ·) with ker(e) = 2πiZ. In
particular, ez+w = ezew and e−z = 1/ez.
(iii) For each b ∈ R, the exponential function is injective on {z ∈ C : b ≤ Im(z) < b+ 2π}.
Proof. (i) Let u = Re(e) and v = Im(e). Check that the partial derivatives of u and v are
continuous and Cauchy-Riemann equations hold. Use [103](ii) to conclude e is holomorphic. It is
easy to see ez = ez. To showdez
dz= ez, use Exercise-20.
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Statement (ii) is left to the student. For (iii), note that if |Im(z)− Im(w)| < 2π and ez = ew,
then z − w ∈ ker(e) = 2πiZ and |Im(z − w)| < 2π, and hence z = w.
Remark: (i) For c ∈ R, the vertical line x = c in C is wound around the circle |z| = ec infinitely
many times with period 2π by the exponential function. Hence, the left half-plane {x+ iy : x < 0}
is mapped onto the bounded set D \ {0} by e. (ii) The real exponential function x 7→ ex from R to
R is injective with range (0,∞). So for every y ∈ (0,∞), we have defined log y (in Real Analysis)
as the unique x ∈ R satisfying ex = y. Since the complex exponential function is not injective, it is
not possible to define complex log function in a unique way. Instead, we have a countably infinite
collection of complex log functions; they are called the branches of logarithm (defined below).
Definition: Let U ⊂ C be a connected open set. A continuous function f : U → C is called a branch
of logarithm on U if ef(z) = z for every z ∈ U . Note that a branch of logarithm is always injective.
It may happen that a connected open set U ⊂ C does not admit any branch of logarithm. For
example, if 0 ∈ U , then U cannot admit a branch of logarithm because the exponential function
never takes the value 0. Thus there is no branch of logarithm on C and D. It is of interest to find
necessary and sufficient conditions for a connected open set to admit a branch of logarithm.
[106] Let U ⊂ C be a connected open set and f : U → C be a branch of logarithm. Then,
(i) f ∈ H(U) and f ′(a) = 1/a for every a ∈ U .
(ii) A continuous function g : U → C is a branch of logarithm iff there is k ∈ Z such that
g(z) = f(z) + 2πik for every z ∈ U .
Proof. (i) limw→0ef(a+w) − ef(a)
f(a+ w)− f(a)=dez
dz|z=f(a) = ef(a) = a = 0 since f is continuous and injective,
and since 0 /∈ U . Hence 1/a = limw→0f(a+ w)− f(a)
ef(a+w) − ef(a)= limw→0
f(a+ w)− f(a)
w.
(ii) ⇒: (g − f)(U) ⊂ ker(e) = 2πiZ. Moreover, (g − f)(U) must be connected since f, g are
continuous and U is connected. Therefore g − f ≡ 2πik for some k ∈ Z.
[107] Let U ⊂ C be a connected open set containing the unit circle. Then U does not admit a
branch of logarithm. In particular, C \ {0} does not admit a branch of logarithm.
Proof. Suppose f : U → C is a branch of logarithm, i.e., f is continous and ef(z) = z for every z ∈ U .
For each eiθ ∈ ∂D ⊂ U , we have ef(eiθ) = eiθ and hence there is kθ ∈ Z with f(eiθ) = i(θ + 2πkθ).
Since f is uniformly continuous on the compact set ∂D, there is δ ∈ (0, π) such that |f(z)−f(w)| < π
for every z, w ∈ ∂D with |z−w| < δ. This implies kθ1 = kθ2 if |eiθ1 − eiθ2 | < δ. Consequently, there
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is a unique k ∈ Z such that f(eiθ) = i(θ + 2πk) for every θ ∈ [0, 2π). Let θ ∈ (π, 2π) be near to 2π
so that |1− eiθ| < δ. Then π > |f(1)− f(eiθ)| = |i(2πk)− i(θ+2πk)| = θ > π, a contradiction.
[108] (i) C \ [0,∞) and C \ (−∞, 0] admit branches of logarithm.
(ii) If B ⊂ C is an open ball with 0 /∈ B, then B admits a branch of logarithm.
Proof. Any ball B ⊂ C \ {0} is included in either C \ [0,∞) or C \ (−∞, 0]. Hence it suffices to
prove (i). If z ∈ C \ [0,∞), then |z| > 0 and there is a unique θ ∈ (0, 2π) with z = |z|eiθ. Define
f : C \ [0,∞) → C as f(z) = log |z|+ iθ. It is easy to verify that f is continuous and ef(z) = z for
every z ∈ C\ [0,∞). Similarly, if z ∈ C\(−∞, 0], then there is a unique θ ∈ (−π, π) with z = |z|eiθ,
and therefore f : C \ (−∞, 0] → C defined as f(z) = log |z|+ iθ is a branch of logarithm.
Remark: Let f : C \ (−∞, 0] → C be f(z) = log |z|+ i arg(z), where arg(z) ∈ (−π, π). We know f
is a branch of logarithm. Moreover, f agrees with the real logarithm on (0,∞). Hence f is often
called the principal branch of logarithm, and is sometimes denoted by the symbol ‘log’. That is,
log z := log |z| + i arg(z), z ∈ C \ (−∞, 0]. It must be kept in mind that this is only a convenient
notation, and there are infinitely many branches of logarithm.
If a, b ∈ (0,∞), then ab = eb log a. We may generalize this using branches of logarithm to define
ab, when a, b are complex numbers with a = 0. It must be noted that since we use branches of
logarithm, in general there in no unique value for ab for a ∈ C \ {0} and b ∈ C.
Definition: Let U ⊂ C be a connected open set and f : U → C be a branch of logarithm. If a ∈ U
and b ∈ C, then ebf(a) is called a branch of ab.
Example: Let U = C \ (−∞, 0] and n ≥ 2 be an integer. Then z1/n is not a unique-valued
function in U ; on the other hand, z1/n has exactly n distinct holomorphic branches in U ; they are
ef0/n, ef1/n, . . . , efn−1/n, where fk(z) = log |z|+ i(arg(z) + 2πk) for z ∈ U for 0 ≤ k ≤ n− 1.
Remark: If a ∈ C \ [0,∞) and b ∈ C \ Z, then we may see from the definition that ab will have
at least two branches and therefore ab is not defined as a unique complex number. In particular,
the identity (ab)c = abc is not true in general for complex numbers a, b, c. If this is not understood
properly, one may get into absurdities. Here is a false proof showing that every integer is zero: if
k ∈ Z, then e2πk = (e2πik)−i = 1−i = 1, and hence 2πk = 0, which implies k = 0.
Definition: Motivated by the identities cos y = (eiy + e−iy)/2 and sin y = (eiy − e−iy)/2i, we
define cos z = (eiz + e−iz)/2 and sin z = (eiz − e−iz)/2i for z ∈ C. Since the exponential function
is holomorphic, it follows that cos z and sin z are holomorphic. It may be verified that (cos z)′ =
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− sin z, (sin z)′ = cos z, and cos(z+2πk) = cos z and sin(z+2πk) = sin z for every k ∈ Z. Moreover,
standard trigonometric identities such as cos2 z + sin2 z = 1 are true.
Exercise-24: The functions cos z and sin z are surjective on C. [Hint : Consider w ∈ C. We have
cos z = w iff u2 − 2wu+1 = 0, where u = eiz. Since the constant term is non-zero, the roots of the
quadratic polynomial are non-zero and hence are of the form eiz.]
Remark: (i) By Exercise-24, cos z and sin z are unbounded (unlike the real functions cosx and
sinx); this can be seen directly also: | cos(in)| → ∞ and | sin(in)| → ∞ as n → ∞. (ii) Let c > 0,
L = {z ∈ C : Im(z) = c}, a = (ec + e−c)/2, and b = (ec − e−c)/2. It may be seen that cos z
maps the line L onto the ellipse {w ∈ C :[Re(w)]2
a2+
[Im(w)]2
b2= 1}, winding L around the ellipse
infinitely many times in a periodic manner.
Seminar topic: Further analysis about complex trigonometric functions.
3 Power series
We will see a little later that all holomorphic functions can be locally expressed as a power series.
Because of this, here we will look at some of the basic properties of a complex power series.
Definition: Let X be a set, fn : X → C be functions for n ∈ N and W ⊂ X. We say that
(i) the series∑∞
n=1 fn converges absolutely on W if∑∞
n=1 |fn(w)| <∞ for each w ∈W ,
(ii) the sequence (fn) converges uniformly on W to a function f :W → C if for every ε > 0, there
is n0 such that sup{|f(w)− fn(w)| : w ∈W} < ε for every n ≥ n0,
(iii) the series∑∞
n=1 fn converges uniformly on W to a function f :W → C if for sn := f1+ · · ·+fn,
the sequence (sn) converges uniformly on W to f .
Exercise-25: (i) [Weierstrass’ M-test] Let X be a set, fn : X → C be functions for n ∈ N and
W ⊂ X. Suppose there are Mn > 0 such that∑∞
n=1Mn <∞ and sup{|fn(w)| : w ∈W} ≤Mn for
each n ∈ N. Then∑∞
n=1 fn converges absolutely and uniformly on W , where the uniform limit is
the function f : W → C given by f(w) =∑∞
n=1 fn(w) for w ∈ W . Moreover, for any permutation
σ of N, the series∑∞
n=1 fσ(n) also converges uniformly on W to the original limit function.
(ii) [Special case] Let∑∞
n=0 an(z− a)n be a formal power series in C, where a, an ∈ C. Let W ⊂ C,
and Mn > 0 be such that∑∞
n=1Mn < ∞ and sup{|an||z − a|n : z ∈ W} ≤ Mn for each n ≥ 0.
Then the series∑∞
n=0 an(z − a)n converges absolutely and uniformly on W .
[Hint : (i) Recall from Real Analysis. To prove uniform convergence, let sn = f1 + · · · + fn, and
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note that sup{|sn+1(w)−sn(w)| : w ∈W} ≤Mn+1, which implies (sn) is uniformly Cauchy on W .]
Definition: The radius of convergence R of a formal power series∑∞
n=0 an(z − a)n in C is defined
as R = sup{r ≥ 0 : (anrn)∞n=0 is bounded} ∈ [0,∞]. For any R ∈ (0,∞), the series
∑∞n=0
zn
Rnhas
radius of convergence equal to R.
Let B(a, r) = {z ∈ C : |z − a| ≤ r} for r ∈ [0,∞].
[109] Let∑∞
n=0 an(z − a)n be a formal power series in C with radius of convergence R. Then,
(i) [Cauchy-Hadamard] R =1
lim supn→∞ |an|1/n(where 1/0 = ∞ and 1/∞ = 0 by convention).
(ii) R = limn→∞ | anan+1
|, provided the limit exists.
Proof. (i) Let L =1
lim supn→∞ |an|1/n. If 0 < r < L, then 1/r > lim supn→∞ |an|1/n, and hence
there is n0 ∈ N such that 1/r > |an|1/n, or equivalently |an|rn < 1 for every n ≥ n0. This means
r ≤ R, and hence L ≤ R. If L < r < ∞, choose t ∈ (L, r). Then 1/t < lim supn→∞ |an|1/n so
that 1/tn < |an| for infinitely many n ∈ N. Then (r/t)n < |an|rn for infinitely many n ∈ N, which
implies R < r and hence R ≤ L.
(ii) Suppose that the limit L := limn→∞ | anan+1
| exists. If 0 < r < L, then there is n0 ∈ N such that
|an/an+1| > r, or equivalently |an+1| < |an|/r for every n ≥ n0. Hence |an|rn ≤ |an0 |rn/rn−n0 =
|an0 |rn0 for every n ≥ n0, which shows r ≤ R and hence L ≤ R. If L < r < ∞, choose t ∈ (L, r).
Then there is n0 ∈ N such that |an/an+1| < t, or equivalently |an+1| > |an|/t for every n ≥ n0.
Hence |an|rn > |an0 |rn/tn−n0 = |an0 |tn0(r/t)n for every n ≥ n0. Since t < r, this means (anrn) is
unbounded and hence R < r. Therefore R ≤ L.
Exercise-26: Find the radius of convergence for the following series:
(i)∑∞
n=1
zn
nk, where k ∈ N ∪ {0}.
(ii)∑∞
n=1 zkn , where (kn) is a strictly increasing sequence in N ∪ {0}.
(iii)∑∞
n=0
zn
n!.
(iv)∑∞
n=0 n!zn.
(v)∑∞
n=0(sinn)zn.
[Hint : (i) R = limn→∞ | anan+1
| = limn→∞(1 + 1/n)k = 1. (ii) lim supn→∞ |an|1/n = 1 and hence
R ≥ 1. By considering z = 1, we see R = 1. (iii) R = limn→∞ |an/an+1| = limn→∞ |n + 1| = ∞.
(iv) R = limn→∞ |an/an+1| = 0. (v) Clearly, lim supn→∞ |an|1/n ≤ 1. Since the interval [π/4, 3π/4]
has length > 1, n (mod 2π) ∈ [π/4, 3π/4] for infinitely many n ∈ N and hence sinn ≥ 1/√2 for
infinitely many n ∈ N. Hence lim supn→∞ |an|1/n ≥ 1 also. So R = 1.]
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Remark: Since R = ∞ for Exercise-26(iii), we deduce limn→∞(1/n!)1/n = 0 by Cauchy-Hadamard.
[110] Let∑∞
n=0 an(z − a)n be a formal power series in C with radius of convergence R. Then,
(i) If z ∈ C and |z − a| > R, then∑∞
n=0 an(z − a)n does not converge.
(ii) If 0 ≤ r < R, then∑∞
n=0 an(z − a)n converges absolutely and uniformly on B(a, r).
(iii) We cannot say anything about the case where |z−a| = R (it depends on the particular series).
(iv) The radius of convergence of the series∑∞
n=1 nan(z − a)n−1 is also equal to R.
Proof. (i) Let sn(z) =∑n
k=0 ak(z − a)k. Assume |z − a| > R. Then (an|z − a|n) is unbounded
by the definition of R, and hence (sn(z)) should also be unbounded since |an+1(z − a)n+1| ≤
|sn+1(z)|+ |sn(z)|. Therefore (sn(z)) cannot converge in C.
(ii) If r = 0, there is nothing to prove. So assume 0 < r < R, and choose t ∈ (r,R). By the
definition of R, there is M > 0 such that supn≥0 |an|tn ≤ M . Then for Mn := M(r/t)n, we have
that∑∞
n=0Mn <∞ and |an|rn ≤Mn for every n ≥ 0. So Weierstrass’ M-test applies.
(iii) Check that the radius of convergence is equal to 1 for the following three series:∑∞
n=1
zn
n2,∑∞
n=0 zn, and
∑∞n=1
(−1)n−1zn
n. By Weierstrass’ M-test,
∑∞n=1
zn
n2converges for every z ∈ ∂D. The
series∑∞
n=0 zn does not converge for any z ∈ ∂D since |sn+1(z)−sn(z)| = 1, where sn(z) =
∑nk=0 z
k.
The series∑∞
n=1
(−1)n−1zn
nconverges for z = 1 but does not converge for z = −1.
(iv) Let R′ be the radius of convergence of∑∞
n=1 nan(z − a)n−1. If 0 < r < R′, then (nanrn−1)
is bounded, which implies (anrn) is bounded, and hence R′ ≤ R. If 0 < r < R, choose t ∈ (r,R).
Then M := supn≥0 |an|tn < ∞. Now n|an|rn−1 ≤ nMrn
rtn→ 0 as n → ∞ since r < t, and hence
(nanrn−1) is bounded. This shows r ≤ R′ and hence R ≤ R′.
Remark: If∑∞
n=0 an(z − a)n is a power series in C with radius of convergence R = ∞, then the
series converges uniformly in the compact ball B(a, r) for every r ∈ (0,∞), but the series may not
converge uniformly on the whole of C. Read [110](ii) carefully.
[111] Let∑∞
n=0 an(z − a)n be a power series in C with radius of convergence R > 0, and let
f(z) =∑∞
n=0 an(z − a)n be the limit function for |z − a| < R. Then, for |z − a| < R, we have:
(i) f is holomorphic and f ′(z) =∑∞
n=1 nan(z − a)n−1.
(ii) f is infinitely often differentiable and f (k)(z) =∑∞
n=k
n!
(n− k)!an(z − a)n−k for every k ∈ N.
(iii) ak =f (k)(a)
k!for every k ≥ 0 so that f(z) =
∑∞n=0
f (n)(a)
n!(z − a)n .
Proof. A repeated application of (i) and [110](iv) will give (ii); and (iii) can be deduced from (ii)
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by putting z = a in the expression for f (k)(z). Therefore it suffices to establish (i). Let g(z) =∑∞n=1 nan(z − a)n−1 for z ∈ B(a,R), where g is well-defined by [110](iv). Consider z0 ∈ B(a,R)
and ε > 0. We need to find δ > 0 such that |f(z)− f(z0)
z − z0− g(z0)| < ε for every z ∈ B(a,R) with
0 < |z − z0| < δ. Choose r > 0 such that |z0 − a| < r < R. Let sn(z) =∑n
k=0 ak(z − a)k and
Rn(z) = f(z)− sn(z) =∑∞
k=n+1 ak(z − a)k for z ∈ B(a,R). Since (s′n) → g uniformly on B(a, r),
there is n1 ∈ N such that |g(z)− s′n(z)| < ε/3 for every n ≥ n1 and z ∈ B(a, r).
Next note that for z ∈ B(a, r) \ {z0}, |Rn(z) − Rn(z0)| ≤∑∞
](a2, b, a1) = π/2. Hence by Exercise-33(i), T (C) is a circle in C for which [T (a1), T (a2)] is a
diameter (draw a picture to understand the argument clearly).
Exercise-34: [Convince yourself that the following is true, at least intuitively] If z1, z2, z3 ∈ C∞
are three distinct points, then there is a unique circle C in C∞ passing through z1, z2, z3. The
complement of C in C∞ consists of two connected open sets. We can call them the left side and
the right side of C if we fix an orientation [z1, z2, z3] for C (i.e., as we travel from z1 to z3 along
the arc in C containing z2). If T is any Mobius map, then the left side (respectively, the right side)
of C is mapped onto the left side (respectively, the right side) of T (C) in a bijective manner with
respect to the orientation [T (z1), T (z2), T (z3)] of the circle T (C). [In practice, we just evaluate T
at a suitable point in one side of C to deduce where this region will be mapped onto by T .]
Example: Suppose C1, C2 are two circles in C∞. Let z1, z2, z3 be three distinct points on C1, and
w1, w2, w3 be three distinct points of C2. By [113](v), there is a unique Mobius map T with T (zj) =
wj for j = 1, 2, 3. Then necessarily T (C1) = C2. By choosing zj ’s and wj ’s appropriately, we may
control which side of C1 is mapped to which side of C2 by T . A specific example is the following.
Let T (z) =(1 + i)(z + 1)
(−1 + i)(z − 1)=
−i(z + 1)
z − 1. Then T is a Mobius map with T (−1) = 0, T (−i) = 1
and T (1) = ∞. Hence T maps the unit circle ∂D onto the circle R ∪ {∞} in C∞. Since T (0) = i,
we deduce that T maps the open unit disc D onto the upper half-plane {w ∈ C : Im(w) > 0}.
Exercise-35: Let T = I be a Mobius map. Then,
(i) Fix(T ) := {z ∈ C∞ : T (z) = z} = {∞} iff T is a translation.
(ii) Fix(T ) = {0,∞} iff T is a dilation.
(iii) T (0) = ∞ and T (∞) = 0 iff T (z) = a/z for some a ∈ C \ {0}.
(iv) If S is a Mobius map, then Fix(S ◦ T ◦ S−1) = S(Fix(T )).
(v) |Fix(T )| = 1 iff T is conjugate to a translation.
(vi) |Fix(T )| = 2 iff T is conjugate to a dilation.
Exercise-36: Let S, T be Mobius maps different from aI, a ∈ C \ {0}. Then S ◦ T = T ◦ S iff
Fix(S) = Fix(T ). [Hint : First suppose Fix(T ) = {z1, z2}, where z1 = z2. If J is a Mobius map
with J(z1) = 0 and J(z2) = ∞, then replacing T with J ◦ T ◦ J−1 and S with J ◦ S ◦ J−1, we
may assume Fix(T ) = {0,∞}. Then Tz = az for some a ∈ C \ {0}. Now if Fix(S) = Fix(T ),
then S(z) = bz for some b ∈ C \ {0}, and hence S ◦ T = T ◦ S. Conversely, if S ◦ T = T ◦ S,
then S({0,∞}) = {0,∞} so that either S(z) = bz or S(z) = b/z; eliminate the second possibility
20
using S ◦ T = T ◦ S. Next suppose Fix(T ) is a singleton. After a conjugation, we may assume
Fix(T ) = {∞}, and then T (z) = z + b for some b ∈ C \ {0}. If Fix(S) = Fix(T ), then S is also
a translation and hence S ◦ T = T ◦ S. Conversely, if S ◦ T = T ◦ S, then S(∞) = ∞. Apply first
case to ensure that S has no other fixed points.]
Exercise-37: Let T (z) =az + b
cz + dbe a Mobius map. Then T (R ∪ {∞}) = R ∪ {∞} iff there is
p ∈ C \ {0} such that pa, pb, pc, pd ∈ R. [Hint : ⇒: First suppose T (∞) = ∞. Then c = 0 and
T (z) = (a/d)z + b/d. Since T (1), T (−1) ∈ R, we have a/d + b/d,−a/d + b/d ∈ R and therefore
a/d, b/d ∈ R. Take p = 1/d. If x = T (∞) ∈ R, then consider S(z) = 1/(T (z)− x), which fixes ∞,
and apply the first case.]
Exercise-38: Let U ⊂ C be a connected open set and f ∈ H(U). If f(U) ⊂ ∂D, then f is a constant.
[Hint : Each z ∈ U has a neighborhood B such that f(B) is a proper subset of ∂D. Since C is
second countable, we may write U =∪∞
n=1Bn, a countable union of open balls, such that f(Bn)
is a proper subset of ∂D for each n ∈ N. Choose Mobius maps Tn such that Tn(∂D) = R ∪ {∞}
and Tn(f(Bn)) ⊂ R. By Exercise-22, Tn ◦ f is a constant on Bn, and hence f is a constant on Bn
(since Tn is bijective). Thus f(U) =∪∞
n=1 f(Bn) must be a countable and connected subset of C,
and hence must be a singleton. Remark: The above reasoning depends on some topological facts.
After learning some more Complex Analysis, we can give an easier argument.]
Exercise-39: Let M be the collection of all Mobius maps. Then,
(i) (M, ◦) is a non-abelian group.
(ii) ϕ : SL(2,C) → M given by
a b
c d
7→ (z 7→ az + b
cz + d) is a surjective group homomorphism
with ker(ϕ) = {±I}, where SL(2,C) = {all 2× 2 complex matrices of determinant 1}.
(iii) Describe all abelian subgroups of M.
[Hint : (ii) From earlier theory, we know ϕ is a surjective group homomorphism. Consider T (z) =az + b
cz + d. If T (z) = z for every z ∈ C∞, then in particular 0, 1,∞ ∈ Fix(∞), and moreover ad− bc =
1. Deduce from this that b = c = 0, a = d and ad = 1; and conclude ker(ϕ) = {±I}. (iii) Let
Hw1,w2 = {T ∈ M : Fix(T ) = {w1, w2}} ∪ {I} for w1, w2 ∈ C∞ (where w1 = w2 is allowed). Then
H is an abelian subgroup of M; and moreover, any abelian subgroup H of M must be a subgroup
of some Hw1,w2 since Fix(S) = Fix(T ) for every S, T ∈ H \ {I} by Exercise-36. Also, Hw1,w2 is
conjugate to either H0,∞ = {all translations} ∼= (C,+) or H∞,∞ = {all dilations} ∼= (C \ {0}, ·).]
Exercise-40: Let N = {I} be a normal subgroup of M := {all Mobius maps}. Then,
(i) N must contain all translations and at least one dilation different from I.
21
(ii) z 7→ −1/z belongs to N .
(iii) We must have N = M.
[Hint : (i) Consider T ∈ N \ {I}. Then |Fix(T )| = 1 or = 2 by [113]. As N is normal, after a
conjugation we may assume that either Fix(T ) = {∞} or Fix(T ) = {0,∞}. In the first case, T
is a translation by Exercise-35, say T (z) = z + b. Conjugating with z 7→ az, we see z 7→ z + ab
belongs to N . Thus all translations belong to N . Conjugating z 7→ z + 1 with z 7→ 1/z, and then
composing with z 7→ z− 1, we see z 7→ −1/(z+1) belongs to N . This Mobius map has exactly two
fixed points, and hence is conjugate to a dilation by Exercise-35. Thus N contains a dilation. If
Fix(T ) = {0,∞}, then T is a dilation, say T (z) = az, a ∈ C\{0, 1}. Conjugating T with z 7→ z+1,
and then composing with T−1, we see the translation z 7→ z+(1− a)/a belongs to N , and then all
translations belong to N as above. (ii) z 7→ −1/(z+1) belongs to N and −1/z = −1/((z− 1)+1).
(iii) Consider T ∈ N \ {I}, T (z) = az, a = 0, 1, and consider b ∈ C \ {0}. Composing T with
z 7→ z + b, conjugating with z 7→ 1/z and then composing with z 7→ z − 1/b, we see z 7→ 1
b2z/a+ bbelongs to N . Composing this with z 7→ −1/z and z 7→ z + b, we get z 7→ −b2z/a belongs to
N . Thus all dilations belong to N . The inversion z 7→ 1/z has exactly two fixed points, hence is
conjugate to a dilation by Exercise-35, and therefore belongs to N . Finally apply [113](iii).]
Definition: Two open sets U, V ⊂ C are said to be holomorphically equivalent if there is a bijection
f : U → V such that both f and f−1 are holomorphic; and in this case, f is called a biholomorphism
from U to V . Note that being holomorphically equivalent is an equivalence relation on the collection
of all open subsets of C.
[115] Any two of the following open sets are holomorphically equivalent:
U1 = D (open unit ball),
U2 = {z ∈ C : Im(z) > 0} (upper half-plane),
U3 = {z ∈ C : Re(z) > 0} (right half-plane),
U4 = {z ∈ C : Im(z − a
b) > 0}, where a ∈ C and b ∈ C \ {0} (an open half-plane),
U5 = {z ∈ C : 0 < Im(z) < π} (a particular horizontal strip),
U6 = {z ∈ C : a < Im(z) < b}, where a < b are real (a horizontal strip),
U7 = {z ∈ C : a < Re(z) < b}, where a < b are real (a vertical strip),
U8 = C \ [0,∞),
U9 = C \ (−∞, 0].
Proof. Verify that each Ti,j below is a biholomorphism from Ui to Uj :
22
T1,2(z) =(1 + i)(z + 1)
(−1 + i)(z − 1)=
−i(z + 1)
z − 1, and
T2,1(z) = T−11,2 (z) =
dz − b
−cz + a=
−z + i
−z − i=z − i
z + i;
T2,3(z) = −iz and T3,2(z) = iz;
T2,4(z) = bz + a and T4,2(z) = (z − a)/b;
T2,5(z) = log |z|+ i arg(z) and T5,2(z) = ez;
T5,6(z) =(b− a)z
π+ ia and T6,5(z) =
π(z − ia)
b− a;
T6,7(z) = −iz and T7,6(z) = iz;
T2,8(z) = z2 and T8,2(z) = e(1/2)(log |z|+i arg(z));
T8,9(z) = −z and T9,8(z) = −z.
Remark: After learning more theory, we can show that no two of the following are holomorphically
equivalent: D, D \ {0}, {z ∈ C : 1 < |z| < 2}, C \ {0}, and C (keep this exercise in mind to be
solved later). However, we have the following:
Exercise-41: There is a surjective holomorphic function from D to each of the following: D \ {0},
{z ∈ C : 1 < |z| < 2}, C \ {0}, and C.
[Hint : D ∼= {z ∈ C : Re(z) < 0} ez−→ D \ {0}; D ∼= {z ∈ C : Im(z) > −1} z2−→ C ez−→ C \ {0};
and D ∼= {z ∈ C : 0 < Re(z) < log 2} ez−→ {z ∈ C : 1 < |z| < 2}.]
5 Integration of a continuous function along a path
Here, as a preparation for Cauchy’s integral theory, we will discuss the integration of complex-valued
continuous functions along paths. The general theory deals with the integration along ‘rectifiable
paths’. For simplicity, we will consider only piecewise smooth paths (defined below).
Definition: Let U ⊂ C be an open set. By a path (or curve) in U , we mean a continuous map
α : [a, b] → U , where a < b are reals. The image α([a, b]) of the path will be denoted as [α], which
is a compact subset of U . A path α : [a, b] → U is said to be (i) closed if α(a) = α(b), (ii) smooth
if α is continuously differentiable (i.e., if α′ is continuous), and (iii) piecewise smooth if there is a
partition a = a0 < a1 < · · · < ak = b of the domain [a, b] of α such that αj := α|[aj−1,aj ] is smooth
for 1 ≤ j ≤ k; and in this case we will use the formal expression α = α1 + · · ·+ αk.
Example: (i) For any k ∈ Z \ {0}, α : [0, 1] → C defined as α(t) = e2πikt is a smooth path and
its image [α] is the unit circle. (ii) A piecewise smooth path parametrizing the boundary of the
23
unit square in C in the anticlockwise direction is α : [0, 4] → C defined as α(t) = t for 0 ≤ t ≤ 1,
α(t) = 1 + i(t− 1) for 1 ≤ t ≤ 2, α(t) = (3− t) + i for 2 ≤ t ≤ 3, and α(t) = i(4− t) for 3 ≤ t ≤ 4.
Definition: (i) If g : [a, b] → C is continuous, we define∫ ba g(t)dt =
∫ ba Re(g(t))dt+ i
∫ ba Im(g(t))dt.
(ii) Let U ⊂ C be open and f : U → C be continuous. If α : [a, b] → U is a smooth path in U ,
then we define the integral of f along α (or over α) as∫α f =
∫α f(z)dz :=
∫ ba f(α(t))α
′(t)dt. This
definition is intuitively justified because if z = α(t), then dz = α′(t)dt. Moreover, the function
t 7→ f(α(t))α′(t) is continuous (and hence integrable as specified by (i)) because α is smooth. If
α is a piecewise smooth path in U , say α =∑k
j=1 αj , where each αj is smooth, then we define∫α f =
∑kj=1
∫αjf . (iii) If U ⊂ C is open, f : U → C is continuous, and α : [a, b] → U is a smooth
path, then∫α |f ||dz| :=
∫ ba |f(α(t))α′(t)|dt. Clearly, this definition implies |
∫α f | ≤
∫α |f ||dz|. If
f ≡ 1, then∫α |f ||dz| =
∫ ba |α′(t)|dt, which is by definition the length of α.
Examples: (i) Let U ⊂ C be open, f : U → C be continuous, and α : [a, b] → U be a smooth path.
If α is constant, then α′ ≡ 0; so∫α f = 0. If f is constant, say f ≡ c, then
∫α f = c(α(b) − α(a)).
(ii) Let α : [0, 1] → C \ {0} be α(t) = e2πit. Then∫α z
kdz = 2πi∫ 10 e
2πi(k+1)tdt = 0 ∀ k ∈ Z \ {−1}.
On the other hand,∫α
dz
z= 2πi
∫ 10 1 dt = 2πi. Similarly, if β : [0, 2π] → C \ {0} is β(t) = eit, then∫
β zkdz = 0 ∀ k ∈ Z \ {−1} and
∫β
dz
z= 2πi. More generally, if k,m ∈ Z, and α : [0, 1] → C \ {0}
is α(t) = e2πimt, then∫α z
kdz = 2πi if k = −1 and m = 0, and∫α z
kdz = 0 otherwise.
Remark: From the examples above, we observe that: (i) the integral of a holomorphic function
over a closed path need not be zero, even though a closed path has the same end points as that
of a constant path, and (ii) the integral of a holomorphic function over two paths with the same
image need not be equal.
Definition and remark: (i) Two smooth paths α : [a, b] → C and β : [c, d] → C are said to be
equivalent if there is an increasing bijection ϕ : [a, b] → [c, d] such that ϕ′ is continuous and α = β◦ϕ.
Note that if α, β are equivalent as defined above, then their images [α] and [β] are the same, and
moreover for any complex-valued f continuous in a neighborhood of [α] = [β], we have that∫α f =∫
β f because∫α f =
∫ ba f(α(t))α
′(t)dt =∫ ba f(β(ϕ(t)))β
′(ϕ(t))ϕ′(t)dt =∫ dc f(β(s))β
′(s)ds =∫β f .
(ii) The natural parametrization of the circle |z− a| = r is given by α : [0, 1] → C, α(t) = a+ re2πit
(or by β : [0, 2π] → C, β(t) = a + reit, which is equivalent to α). When we write∫|z−a|=r f , it
is understood that the integral is to be considered with respect to the natural parametrization of
the circle. Similarly, if w1, w2 ∈ C, then∫[w1,w2]
f is to be considered with respect to the natural
parametrization of the line segment, i.e., w.r.to α : [0, 1] → C given by α(t) = w1 + t(w2 − w1).
24
However, at this stage, it is meaningless to write∫ w2
w1f since there are many paths from w1 to w2
and the integrals over these paths need not be the same.
(iii) If α : [a, b] → C is a path, we denote by −α the path going in the ‘opposite direction’, i.e.,
−α : [a, b] → C is defined −α(a+ t(b−a)) = α(b+ t(a− b)) for t ∈ [0, 1]. If α is a piecewise smooth
path in an open set U ⊂ C, and f : U → C is continuous, verify that∫−α f = −
∫α f .
(iv) If α : [a, b] → C and β : [b, c] → C are paths with α(b) = β(b), then we define the path
α + β : [a, c] → C as (α + β)(t) = α(t) for a ≤ t ≤ b, and (α + β)(t) = β(t) for b ≤ t ≤ c. If α, β
are piecewise smooth paths in an open set U ⊂ C such that α + β is defined, and if f : U → C is
continuous, then clearly∫α+β f =
∫α f +
∫β f .
[116] Let a ∈ C, r > 0. Then∫|z−a|=r(z−a)
kdz = 0 for every k ∈ Z\{−1}, and∫|z−a|=r
dz
z − a= 2πi.
Proof. We have already noted this for the unit circle. Argue similarly.
Definition: Let U ⊂ C be open, and f : U → C be continuous. A function g ∈ H(U) is said to
be a primitive of f if g′ = f . The main advantage of having a primitive is the path independence
property stated in Exercise-42(ii) below.
Exercise-42: Let U ⊂ C be a connected open set, and f : U → C be continuous.
(i) If g1, g2 ∈ H(U) are primitives of f , then (g1 − g2)′ ≡ 0 and hence g1 − g2 is a constant.
(ii) [Path independence] Suppose f has a primitive g ∈ H(U) and let z1, z2 ∈ U . Then for any
piecewise smooth path α in U from z1 to z2, we have that∫α f = g(z2)− g(z1), and thus this value
is independent of the particular path α. This value is also independent of the particular choice of
a primitive g of f because of (i).
(iii) Let α be a piecewise smooth path in U . Suppose fn : U → C are continuous functions
converging to f uniformly on the image [α] of α. Then limn→∞∫α fn =
∫α f .
[Hint : (ii) Assume α : [a, b] → U is smooth. Then∫α f =
∫ ba f(α(t))α
′(t)dt =∫ ba g
′(α(t))α′(t)dt =∫ ba (g ◦α)
′(t)dt = g(α(b))− g(α(a)) = g(z2)− g(z1). (iii) We may assume α is not a constant so that
its length l(α) > 0. Given ε > 0, choose n0 ∈ N such that |f(z)− fn(z)| < ε/l(α) for every n ≥ n0
and z ∈ [α]. Then∫α |f − fn| ≤
∫ ba
ε|α′(t)|l(α)
dt = ε for every n ≥ n0.]
Definition: Let U ⊂ C be open and f : U → C be continuous. We say f is integrable if for
any two points z1, z2 ∈ U and any two piecewise smooth paths α, β in U from z1 to z2, we have
that∫α f =
∫β f (so that
∫ z2z1f(z)dz has a well-defined meaning). Note that f is integrable iff the
restriction of f to each connected component of U is integrable (since a path in U should be entirely
25
contained in a connected component of U).
Exercise-43: [Topological fact] If U ⊂ C is a nonempty connected open set, then U is polygonally
connected, i.e., for any two w, z ∈ U , there is a polygonal path in U from w to z. Here, a polygonal
path means a path whose image consists of finitely many line segments. [Hint : Fix w ∈ U , and let
A be the collection of all z ∈ U such that there is a polygonal path in U from w to z. Show that
A is both closed and open in U .]
[117] [Integrability criterion] Let U ⊂ C be open and f : U → C be continuous. Then the following
are equivalent: (i) f has a primitive g ∈ H(U).
(ii) f is integrable.
(iii)∫α f = 0 for every piecewise smooth closed path α in U .
Proof. (i) ⇒ (ii): We may suppose U is connected. For any two points z1, z2 ∈ U and any two
piecewise smooth paths α, β in U from z1 to z2, we have that∫α f = g(z2) − g(z1) =
∫β f by
Exercise-42(ii).
(ii) ⇒ (iii): Let α : [a, b] → U be a piecewise smooth closed path and β : [a, b] → U be the constant
path β ≡ α(a) = α(b). The integrability of f implies that∫α f =
∫β f = 0.
(iii) ⇒ (i): We may assume U is connected. Fix c ∈ U , and define g : U → C as follows: if p ∈ U ,
then g(p) :=∫α f , where α is any piecewise smooth path in U from c to p. Note that the existence
of such a path α is guaranteed by Exercise-43; and the value∫α f is independent of the particular
choice of a path because if β is another such path, then α− β is a piecewise smooth closed path in
U and hence 0 =∫α−β f =
∫α f −
∫β f by (iii). We claim that g is a primitive of f in U .
Consider p ∈ U and let ε > 0. Since U is open and f is continuous, we may choose δ > 0 such
that B(p, δ) ⊂ U and such that |f(p)− f(q)| < ε for every q ∈ B(p, δ). If h ∈ C is with 0 < |h| < δ,
let α, β be piecewise smooth paths in U from c to p and from c to p+h respectively. Then g(p) =∫α f
and g(p+h) =∫β f . The integral of f over the piecewise smooth closed path α+[p, p+h]−β is zero
by (iii), and hence g(p + h) − g(p) =∫β−α f =
∫[p,p+h] f(z)dz. Moreover, hf(p) =
∫[p,p+h] f(p)dz.
Therefore, for 0 < |h| < δ, we have that |g(p+ h)− g(p)− hf(p)| ≤∫[p,p+h] |f(z)− f(p)||dz| < ε|h|.
This shows that g is holomorphic at p with g′(p) = f(p).
Remark: In Real Analysis, every continuous function f : [a, b] → R is Riemann integrable, and has
a ‘primitive’; indeed, g : [a, b] → R defined as g(x) =∫ xa f(t)dt satisfies g
′ = f . Contrast this with
the following example in Complex Analysis. Let U = {z ∈ C : 1/2 < |z| < 2} and f : U → C be
26
f(z) = 1/z. Then, U is a bounded open set and f ∈ H(U) is bounded, but we note by [117] that
f is not integrable and f does not have a primitive in U because∫|z|=1 f = 2πi = 0.
Definition: Let U ⊂ C be open and f : U → C be continuous. We say f is locally integrable in U if
for each z ∈ U , there is δ > 0 such that B(z, δ) ⊂ U and the restriction of f to B(z, δ) is integrable.
Exercise-44: The holomorphic function f(z) = 1/z is locally integrable in any open subset U of
C \ {0}. Hence local integrability does not imply (global) integrability. [Hint : Given z ∈ U , we can
find δ > 0 with B(z, δ) ⊂ U ⊂ C \ {0}. There is a branch of logarithm g : B(z, δ) → C by [108](ii).
Since g′(z) = 1/z, g is a primitive of f in B(z, δ). Now apply [117].]
Remark: Let U ⊂ C be open and f : U → C be continuous. With more theory, it can be shown
that (i) f is locally integrable ⇔ f ∈ H(U), and (ii) if U is connected and has no ‘holes’, then f is
locally integrable ⇔ f is integrable.
Definition: An open set U ⊂ C is called a star region if there is c ∈ U such that [c, z] ⊂ U for every
z ∈ U (where [c, z] is the line segment joining c and z). Here c is called a center of U .
Example: Every convex open subset U of C is a star region, and in this case, every point of U is a
center of U . The sets C \ [0,∞) and C \ (−∞, 0] are not convex, but they are star regions: every
c ∈ (−∞, 0) is a center of C \ [0,∞), and every c ∈ (0,∞) is a center of C \ (−∞, 0]. Another
example of a star regions which is non-convex is {x+ iy : |x| < 1 and |xy| < 1}. On the other hand,
the following connected open sets are not star regions: D \ {0}, C \ {0}, {z ∈ C : 1 < |z| < 2}, and
{z ∈ C : Im(z) > 0 and |z| > 1}.
Remark: Being a star region is a geometric property. In general this property is not preserved
under topological or holomorphic equivalence. Let δ > 0 be very small, U be the open rectangle
with vertices 0, δ, δ+ iπ, iπ, and V be the image of U under ez. Then it can be verified that the star
region U is holomorphically equivalent to V via the exponential map, but V is not a star region
since V = {z ∈ C : 1 < |z| < eδ and Im(z) > 0} is the upper-half of a thin open annulus.
[118] Let U ⊂ C be a star region and f : U → C be continuous. If∫∂∆ f = 0 for every solid triangle
∆ ⊂ U , then f has a primitive g ∈ H(U) and consequently f is integrable (here the integral∫∂∆ f
is taken with respect to the natural parametrization of ∂∆ in the anticlockwise direction).
Proof. Let c ∈ U be a center of U . Define g : U → C as g(p) =∫[c,p] f(z)dz. Consider p ∈ U and
ε > 0. Choose δ > 0 such that B(p, δ) ⊂ U and such that |f(p) − f(q)| < ε for every q ∈ B(p, δ).
For h ∈ C with 0 < |h| < δ, note that ∆ := [c, p, p + h] ⊂ U and hence∫∂∆ f = 0 by hypothesis,
27
which implies that g(p+ h)− g(p) =∫[p,p+h] f(z)dz; moreover, hf(p) =
∫[p,p+h] f(p)dz. Therefore,
for h ∈ C with 0 < |h| < δ, we see that |g(p+ h)− g(p)− hf(p)| ≤∫[p,p+h] |f(z)− f(p)||dz| < ε|h|,
and this shows g′(p) = f(p). Thus g is a primitive of f , and then f is integrable by [117].
6 Cauchy’s integral formula and power series representation
The relevance of a star region in the context of holomorphic functions can be first observed by
noting that the hypothesis of [118] is satisfied by holomorphic functions:
[119] Let U ⊂ C be open and f ∈ H(U). Then,
(i) [Goursat’s integral lemma]∫∂∆ f = 0 for every solid triangle ∆ contained in U .
(ii) [Cauchy’s integral theorem for star regions] If U is a star region, then f has a primitive in U ,
f is integrable, and∫α f = 0 for every piecewise smooth closed path α in U .
Proof. Since (ii) is a direct consequence of part (i), [118] and [117], it suffices to prove (i). Let
l(∂∆) denote the length of the boundary of a solid triangle ∆. Note that if ∆′ is one of the four
solid triangles obtained by joining the midpoints of ∂∆, then l(∂∆′) = l(∂∆)/2. Below, all integrals
over triangles are considered with respect to the anticlockwise orientation.
Consider a solid triangle ∆ ⊂ U and ε > 0. We will show (and it is enough to show) that
|∫∂∆ f | ≤ ε. Divide ∆ into four solid triangles ∆1,∆2,∆3,∆4 of equal size by joining the midpoints
of the sides of ∂∆. Observe that∑4
j=1
∫∂∆j
f =∫∂∆ f since the integrals along common edges will
get cancelled due to opposite orientations. Hence one of the ∆j ’s, call it ∆(1), satisfies |
∫∂∆(1) f | ≥
|∫∂∆ f |/4. Also, l(∂∆
(1)) = l(∂∆)/2, as noted above.
At the n the step, divide ∆(n) into four solid triangles ∆(n)1 ,∆
(n)2 ,∆
(n)3 ,∆
(n)4 of equal size by
joining the midpoints of the sides of ∂∆(n). Since∑4
j=1
∫∂∆
(n)j
f =∫∂∆(n) f , there exists ∆(n+1) ∈
{∆(n)1 ,∆
(n)2 ,∆
(n)3 ,∆
(n)4 } with |
∫∂∆(n+1) f | ≥ |
∫∂∆(n) f |/4 = |
∫∂∆ f |/4
n+1. Moreover, l(∂∆(n+1)) =
l(∂∆(n))/2 = l(∂∆)/2n+1.
Since (∆(n)) is a decreasing sequence of nonempty compact sets with diam(∆(n)) → 0, there is
p ∈ U such that∩∞
n=1∆(n) = {p}. Since f is holomorphic, there is an open ball B ⊂ U centered at
p such that |f(z)− f(p)− (z − p)f ′(p)| < ε|z − p|l(∂∆)2
for every z ∈ B.
Choose n large enough with ∆(n) ⊂ B. Note that z 7→ f(p) − (z − p)f ′(p) is a poly-
nomial and hence has a primitive. Consequently,∫∂∆(n)(f(p) − (z − p)f ′(p))dz = 0 by [117],
28
and therefore |∫∂∆(n) f(z)dz| ≤
∫∂∆(n) |f(z) − f(p) − (z − p)f ′(p)||dz| ≤
∫∂∆(n)
ε|z − p|l(∂∆)2
|dz| ≤∫∂∆(n)
εl(∂∆(n))
l(∂∆)2|dz| ≤ εl(∂∆(n))2
l(∂∆)2= ε/4n. Hence |
∫∂∆ f | ≤ 4n|
∫∂∆(n) f | ≤ ε.
Remark: [119](ii) implies that if f is holomorphic in a star region U , then ‘∫ z2z1f(z)dz’ has a
well-defined meaning (independent of the path) for any two z1, z2 ∈ U .
Exercise-45: Let f : D → C be continuous and suppose f is holomorphic on D. Then∫|z|=1 f = 0.
[Hint : Define fn : B(0, n+1n ) → C as fn(z) = f( nz
n+1). Then fn is holomorphic on B(0, n+1n ) and
hence∫|z|=1 fn = 0 by [119](ii). The uniform continuity of f on the compact set ∂D implies that
(fn) → f uniformly on ∂D. Therefore,∫|z|=1 f = limn→∞
∫|z|=1 fn = 0 by Exercise-42(iii).]
[120] (i) Let a ∈ C and r > 0. Then∫|z−a|=r
dz
z − b= 2πi for every b ∈ B(a, r).
(ii) Let V ⊂ C be an open set such that ∂V is a polygon. Then∫∂V
dz
z − b= 2πi for every b ∈ V .
Proof. (i) Let f(z) = 1/(z − b), which is holomorphic on C \ {b}. Choose ε > 0 small enough with
B(b, ε) ⊂ B(a, r). See the figure with piecewise smooth closed paths β1 and β2. Since β1 is contained
in the star region C\{b+t : t ≤ 0} where f is holomorphic,∫β1f = 0 by [119](ii). Similarly,
∫β2f = 0
by [119](ii) since β2 is contained in the star region C \ {b+ t : t ≥ 0} where f is holomorphic. Thus
0 =∫β1f +
∫β2f =
∫|z−a|=r
dz
z − b−∫|z−b|=ε
dz
z − b. Also we know
∫|z−b|=ε
dz
z − b= 2πi by [116].
(ii) Similar to the proof of (i).
[121] [An improvement of Cauchy’s integral theorem for star regions] Let U ⊂ C be a star region,
p ∈ U be a center of U , let f : U → C be continuous, and suppose f is holomorphic on U \ {p}.
Then f has a primitive in U , f is integrable in U , and∫α f = 0 for every piecewise smooth closed
path α in U .
Proof. By [118], it suffices to show that∫∂∆ f = 0 for an arbitrary solid triangle ∆ ⊂ U .
29
Case-1 : p /∈ ∆. Then∫∂∆ f = 0 by Goursat’s integral lemma applied to U \ {p}.
Case-2 : p is a vertex of ∆. Let q1, q2 be the other two vertices of ∆. Choose points p1 ∈ (p, q1)
and p2 ∈ (p, q2) very near to p. Now,∫∂∆ f =
∫[p,p1,p2]
f +∫[p1,q1,p2]
f +∫[p2,q1,q2]
f , where the last
two integrals are zero by case-1. Moreover, |∫[p,p1,p2]
f | can be made arbitrarily small by taking p1
and p2 sufficiently close to p. Hence∫∂∆ f = 0.
Case-3 : p is an interior point of an edge of ∆. Divide ∆ into two solid triangles ∆1,∆2 by
joining p with the opposite vertex of ∆. Then∫∂∆j
f = 0 for j = 1, 2 by case-2, and hence∫∂∆ f =
∫∂∆1
f +∫∂∆2
f = 0.
Case-4 : p ∈ int(∆). Divide ∆ into three solid triangles ∆1,∆2,∆3 by joining p with the vertices
of ∆. Then∫∂∆j
f = 0 for j = 1, 2, 3 by case-2, and hence∫∂∆ f =
∑3j=1
∫∂∆j
f = 0.
Now we are ready to prove one of the most important results about holomorphic functions.
The result says in particular that a holomorphic function f is completely determined in an open
ball B(a, r) by the values of f on the boundary circle |z − a| = r provided f is holomorphic in a
neighborhood the closed ball B(a, r)
[122] [Cauchy’s integral formula for discs] Let U ⊂ C be open, f : U → C be holomorphic and
suppose B(a, r) ⊂ U . Then f(b) =1
2πi
∫|z−a|=r
f(z)
z − bdz for every b ∈ B(a, r).
Proof. Let s > r be with B(a, s) ⊂ U (such a choice is possible since the compact set B(a, r) is
contained in U). Fix b ∈ B(a, r) and define g : B(a, s) → C as g(b) = f ′(b) and g(z) =f(z)− f(b)
z − bfor z = b. Then g is continuous on B(a, s) and holomorphic on B(a, s) \ {b}. By [121] and [120](i),
we get 0 =∫|z−a|=r g(z)dz =
∫|z−a|=r
f(z)
z − bdz −
∫|z−a|=r
f(b)
z − bdz =
∫|z−a|=r
f(z)
z − bdz − 2πif(b).
[123] [Power series representation] Let U ⊂ C be open, f : U → C be holomorphic and let
B(a,R) ⊂ U . Then there is a complex power series∑∞
n=0 an(z − a)n with radius of convergence
≥ R such that f(z) =∑∞
n=0 an(z − a)n for every z ∈ B(a,R). The coefficients an are given by
an =f (n)(a)
n!=
1
2πi
∫|z−a|=r
f(z)
(z − a)n+1dz for any r ∈ (0, R).
Proof. Let 0 < r < R. Then B(a, r) ⊂ B(a,R) ⊂ U . Let b ∈ B(a, r). By [122], we have
f(b) =1
2πi
∫|z−a|=r
f(z)
(z − b)dz. Since
1
z − b=
1
(z − a)− (b− a)=
1
z − a
∑∞n=0
(b− a
z − a
)n
, we have
that∫|z−a|=r
f(z)
(z − b)dz =
∫|z−a|=r
∑∞n=0
(f(z)(b− a)n
(z − a)n+1
)dz. Here, we would like to interchange the
integration and summation. Define g, gn : ∂B(a, r) → C as g(z) =f(z)
z − b=∑∞
n=0
f(z)(b− a)n
(z − a)n+1and
30
gn(z) =∑n
k=0
f(z)(b− a)k
(z − a)k+1, which are continuous functions. Let M = sup{|f(z)| : |z − a| = r}.
For |z − a| = r, we see that |g(z) − gn(z)| ≤∑∞
k=n+1
|f(z)||b− a|k
|z − a|k+1≤∑∞
k=n+1
M |b− a|k
rk+1→ 0
independent of z as k → ∞. That is, (gn) → g uniformly on ∂B(a, r). Hence by Exercise-42(iii),∫|z−a|=r
f(z)
(z − b)dz =
∫|z−a|=r g = limn→∞
∫|z−a|=r gn = limn→∞
∫|z−a|=r
∑nk=0
f(z)(b− a)k
(z − a)k+1dz =∑∞
n=0
(∫|z−a|=r
f(z)
(z − a)n+1dz
)(b − a)n. So, f(b) =
1
2πi
∫|z−a|=r
f(z)
(z − b)dz =
∑∞n=0 an(b − a)n,
where an =1
2πi
∫|z−a|=r
f(z)
(z − a)n+1dz. Note that an is independent of the point b ∈ B(a, r). Thus
f(z) =∑∞
n=0 an(z − a)n for every z ∈ B(a, r). Hence the radius of convergence of the power series∑∞n=0 an(z − a)n is ≥ r. Now from the theory of power series, it follows that f is infinitely often
(complex) differentiable in B(a, r) and an =f (n)(a)
n!. Thus an’s are independent of the particular
choice of r ∈ (0, R) also. Since r ∈ (0, R) was arbitrary, the proof is complete.
[124] [Corollary] Let U ⊂ C be open and f ∈ H(U). Then,
(i) f is infinitely often complex differentiable in U .
(ii) If B(a, r) ⊂ U , then f (n)(a) =n!
2πi
∫|z−a|=r
f(z)
(z − a)n+1dz for every n ∈ N.
(iii) [Cauchy’s estimate] If B(a,R) ⊂ U and M := sup{|f(z)| : |z − a| < R}, then |f (n)(a)| ≤ n!M
Rn
for every n ∈ N.
(iv) If B(a, r) ⊂ U , then f (n)(b) =n!
2πi
∫|z−a|=r
f(z)
(z − b)n+1dz for every n ∈ N and b ∈ B(a, r).
(v) [Cauchy’s estimate generalized] If B(a,R) ⊂ U and M := sup{|f(z)| : |z − a| < R}, then
|f (n)(b)| ≤ n!MR
(R− |b− a|)n+1for every n ∈ N and b ∈ B(a,R).
Proof. (i) This is clear from [123].
(ii) By [123], f(z) =∑∞
n=0 an(z − a)n in B(a, r) and an =f (n)(a)
n!=
1
2πi
∫|z−a|=r
f(z)
(z − a)n+1dz.
(iii) Let r ∈ (0, R). By (ii), |f (n)(a)| ≤ n!
2π
∫|z−a|=r
M
rn+1|dz| = n!M
2πrn+1× 2πr =
n!M
rn. Now let
r → R.
(iv) Fix b ∈ B(a, r). Let ε > 0 be small enough with B(b, ε) ⊂ B(a, r). We have that f (n)(b) =n!
2πi
∫|z−b|=ε
f(z)
(z − b)n+1dz by (ii). Let g : U \ {b} → C be g(z) =
f(z)
(z − b)n+1, which is holomorphic.
Consider piecewise smooth paths β1, β2 in U \ {b} as in the proof of [120](i), and establish the
following:∫βjg = 0 for j = 1, 2 and
∫β1g +
∫β2g =
∫|z−a|=r g −
∫|z−b|=ε g.
(v) Fix b ∈ B(a,R). Let r ∈ (0, R) be such that b ∈ B(a, r). By (iii), we get |f (n)(b)| ≤n!
2π
∫|z−a|=r
M
(r − |b− a|)n+1|dz| = n!M
2π(r − |b− a|)n+1×2πr =
n!Mr
(r − |b− a|)n+1. Now let r → R.
31
Exercise-46: Let f(z) =∑∞
n=0 an(z − a)n be holomorphic on B(a,R).
(i) If M = sup{|f(z)| : |z − a| < R} <∞, then |an| ≤M/Rn for every n ≥ 0.
(ii) If M ′ = sup{|f ′(z)| : |z − a| < R} <∞, then n|an| ≤M/Rn−1 for every n ∈ N.
[Hint : (i) For r ∈ (0, R), we get using [123] that |an| ≤1
2π
∫|z−a|=r
M
rn+1|dz| = M
rn. Now let r → R.
(ii) Since f ′(z) =∑∞
n=1 nan(z − a)n−1, we may apply (i).]
Exercise-47: Evaluate the following: (i)∫|z|=1
cos z
zdz, (ii)
∫|z|=1
sin z
z4dz, (iii)
∫|z|=1
sin z
(z − 3)5dz,
(iv)∫|z|=2
1
z2 + 1dz, (v)
∫|z|=2
9z + 1
(z − 1)2(z + 1)dz.
[Hint : (i) Let f(z) = cos z. Then∫|z|=1
cos z
zdz = 2πif(0) = 2πi. (ii) Let f(z) = sin z. Then∫
|z|=1
sin z
z4dz = 2πif (3)(0)/3! = −πi/3. (iii) f(z) :=
sin z
(z − 3)5is holomorphic in the star region
{z ∈ C : Re(z) < 2}. Hence∫|z|=1 f = 0. (iv)
1
z2 + 1=
i/2
z + i− i/2
z − i. Hence
∫|z|=2
1
z2 + 1dz =∫
|z|=2
i/2
z + idz −
∫|z|=2
i/2
z − idz = (i/2)2πi − (i/2)2πi = 0. (v)
9z + 1
(z − 1)2(z + 1)=
2z + 3
(z − 1)2− 2
z + 1.
Let f(z) = 2z + 3 and g ≡ 2. Then∫|z|=2
9z + 1
(z − 1)2(z + 1)dz =
2πi
1!f ′(1)− 2πig(−1) = 0.]
Exercise-48: (i) If b ∈ D is such that∫|z|=1
ez
(z − b)4dz =
2πi
3, determine b.
(ii) If c ∈ C is such that∫|z|=1
c
z3 − 4z2 + 4zdz = 2πi, determine c.
[Hint : (i) Let f(z) = ez. By hypothesis,2πi
3=
2πif (3)(b)
3!, and hence 2 = eb, which im-
plies b = log 2. (ii) Let f(z) = c/(z − 2)2, which is holomorphic in B(0, 2). Since 2πi =∫|z|=1
c
z3 − 4z2 + 4zdz =
∫|z|=1
f(z)
zdz = 2πif(0) = 2πic/4, we get c = 4.]
Exercise-49: (i) If f(z) =∑∞
n=0 anzn is holomorphic in a neighborhood of 0, then limn→∞
ann!
= 0.
(ii) There do not exist any R > 0 and f ∈ H(B(0, R)) with the property that |f (n)(0)| ≥ (n!)2 for
infinitely many n ∈ N.
(iii) Let k ∈ N and f : C \ {0} → C be f(z) =sin z
z2k. Then f does not have a primitive in C \ {0}.
[Hint : (i) Otherwise, there is δ > 0 such that lim supn→∞|an|n!
≥ δ > 0. Then lim sup |an|1/n ≥
lim δ1/n(n!)1/n = ∞ (since lim δ1/n = 1 and lim(n!)1/n = ∞), implying that the radius of con-
vergence of f is zero, a contradiction. (ii) This follows from (i) since an = f (n)(0)/n!. A slightly
different argument is as follows. Suppose f is holomorphic in a neighborhood B(0, R), and let
M = sup{|f(z)| : |z| ≤ R} < ∞. By Cauchy’s estimate, |f (n)| ≤ n!M
Rnfor every n ∈ N, and hence
|f (n)| ≥ (n!)2 for at most finitely many n ∈ N. (iii) If f has a primitive in C \ {0}, then we must
have∫|z|=1 f = 0 by [117]. But
∫|z|=1
sin z
z2kdz =
±2πi cos 0
(2k − 1)!= 0.]
Exercise-50: Let α1, α2 be smooth paths in C with α1(t1) = α2(t2) =: w, and α′1(t1) = 0, α′
2(t2) = 0.
32
Then the angle between α1 and α2 at w is defined as arg(α′1(t1))− arg(α′
2(t2)). If U ⊂ C is open,
f ∈ H(U) and w ∈ U is with f ′(w) = 0, then f preserves angles at w in the following sense: if
α1, α2 are smooth paths in U such that the angle between α1 and α2 at w is defined, then this angle
is same as the angle between the smooth paths f ◦ α1 and f ◦ α2 at f(w). Here f ◦ αj is smooth
since f is infinitely often differentiable. [Hint : (f ◦ αj)′(tj) = f ′(αj(tj))α
′j(tj) = f ′(w)α′
j(tj) and
hence arg((f ◦ αj)′(tj)) = arg(f ′(w)) + arg(α′
j(tj)).]
Exercise-51: [Prove this using power series and not using [119](ii)] Let U = D or U = C. Then,
(i) If f ∈ H(U), then f has a primitive g ∈ H(U). Consequently, f is integrable and∫α f = 0 for
every piecewise smooth closed path α in C.
(ii) If f, g ∈ H(U) are such that f (n)(0) = gn(0) for every n ≥ 0, then f = g.
[Hint : Write f(z) =∑∞
n=0 anzn. (i) Take g(z) =
∑∞n=0
ann+ 1
zn+1. (ii) an =f (n)(0)
n!by [123].]
A few more interesting corollaries of [123] and [124] are the following:
[125] [Morera’s theorem - converse of Goursat’s integral lemma] Let U ⊂ C be open and f : U → C
be continuous. If∫∂∆ f = 0 for every solid triangle ∆ ⊂ U , then f ∈ H(U).
Proof. Fix a ∈ U , and let B ⊂ U be an open ball centered at a. Since B is a star region, f |B is
integrable by [118], and then by [117], f |B has a primitive g ∈ H(B). Since g is infinitely often
differentiable by [124], it follows that f |B = g′ is also infinitely often differentiable. In particular f
is holomorphic at a.
[126] [Comparing integrability and differentiability] Let U ⊂ C be open and f : U → C be
continuous. Then (i) f ∈ H(U) ⇔ f is locally integrable in U .
(ii) If U is a star region, then f ∈ H(U) ⇔ f is integrable in U .
Proof. Since being holomorphic is a local property, the implication ‘⇐’ follows by [125] in both (i)
and (ii). Moreover, the implication ‘⇒’ in (ii) is nothing but Cauchy’s integral theorem for star
regions. Therefore, it remains to show ‘⇒’ in (i). Assume f ∈ H(U), and write U =∪∞
n=1Bn,
a countable union of open balls. Since Bn is a star region, f |Bn is integrable as noted above, by
Cauchy’s integral theorem for star regions. Hence f is locally integrable in U .
[127] [Riemann continuation theorem] Let U ⊂ C be open, A ⊂ U be discrete and closed in U and
f : U \A→ C be holomorphic. Then the following are equivalent:
(i) f has a holomorphic extension to the whole of U .
(ii) f has a continuous extension to the whole of U .
33
(iii) For each a ∈ A, there is an open ball B ⊂ U centered at a such that f |B\{a} is bounded.
(iv) For each a ∈ A, limz→a(z − a)f(z) = 0.
Proof. The implications (i) ⇒ (ii) ⇒ (iii) ⇒ (iv) are clear. To prove (iv) ⇒ (i), we may assume
A = {a} and U = B := B(a, r) for some r > 0 since being holomorphic is a local property. Define
g : B → C as g(a) = 0 and g(z) = (z − a)f(z) for z = a. Then g is continuous in the star region
B by (iv), and g is holomorphic in B \ {a}. By [121], g is integrable in B and hence∫∂∆ g = 0
for every solid triangle ∆ ⊂ B, which implies by Morera’s theorem [125] that g ∈ H(B). By [123],
we may write g(z) =∑∞
n=0 an(z − a)n for z ∈ B. Since g(a) = 0, we have a0 = 0, and hence
g(z) = (z − a)[∑∞
n=1 an(z − a)n−1]. Define f : B → C as f(z) =∑∞
n=1 an(z − a)n−1. By [110](v)
and [111](i), f ∈ H(B). Moreover, f(z) = g(z)/(z − a) = f(z) for every z ∈ B \ {a}.
Usually, we will use the following special case of [127]:
[127′] Let U ⊂ C be open, a ∈ U , f : U → C be continuous, and suppose f is holomorphic in
U \ {a}. Then f ∈ H(U).
Remark: Result [127′] indicates the strong nature of complex differentiability. In contrast note that
in Real Analysis, f : R → R given by f(x) = |x| is continuous on R and differentiable on R \ {0},
but not differentiable at 0.
7 Liouville’s theorem and Zeroes theorem
Definition: Any f ∈ H(C) is called an entire function. If f is an entire function, then we may write
f(z) =∑∞
n=0 anzn, where the power series converges for every z ∈ C. We will see that an entire
function behaves somewhat like a polynomial.
[128] (i) [Liouville’s theorem] A bounded entire function is a constant.
(ii) Let f ∈ H(C). If there is c ∈ C such that limz→∞ f(z) = c, then f ≡ c.
(iii) If f ∈ H(C) is non-constant, then f(C) is dense in C.
Proof. (i) Let f ∈ H(C) be such that M := sup{|f(z)| : z ∈ C} < ∞. For any z ∈ C and r > 0,
applying Cauchy’s estimate to B(z, r) we get |f ′(z)| ≤M/r, and consequently f ′(z) = 0 since r > 0
is arbitrary. Thus f ′ ≡ 0, and therefore f is a constant by [104](i).
(ii) There is r > 0 such that |f(z)− c| < 1 for |z| > r, and f is bounded on the compact set B(0, r).
So f is bounded on C and hence constant by (i). The constant has to be c since limz→∞ f(z) = c.
34
(iii) If f(C) is not dense in C, there exist a ∈ C and r > 0 such that |f(z)− a| ≥ r for every z ∈ C.
Then g : C → C defined as g(z) = 1/(f(z)− a) is holomorphic and |g| ≤ 1/r. By (i), g is constant,
and then f should also be constant, a contradiction.
Exercise-52: (i) If f : C → D is holomorphic, then f is constant.
(ii) Let W be equal to one of the following: {z ∈ C : Re(z) > 0}, {z ∈ C : Im(z) > 0}, C \ (−∞, 0],
or C \ [0,∞). If f : C →W is holomorphic, then f is constant.
(iii) If f : C \ {0} → C is bounded and holomorphic, then f is constant.
(iv) If f ∈ H(C) and Re(f) is bounded, then f is constant.
[Hint : (ii) By [115], there is a Mobius map T taking W biholomorphically onto D. Then T ◦ f :
C → D is holomorphic and hence constant by (i). Since T is injective, f is also constant. (iii) Since
f is bounded on C \ {0}, f extends to an entire function by [127] so that [128] can be applied. (iv)
Note that f(C) cannot be dense and apply [128](iii). Another argument is: g := ef is bounded
and hence constant by Liouville’s theorem, which implies f(C) is discrete, and consequently f is
constant since f(C) must be connected.]
Exercise-53: Let p(z) =∑n
j=0 ajzj , where n ∈ N, aj ∈ C, and an = 0. Then,
(i) There is r > 0 such that 12 |anz
n| ≤ |p(z)| ≤ 32 |anz
n| for every z ∈ C with |z| ≥ r.
(ii) limz→∞ |p(z)| = ∞.
(iii) [Fundamental theorem of algebra] There is a ∈ C with p(a) = 0.
[Hint : (i) Let M =∑n−1
j=0 |aj |. Then for |z| ≥ 1, |anzn| −M |zn−1| ≤ |p(z)| ≤ |anzn| +M |zn−1|.
Let r = max{1, 2M/|an|}. Then for |z| ≥ r, we have M |zn−1| ≤ |anzn−1|r/2 ≤ |anzn|/2. Use this
in the earlier inequalities. (iii) Assume p is non-vanishing. Then f(z) := 1/p(z) defines an entire
function, and limz→∞ f(z) = 0 by (ii). So f (= 1/p) must be constant by [128](ii), a contradiction.]
Exercise-54: Let f ∈ H(C). (i) If there is k ∈ N such that limz→∞|f(z)||z|k
= 0, then f a polynomial
of degree ≤ k − 1. In particular, if limz→∞|f(z)||z|
= 0, then f is constant.
(ii) If there is g ∈ H(C) such that g is non-vanishing and |f(z)| ≤ |g(z)| for every z ∈ C, then there
is c ∈ C such that f = cg. In particular, if there is C > 0 such that |f(z)| ≤ C|ez| for every z ∈ C,
then there is c ∈ C such that f(z) = cez for every z ∈ C.
[Hint : (i) Write f(z) =∑k−1
n=0 anzn + zk
∑∞n=k anz
n−k = g(z) + zkh(z). By Exercise-53(i), we have
limz→∞|g(z)||z|k
= 0. Hence by hypothesis, 0 = limz→∞|f(z)||z|k
= 0 + limz→∞ |h(z)|. By [128](ii),
h ≡ 0. Thus f = g. (ii) h := f/g ∈ H(C) and is bounded.]
Exercise-55: Let f ∈ H(C) be doubly periodic in the following sense: there exist R-linearly inde-
35
pendent vectors a, b ∈ C such that f(z+ a) = f(z) = f(z+ b) for every z ∈ C. Then f is constant.
[Hint : Let K be the parallelogram with vertices 0, a, b, a + b. Then f(C) = f(K) because f is
doubly periodic. Also f(K) is bounded since K is compact. By Liouville’s theorem, f is constant.]
[129] [Zeroes theorem] Let U ⊂ C be a connected open set and f ∈ H(U). Then the following are
equivalent: (i) f ≡ 0, (ii) {z ∈ U : f(z) = 0} has a limit point in U , (iii) There is a ∈ U such that
f (n)(a) = 0 for every n ≥ 0.
Proof. (ii) ⇒ (iii): Let a ∈ U be a limit point of {z ∈ U : f(z) = 0}. By continuity, f(a) = 0. Let
B(a, r) ⊂ U , and write f(z) =∑∞
n=0 an(z − a)n for z ∈ B(a, r) by [123]. Since f (n)(a) = n!an by
[123], it suffices to show an = 0 for every n ≥ 0. Note that a0 = f(a) = 0. If possible, let an = 0
for some n ∈ N, and choose the smallest such n. Then f(z) = (z − a)ng(z) for z ∈ B(a, r), where
g(z) =∑∞
k=n ak(z − a)k−n. Since g(a) = an = 0, there is 0 < δ < r such that f(z) = 0 for every
z ∈ B(a, δ) \ {a}, a contradiction since a is a limit point of {z ∈ U : f(z) = 0}.
(iii) ⇒ (i): Let A = {z ∈ U : f (n)(z) = 0 for every n ≥ 0}. Then A = ∅ by (iii), and A is closed in
U since each f (n) is continuous. If a ∈ A and B(a, r) ⊂ U , then f(z) =∑∞
n=0
f (n)(a)
n!(z − a)n = 0
for every z ∈ B(a, r), which means B(a, r) ⊂ A. Thus A is also open in U . Since U is connected,
it follows that A = U , and in particular f ≡ 0.
Remark: Let U = B(1, 1) = {z ∈ C : |z − 1| < 1} and f : U → C be f(z) = sin(1/z). Then
f ∈ H(U), the set A := {z ∈ U : f(z) = 0} is infinite (consider the points 12πn) and 0 ∈ C \ U is a
limit point of A, but f is not identically zero.
Definition: Let U ⊂ C be open and connected, and suppose f ∈ H(U) is not identically zero. If
a ∈ U is with f(a) = 0, then m := min{n ∈ N : f (n)(a) = 0} (which exists by [129]) is called the
order of the zero a for f .
[130] [Corollaries of [129]] Let U ⊂ C be open and connected. Then,
(i) If f, g ∈ H(U) are such that {z ∈ C : f(z) = g(z)} has a limit point in U , then f = g.
(ii) If f ∈ H(U) is not identically zero, then |{z ∈ K : f(z) = 0}| <∞ for any compact set K ⊂ U .
(iii) [Factorization] Assume f ∈ H(U) is not identically zero. If a ∈ U is a zero of f of order m ∈ N,
then there is g ∈ H(U) such that f(z) = (z− a)mg(z) for every z ∈ U and g(a) = 0. Consequently,
we can find δ > 0 such that B(a, δ) ⊂ U and f(z) = 0 for every z ∈ B(a, δ) \ {a}.
(iv) [Structure of the set of zeroes] Assume f ∈ H(U) is not identically zero. Then f−1(0) = {z ∈
U : f(z) = 0} is discrete and closed in U (possibly empty), and consequently f−1(0) is countable.
36
(v) [Structure of the pre-image set] Assume f ∈ H(U) is non-constant. Then for every b ∈ C, the
set f−1(b) is discrete and closed in U (possibly empty), and consequently f−1(b) is countable.
(vi) H(U) is an integral domain.
Proof. (i) Apply [129] to f − g.
(ii) Note that any infinite subset of a compact K set must have a limit point in K.
(iii) Let r > 0 be with B(a, r) ⊂ U , and write f(z) =∑∞
n=0 an(z − a)n for z ∈ B(a, r). Since a is a
zero of order m, we have that an = f (n)(a)/n! = 0 for 0 ≤ n < m and am = f (m)(a)/m! = 0. Hence
f(z) = (z − a)mg(z) for z ∈ B(a, r), where g(z) :=∑∞
k=m ak(z − a)k−m. Then g ∈ H(B(a, r)) and
g(a) = am = 0. Extend g holomorphically to U by putting g(z) = f(z)/(z−a)m for z ∈ U \B(a, r).
(iv) f−1(0) is closed in U by the continuity of f , and f−1(0) is discrete in U either by part (iii) or
by [129]. To see f−1(0) is countable, argue as follows. For each z ∈ U , we can find an open ball Bz
centered at z such that Bz contains at most one zero of f (since the zeroes of f are isolated). We
may cover U with countably many such balls since U is second countable.
(v) Fix b ∈ C and apply part (iv) to the function g(z) := f(z)− b.
(vi) Verify that H(U) is a commutative ring with unity under pointwise operations. If f, g ∈ H(U)
are such that f(z)g(z) = 0 for every z ∈ U , then either f−1(0) or g−1(0) should be uncountable,
and consequently either f ≡ 0 or g ≡ 0 by (iv).
Exercise-56: (i) There does not exist any f ∈ H(D) such that f(1/n) = (−1)n/n for every n ≥ 2.
(ii) If f ∈ H(C) is such that f(f(z)) = f(z) for every z ∈ ∂D, then either f = I or f is constant.
(iii) Let U ⊂ C be open and connected, and f ∈ H(U). If there exist a ∈ U and k ∈ N such that
f (n)(a) = 0 for every n ≥ k, then f is a polynomial of degree ≤ k − 1.
[Hint : (i) f( 12n) = 1
2n for every n ∈ N and hence f = I by [130], a contradiction. (ii) f(∂D) is
compact and connected, and hence f(∂D) is either uncountable or a singleton. In the first case,
f = I by [130] since f(w) = w for every w ∈ f(∂D). In the second case, f is constant by [130] since
∂D has a limit point in C. (iii) f (k) ≡ 0 by Zeroes theorem.]
Exercise-57: Let U ⊂ C be open and connected. (i) Let f : U → C be continuous, and define
g : U → C as g(z) = f(z)2. If g ∈ H(U), then f ∈ H(U).
(ii) If f ∈ H(U), g ∈ H(U) \ {0} are such that fg ∈ H(U), then f is constant.
(iii) Let B ⊂ U be an open ball, and Φ : H(U) → H(B) be defined as Φ(f) = f |B. When is Φ
injective? When is Φ surjective?
37
[Hint : (i) Fix a ∈ U . Assume f is not identically zero, and choose an open ball B ⊂ U centered
at a by [130](iv) such that (g and hence) f is non-vanishing in B \ {a}. Since g(z) − g(w) =
(f(z) + f(w))(f(z) − f(w)), we get that limz→wf(z)− f(w)
z − w=
g′(w)
2f(w)for every w ∈ B \ {a}.
Hence f is holomorphic in B \ {a}, and hence by [127′] holomorphic in B. (ii) Choose an open ball
B ⊂ U such that g does not vanish in B. Since 1/g ∈ H(B), we get f = (fg)/g ∈ H(B). Then
Re(f) = (f + f)/2 and Im(f) = (f − f)/2i are holomorphic in B, and hence constants. Thus f |Bis constant. Now [130] may be used. (iii) Φ is always injective by [130]. If there is a ∈ U \ B, put
g(z) = 1/(z − a). If there is f ∈ H(U) with f |B = g|B, then f = g on U \ {a} by [130]; but this is
not possible since g is unbounded near a. Thus Φ is surjective iff U = B.]
Exercise-58: If f is a non-constant complex polynomial, then the zeroes of f lie in the convex hull of
the zeroes of f ′. [Hint : Let f(z) = c(z − a1) · · · (z − an), where c = 0. Thenf ′(z)
f(z)=∑n
j=1
1
z − aj.
If w ∈ C is such that f ′(w) = 0 and f(w) = 0, then 0 =f ′(w)
f(w)=∑n
j=1
1
w − aj=∑n
j=1
w − aj|w − aj |2
.
Letting αj =1
|w − aj |2, we see w = (
∑nj=1 αjaj)/(
∑nj=1 αj).]
Exercise-59: Let U ⊂ C be open and connected, and F : U × U → C be a function which is
holomorphic in each variable. Let A ⊂ U be a subset having a limit point in U and suppose
F (z, w) = 0 for every z, w ∈ A. Then F ≡ 0 on U × U . [Hint : For each z0 ∈ A, w 7→ F (z0, w) is
identically zero in U by [129]. Now fix w0 ∈ U and consider f : U → C defined as f(z) = F (z, w0).
Then f(z) = 0 for each z ∈ U by the first sentence. Since w0 is arbitrary, we get F ≡ 0.]
Remark: Considering F : C × C → C given by F (z, w) = ez+w − ezew and taking A = R, we
may deduce using Exercise-59 that ez+w = ezew for every z, w ∈ C. Similarly, we may also deduce
various trigonometric identities such as cos(z + w) = cos z cosw − sin z sinw for z, w ∈ C.
8 Some more fundamental theorems about holomorphic functions
[131] [Maximum modulus principle] Let U ⊂ C be open and connected, and f ∈ H(U) be non-
constant. Then, (i) |f | has no maximum in U .
(ii) If K ⊂ U is compact and a ∈ K is with |f(a)| = max{|f(z)| : z ∈ K}, then a ∈ ∂K.
Proof. (i) Suppose there is a ∈ U with |f(a)| = sup{|f(z)| : z ∈ U}. Choose R > 0 with
B(a,R) ⊂ U and let 0 < r < R. By Cauchy’s integral formula, f(a) =1
2πi
∫|z−a|=r
f(z)
z − adz, and
hence 2πr|f(a)| ≤∫|z−a|=r |f(z)||dz|. On the other hand, 2πr|f(a)| =
∫|z−a|=r |f(a)||dz|. Thus
we get∫|z−a|=r(|f(a)| − |f(z)|)|dz| ≤ 0. Since the integrand |f(a)| − |f(z)| is ≥ 0, we must have
38
|f(a)| = |f(z)| for every z with |z − a| = r. Since r ∈ (0, R) is arbitrary, we get |f(a)| = |f(z)|
for every z ∈ B(a,R). Thus f(B(a,R)) is contained in the circle ∂B(0, |f(a)|). By Exercise-38,
f |B(a,R) must be a constant. Since B(a,R) is a set having a limit point in U , [129] implies that f
is a constant in U , a contradiction.
(ii) If int(K) = ∅, then ∂K = K, and there is nothing to prove. So assume that V := int(K) = ∅.
Any connected component W of V is a connected open set in C and hence by (i), |f | restricted to
W has no maximum in W , which implies a /∈W . Therefore, a ∈ K \ V = ∂K.
Remark: Let a ∈ C, r > 0, and f ∈ H(B(a, r)) be non-constant. Using [131], it may be shown that
ϕ : [0, r) → R defined as ϕ(t) = max{|f(z)| : |z − a| = t} is (continuous and) strictly increasing.
Exercise-60: Prove Fundamental theorem of Algebra using Maximum modulus principle. [Hint :
Let p(z) =∑n
j=0 ajzj , where n ∈ N and an = 0. By Exercise-53, there is r > 0 such that
|anzn|/2 ≤ |p(z)| whenever |z| ≥ r. Here, r may be chosen large enough so that |p(0)| < |an|rn/2.
If p does not vanish in C, then for f := 1/p, we get |f(z)| < |f(0)| whenever |z| = r, a contradiction
to Maximuum modulus principle.]
Exercise-61: If a1, . . . , an ∈ ∂D, then there is z ∈ ∂D with∏n
j=1 |z − aj | > 1. [Hint : Since
p(z) :=∏n
j=1(z − aj) is non-constant, ∃ z ∈ ∂D with |p(z)| > |p(0)| =∏n
j=1 |aj | = 1 by [131](ii).]
Exercise-62: Let U ⊂ C be open and connected.
(i) Let f ∈ H(U). If there is a ∈ U with Re(f(a)) ≥ Re(f(z)) for every z ∈ U , then f is constant.
(ii) Let f ∈ H(U). If there is a ∈ U with |Re(f(a))| ≥ |Re(f(z))| ∀ z ∈ U , then f is constant.
(iii) Let f, g ∈ H(U) be non-vanishing. Suppose there is a closed ball B(a, r) ⊂ U such that
|f(z)| = |g(z)| whenever |z − a| = r. Then there is c ∈ ∂D with f(z) = cg(z) for every z ∈ U .
[Hint : (i) Let L = {z ∈ C : Re(z) = Re(f(a))}, W = {z ∈ C : Re(z) < Re(f(a))}, and let T be a
Mobius map such that T (L∪{∞}) = ∂D and T (W ) = D. Then |(T ◦ f)(z)| ≤ |(T ◦ f)(a)| for every
z ∈ U , and hence by [131], T ◦ f is constant. Since T is injective, f is also constant. (ii) Replacing
f with −f if necessary, assume Re(f(a)) ≥ 0. Then use (i) or argue as follows. Letting g = ef , we
have |g(z)| = eRe(f(z)) ≤ eRe(f(a)) = |g(a)|, and hence g is constant by [131]. So the connected set
f(U) is discrete and hence must be a singleton. (iii) h1 := f/g and h2 := g/f are holomorphic in
U . Since max{|hj(z)| : |z − a| = r} = 1 for j = 1, 2, we get |f(z)| = |g(z)| for every z ∈ B(a, r) by
[131]. Then h1(B(a, r)) ⊂ ∂D, and hence by Exercise-38, there is c ∈ ∂D with h1 ≡ c on B(a, r).
Then by [130], f/g = h1 ≡ c on U .]
Exercise-63: (i) Let a ∈ C, R > 0, and f ∈ H(B(a,R)) be f(z) =∑∞
n=0 an(z − a)n. If 0 < r < R
39
and M = max{|f(z)| : |z − a| = r}, then∑∞
n=0 |an|2r2n ≤M2.
(ii) Use (i) to give a proof of the Maximum modulus principle.
[Hint : (i) Let J =∫|z−a|=r
|f(z)|2
z − adz. Since |f(z)|2 = f(z)f(z) = f(z)
∑∞n=0 an(z − a)
nand since
the uniform convergence of the series on the circle |z−a| = r allows us to interchange the integral and
summation, J =∑∞
n=0 an
(∫|z−a|=r
f(z)(z − a)n
z − adz
)=∑∞
n=0 anr2n
(∫|z−a|=r
f(z)
(z − a)n+1dz
)=
2πi∑∞
n=0 |an|2r2n. On the other hand, |J | ≤∫|z−a|=r
M2
r|dz| = 2πM2. (ii) Let U ⊂ C be open
and connected and f ∈ H(U). Suppose there is a ∈ U is with |f(a)| ≥ |f(z)| for every z ∈ U . Let
B(a,R) ⊂ U , r ∈ (0, R) and M = max{|f(z)| : |z − a| = r}. If f(z) =∑∞
n=0 an(z − a)n in B(a,R),
then M ≤ |f(a)| = |a0|, and hence an = 0 for every n ≥ 1 by (i), which means f ≡ a0 in B(a,R).
Then by [130], f ≡ a0 in U .]
[132] [Weierstrass’ uniform convergence theorem] Let U ⊂ C be open, f : U → C be a function,
and (fn) be a sequence in H(U) converging to f uniformly on compact subsets of U . Then,
(i) f ∈ H(U).
(ii) For each k ∈ N, (f (k)n ) → f (k) as n→ ∞ uniformly on compact subsets of U .
Proof. (i) The hypothesis implies that for every open ball B with B ⊂ U , (fn) → f uniformly
on B, and this implies f is continuous in U . If ∆ ⊂ U is a solid triangle, then∫∂∆ fn = 0 for
every n ∈ N by Goursat’s lemma [119](i), and then∫∂∆ f = limn→∞
∫∂∆ fn = 0 by Exercise-42(iii).
Hence f ∈ H(U) by Morera’s theorem [125].
(ii) It suffices to show (f ′n) → f ′ uniformly on compact subsets of U (for then we may argue
inductively). Consider a compact subset A of U , and let ε > 0 be given. We need to find n0 ∈ N
such that |f ′(a) − f ′n(a)| ≤ ε for every n ≥ n0 and every a ∈ A. Choose r > 0 such that
K := {z ∈ C : dist(z,A) ≤ r} ⊂ U . SinceK is compact, there is n0 ∈ N such that |f(z)−fn(z)| ≤ εr
for every n ≥ n0 and every z ∈ K. Then by Cauchy’s estimate, for n ≥ n0 and a ∈ A we get that
|f ′(z)− f ′n(z)| ≤ max{|f(z)− fn(z)|r
: z ∈ B(a, r)} ≤ max{|f(z)− fn(z)|r
: z ∈ K} ≤ εr
r= ε.
Remark: Result [132] is another illustration of the strength of complex differentiability since [132]
says essentially that the operation f 7→ f ′ of complex differentiation on H(U) is continuous when
U ⊂ C is open. In contrast, differentiation of real functions is not a continuous operation: we
can construct a sequence (gn) of continuously differentiable functions from [0, 1] to R such that
∥gn∥∞ → 0 as n→ ∞ but ∥g′n∥∞ ≥ 1 for every n ∈ N, where ∥ · ∥∞ is the supremum norm.
Exercise-64: Let U ⊂ C be open, (fn) be a sequence in H(U), and f : U → C be a function.
40
(i) Let (f1+ · · ·+ fn) → f uniformly on compact subsets of U . Then f ∈ H(U), f =∑∞
n=1 fn, and
f (k) =∑∞
n=1 f(k)n for every k ∈ N, where the series converges uniformly on compact subsets of U .
(ii) Suppose f is continuous and let (fn) → f uniformly on the image [α] for each piecewise smooth
path α in U . Then f ∈ H(U) and (fn) → f uniformly on compact subsets of U .
[Hint : (ii) For every solid triangle ∆ ⊂ U , (fn) → f uniformly on ∂∆ and hence∫∂∆ f =
limn→∞∫∂∆ fn = 0 by Goursat’s lemma. Then f ∈ H(U) by Morera’s theorem. For any B(a, r) ⊂
U , (fn) → f uniformly on the circle |z− a| = r, and hence by Maximum modulus principle applied
to f − fn, we get (fn) → f uniformly on B(a, r). If K ⊂ U is compact, we may find finitely many
open balls Bj such that K ⊂∪m
j=1Bj ⊂ U , and hence (fn) → f uniformly on K.]
Exercise-65: Let g : C → C be a function, f ∈ H(C), and suppose (f (n)) → g uniformly on compact
subsets of C. Then there is c ∈ C such that g(z) = cez for every z ∈ C. [Hint : By [132], g ∈ H(C)
and g′ = limn→∞ f (n+1) = g. Writing g(z) =∑∞
n=0 anzn and using g′ = g, we get an = a0/n! for
every n and hence g(z) = a0ez.]
[133] [Open mapping theorem] Let U ⊂ C be open and connected, and f ∈ H(U) be non-constant.
Then f(U) is open in C.
Proof. Fix a ∈ U . We need to find ε > 0 such that B(f(a), ε) ⊂ f(U). Let R > 0 be with
B(a,R) ⊂ U . If for each r ∈ (0, R), there is zr ∈ U with |zr − a| = r and f(zr) = f(a), then
it will follow by Zeroes theorem that f ≡ f(a) on U , which is a contradiction to the hypothesis.
Therefore, there exist r ∈ (0, R) and ε > 0 such that |f(z) − f(a)| ≥ 2ε for every z ∈ U with
|z − a| = r by the compactness of the circle |z − a| = r. We claim that B(f(a), ε) ⊂ f(U).
Suppose the claim is false. Then there is w ∈ B(f(a), ε) such that w = f(z) for every z ∈ U .
Then g : U → C defined as g(z) = 1/(f(z)−w) is holomorphic. If |z−a| = r, then |f(z)−f(a)| ≥ 2ε
and |w − f(a)| < ε so that |f(z) − w| > ε, which implies |g(z)| < 1/ε. This is a contradiction to
the Maximum modulus principle since |g(a)| = 1/|f(a)− w| > 1/ε as w ∈ B(f(a), ε).
Remark: In the proof of Open mapping theorem, we used the Maximum modulus principle. It is
also possible to deduce the Maximum modulus principle from the Open mapping theorem - it is
easy to see this geometrically (do it).
[134] Let U ⊂ C be a star region (or more generally any connected open set such that every
f ∈ H(U) is integrable), and f ∈ H(U) be non-vanishing. Then,
(i) There is g ∈ H(U) such that eg = f .
41
(ii) [Existence of nth root] For each n ∈ N, there is h ∈ H(U) such that hn = f .
Proof. (i) For g ∈ H(U) to satisfy eg = f , it is necessary that we should have g′ = f ′/f . This gives
us a hint. Since U is a star region and f ′/f ∈ H(U), f ′/f has a primitive g1 ∈ H(U) by [119](ii),
i.e., g′1 = f ′/f . Then fe−g1 ∈ H(U) and (fe−g1)′ = (−fg′1+f ′)e−g1 ≡ 0. Hence there is c ∈ C with
fe−g1 ≡ c. Note that c = 0 since f and e−g1 are non-vanishing. Let b ∈ C be such that eb = c.
Then f(z) = eg1(z)+b for every z ∈ U . Take g(z) = g1(z) + b for z ∈ U .
(ii) Let g be as in (i) and put h = eg/n.
Exercise-66: Let f, g ∈ H(C) be with f(z)2+g(z)2 = 1 for every z ∈ C. Then there is ϕ ∈ H(C) such
that f(z) = cosϕ(z) and g(z) = sinϕ(z) for every z ∈ C. [Hint : 1 = (f(z)+ig(z))(f(z)−ig(z)) and
hence f(z) + ig(z) = 0. By [134](i), there is ϕ ∈ H(C) with eϕ(z) = f(z) + ig(z) = 1/(f(z)− ig(z))
for every z ∈ C. Then f(z) =eiϕ(z) + e−iϕ(z)
2= cosϕ(z) and g(z) =
eiϕ(z) − e−iϕ(z)
2i= sinϕ(z).]
[135] Let U ⊂ C be open and f ∈ H(U).
(i) If f is injective, then (f(U) is open in C) and f : U → f(U) is a biholomorphism.
(ii) If f(U) is open in C and f : U → f(U) is a biholomorphism, then f ′(z) = 0 for every z ∈ U .
(iii) If f ′(z) = 0 for every z ∈ U , then f is locally injective, i.e., for each a ∈ U , there is an open
ball B ⊂ U centered at a such that f |B is injective (however, we cannot expect f to be injective
on U ; for example, consider f(z) = ez on C).
Proof. (i) We may assume U is connected. By Open mapping theorem, f(U) is open, and g :=
f−1 : f(U) → U is continuous. Since f is injective, f ′ is not identically zero, and hence the set
A := {z ∈ U : f ′(z) = 0} is discrete and closed in U by [130]. As f is an injective open map,
f(A) is discrete and closed in f(U). So by Riemann continuation theorem [127], to show g is
holomorphic in f(U), it suffices to show g is holomorphic in f(U) \ f(A). Fix b ∈ U \ A, let
c = f(b), and (wn) be an arbitrary sequence in f(U) with (wn) → c. Then zn := g(wn) → g(c) = b,
and limn→∞g(wn)− g(c)
wn − c= limn→∞
zn − b
f(zn)− f(b)= 1/f ′(b) since f ′(b) = 0. This shows that g is
holomorphic at c with g′(c) = 1/f ′(b).
(ii) Letting g = f−1 : f(U) → U , we see 1 = (g ◦ f)′(z) = g′(f(z))f ′(z) and hence f ′(z) = 0.
(iii) Let a ∈ U . Since |f ′(a)| > 0 and f ′ is continuous at a, there is an open ball B ⊂ U
centered at a such that |f ′(b) − f ′(a)| < |f ′(a)|/2 for every b ∈ B. Now for z1, z2 ∈ B, observe
that |f(z2) − f(z1) − f ′(a)(z2 − z1)| = |∫[z1,z2]
(f ′(z) − f ′(a))dz| ≤∫[z1,z2]
|f ′(z) − f ′(a)||dz| ≤
|f ′(a)||z2 − z1|/2, from which it follows that f(z1) = f(z2) if z1 = z2. Thus f |B is injective.
42
Exercise-67: Let U, V ⊂ C be open and connected, f : U → V be injective and continuous,
g ∈ H(V ) be non-constant, and suppose g ◦ f ∈ H(U). Then f ∈ H(U). [Hint : K := {w ∈ V :
g′(w) = 0} is discrete and closed in V by hypothesis, and hence A := f−1(K) is discrete and closed
in U by the injectivity and continuity of f . By Riemann continuation theorem [127], it suffices to
show f is holomorphic in U \A. Let z ∈ U \A. Then f(z) /∈ K and hence g′(f(z)) = 0. By [135],
there is an open ball B ⊂ V centered at f(z) such that g : B → g(B) is a biholomorphism. Hence
f = g−1 ◦ (g ◦ f) on the neighborhood f−1(B) of z, and thus f is holomorphic at z.]
[136] [Non-constant holomorphic functions look like z 7→ zn locally] Let U ⊂ C be open and
connected, and f ∈ H(U) be non-constant. Then for every a ∈ U , there exist an open ball B ⊂ U
centered at a and a biholomorphism h : B → h(B) such that f(z) = f(a)+(h(z))m for every z ∈ B,
where m ∈ N is the order of the zero a for f(z)− f(a).
Proof. By [130](iii), there exist an open ball B1 ⊂ U centered at a and g ∈ H(B1) with g(a) = 0
such that f(z)−f(a) = (z−a)mg(z) for every z ∈ B1. By taking B1 small enough, we may suppose
that g is non-vanishing in B1. Then by [134](ii), there is h1 ∈ H(B1) with hm1 = g. Let h ∈ H(B1)
be defined as h(z) = (z−a)h1(z). Since h′(a) = h1(a) = 0, we may choose by [135](iii) a concentric
open ball B ⊂ B1 such that h|B is injective. Then by [135](i), h : B → h(B) is a biholomorphism
and f(z) = f(a) + (z − a)mg(z) = f(a) + (h(z))m for every z ∈ B.
9 Winding number and Cauchy’s general theory
Cauchy’s general theory of integration addresses the following questions: (i) Given an open set
U ⊂ C and piecewise smooth closed paths α, β in U , when can we say that∫α f =
∫β f for every
f ∈ H(U)? In particular, when can we say that∫α f = 0 for every f ∈ H(U)? (ii) Is it possible
to modify Cauchy’s integral formula for discs by replacing ‘∫|z−a|=r’ by ‘
∫α’, where α is a piecewise
smooth closed path?
[137] (i) Let U ⊂ C be open, f ∈ H(U), and α : [a, b] → U be a piecewise smooth closed path in
U . If f is non-vanishing on [α], then1
2πi
∫α
f ′(z)
f(z)dz ∈ Z.
(ii) Let α be a piecewise smooth closed path in C and p ∈ C \ [α]. Then 1
2πi
∫α
1
z − pdz ∈ Z.
Proof. (i) Since f is non-vanishing on the compact set [α], we can find an open neighborhood
V ⊂ U of [α] such that f is non-vanishing in V . Then, again by the compactness of [α], we may
choose finitely many points a = t0 < t1 < · · · < tn = b and open balls B1, . . . , Bn ⊂ V such that
43
αj := α|[tj−1,tj ] is smooth and [αj ] ⊂ Bj for 1 ≤ j ≤ n. Let zj = α(tj) for 0 ≤ j ≤ n. Since f is non-
vanishing in the star region Bj , there is gj ∈ H(Bj) with egj = f |Bj for 1 ≤ j ≤ n by [134](i). Then∫
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