(1)The Closed Curve Theorem and Cauchy’s Integral Formula Question Given a holomorphic function f (z), can we find a function F (z) holomorphic such that F0(z

The Closed Curve Theorem and Cauchy’s Integral Formula

Question Given a holomorphic function f (z), can we find a function F (z) holomorphic such that

F⁰(z) = f (z) ?

To answer this question, we begin with the deinifiton of the integral of f along a contour C, denoted by

Z

C

f (z) dz.

Definition The image of γ(t) : [a, b] → C is called a contour curve if (i) γ is continuously differentiable except at finitely many points, and

(ii) when γ is differentiable at t, then γ⁰(t) 6= 0 except at finitely many points.

Let C be a contour and let f be a (continuous) function defined on C. Then the integral of f along contour C is defined by

Z

C

f (z) dz = Z b

a

f (γ(t)) γ⁰(t) dt,

where γ(t) : [a, b] → C is a parametrization of the contour C.

Remarks

• If γ₁ : [a, b] → C and γ2 : [c, d] → C are parametrizations, then γ2(t) is called areparametriza- tion of γ₁(t), denoted by γ₁(t) ∼ γ₂(t), if there exists λ(t) : [a, b] → [c, d] such that

γ₂(t) = γ₁(λ(t)).

• The integral of f along contour C is well defined since if γ2(t) is a reparametrization of γ₁(t), then

Z d c

f (γ₂(t)) γ₂⁰(t) dt = Z d

c

f (γ₁(λ(t))) γ₁⁰(λ(t)) λ⁰(t) dt = Z b

a

f (γ₁(t)) γ₁⁰(t) dt.

• If C : γ(t) : [a, b] → C is a contour, then define the contour −C by

−C : η(u) = γ(a + b − u); [a, b] → C such that

{η(u) | u ∈ [a, b]} = {γ(t) | t ∈ [a, b]}, η⁰(u) = −γ⁰(a + b − u) Z

−C

f (z) dz = Z b

a

f (η(u)) η⁰(u) du

= − Z b

a

f (γ(a + b − u)) γ⁰(a + b − u) du

= − Z a

b

f (γ(t)) γ⁰(t) dt

dudu by setting t = a + b − u

= Z a

b

f (γ(t)) γ⁰(t) du

= − Z b

a

f (γ(t)) γ⁰(t) du= − Z

C

f (z) dz

Example Let k ∈ Z, and S¹ be the unit circle in C oriented counterclockwise. Compute the (contour) integral

Z

S¹

z^kdz.

Let S¹ be parametrized by

γ(t) = e^it = cos t + i sin t for t ∈ [0, 2π].

Since γ⁰(t) = ie^it, we have Z

S¹

z^kdz = Z 2π

0

e^iktie^itdt = i Z 2π

0

e^i(k+1)tdt =

(2πi if k = −1, 0 if k 6= −1.

Note that

• when k ∈ Z, k 6= −1, z^k has an antiderivative z^k+1

k + 1 defined in C \ {0},

• when k = −1, 1

z does not have an antiderivative defined in C \ {0}. In fact, since

∂

∂z log z
= 1

z for all z ∈ C \ {x | x ≤ 0}, the principal branch of log z is an antiderivative of 1

z on C \ {x | x ≤ 0}.

Proposition If f (z) and F (z) are holomorphic with F⁰(z) = f (z) for all z ∈ C, then Z

C

f (z) dz = F (end) − F (start), where C is a path from start to end.

Proof Use the complex chain rule.

In the following, let f be holomorphic on C (which is also called an entire function), we’ll show

• thereexists an antiderivative F (z) such that F⁰(z) = f (z) for all z ∈ C,

• the closed curve theorem holds, i.e.

Z

C

f (z) dz = 0, for any closed contour C in C.

Fora motivation of the closed curve theorem, we recall the following Green’s Theorem Theorem

• Let γ(t) : [a, b] → C be the closed, counterclockwise boundary curve of a (polygonally connected, open) region D.

• Suppose that P, Q, Q_x, P_y are continuous functions defined on ¯D.

Then

Z b a

(P, Q) · γ⁰(t) dt = Z Z

D

∂Q

∂x −∂P

∂y
dxdy.

Proposition Let f be holomorphic on C and let C : γ(t) : [a, b] → C be the closed, counter- clockwise boundary of a region D. Then

Z

C

f (z) dz = Z b

a

f (γ(t)) γ⁰(t) dt = 0.

Proof (Incomplete) Let

f (z) = f (x, y) = u(x, y) + iv(x, y), and γ(t) = x(t) + iy(t), where u, v, x(t), y(t) are real-valued functions. Then

Z b a

f (γ(t)) γ⁰(t) dt = Z b

a

(u + iv) (x⁰(t) + y⁰(t)) dt

= Z b

a

(ux⁰− vy⁰) + i(uy⁰+ vx⁰) dt

= Z b

a

(u, −v) · (x⁰, y⁰) dt + i Z

(v, u) · (x⁰, y⁰) dt.

By the Green’s theorem (which applies when all partials are continuous in D) and the Cauchy Riemann equations, we have

Z b a

f (γ(t)) γ⁰(t) dt ^{if f}

0 is continuous

=

Green’s Thm

Z Z

D

− ∂v

∂x − ∂u

∂y
dxdy + i Z Z

D

∂u

∂x − ∂v

∂y
dxdy ^(∗)=

CR eqns0 Remarks

• In the language of vector fields,(∗) is equivalent to Z Z

D

curl(f ) dA + i Z Z

D

div(f ) dx dy = 0, where f = (u, −v)

and CR equations are equivalent to curl(f ) = 0 and div(f ) = 0 as a vector filed on R².

• The above conclusion(∗) does not require the first partials to be continuous.

• The closed curve theorem will be used to prove: If f is holomorphic on the interior of a disk, then f has a convergent power series expansion on that disk.

Lemma The closed curve theorem is true for linear functions f (z) = a + bz, i.e.

Z

C

f (z) dz = 0, for any closed contour C in C.

Proof Let F (z) = az + 1

2bz² for z ∈ C. Then F⁰(z) = f (z) for all z ∈ C and thus Z

C

f (z) dz = F (final) − F (initial) = 0, for any closed contour curve C in C.

Theorem (Closed Curve Theorem for disks) Suppose f (z) is holomorphic in an open disk D. Then

Z

C

f (z) dz = 0, for any closed contour C in D.

Remark The closed curve theorem follows from the following Claims:

(1) Z

Γ

f (z) dz = 0 for any polygonal curve Γ in D.

(2) f (z) has an antiderivative on D;

(3) Z

C

f (z) dz = 0 for any closed contour C in D.

Proof of Claim (1)

• Let T be a triangle with counterclockwise oriented boundary (a closed triangle contour) Γ.

• Suppose f (z) is holomorphic on an open neighborhood of T ∪ Γ,

• and suppose that

    Z

Γ

f (z) dz    

= c.

First, we use 3 side medians to subdivide the triangle into 4 congruent subtriangles Ti with counterclockwise oriented boundaries Γ_i for 1 ≤ i ≤ 4.

Since

c =     Z

Γ

f (z) dz    

=      Z

S4 i=1Γi

f (z) dz     

=     

4

X

i=1

Z

Γi

f (z) dz     

≤

4

X

i=1

    Z

Γi

f (z) dz     ,

there exists a subtriangle T_i such that     Z

Γi

f (z) dz    

≥ c 4.

By relabeling, we shall call this subtriangle T¹ with boundary Γ¹.

Keep subdividing to get a sequence of triangles T¹, T², T³, . . . with boundaries Γ¹, Γ², Γ³, . . . , respectively, such that

•     Z

Γ^k

f (z) dz    

≥ c

4^k, for each k = 1, 2, . . . .

• if L is the perimeter of T, then the perimeter of T^k is L

2^k and the diameter of T^k is ≤ L 2^k.

• { ¯T^k} is a nested sequence of nonempty (compact) triangles, and there exists a unique point z0 ∈ D such that

∞

\

k=1

T¯^k = {z0}.

Since f (z) is holomorphic at z₀,we have

z→zlim0

f (z) − f (z₀)

z − z₀ − f⁰(z₀) = 0,

i.e.

f (z) = f (z₀) + f⁰(z₀)(z − z₀) + ε(z)(z − z₀), where lim

z→z0

ε(z) = 0.

By the Lemma above, we have Z

Γ^k

f (z) dz = Z

Γ^k

f (z0) + f⁰(z0)(z − z0) + ε(z)(z − z0) dz = Z

Γ^k

ε(z)(z − z0) dz.

For each ε > 0,since lim

z→z0

ε(z) = 0,there exists an N ∈ N such that if k ≥ N, then |ε(z)| < ε for all z ∈ ¯T^k.Thus we have

c 4^k ≤

    Z

Γ^k

f (z) dz    

=     Z

Γ^k

ε(z)(z − z₀) dz    

≤ Z

Γ^k

|ε(z)| |z − z₀| |dz| ≤ ε L 2^k

Z

Γ^k

|dz|≤ εL² 4^k

which implies that c ≤ εL² and 

   Z

Γ

f (z) dz    

= c ≤ εL² ∀ ε > 0 =⇒

    Z

Γ

f (z) dz    

= c = 0 for any triangle contour Γ.

Corollary (extension to polygons) If P is a polygon and f (z) is holomorphic on an open neighborhood of P with polygonal boundary Γ, then

Z

Γ

f (z) dz = 0.

Proof Divide P into triangles.

Claim (2) (Antiderivative Theorem) If f (z) isholomorphic on an open disk D (or an open polygonally simply connected region Ω ⊆ C), then there exists a holomorphic function F (z) on D (or Ω) such that F⁰(z) = f (z) for all z ∈ D (or Ω).

Definition

• An open region Ω ⊆ C is polygonally connected (in topology, connected) if ∀ z, w ∈ Ω, ∃ piecewise linear curve in Ω connecting z, w.

• A region is polygonally simply connected (simply connected) if any closed polygonal curve is the boundary of a union of polygons in Ω.

Proof of Claim (2) (Antiderivative Theorem)

Let p ∈ D (or Ω ⊆ C). For each z ∈ D (or Ω), let F : D (or Ω) → C be defined by F (z) =

Z z p

f (ζ) dζ = the line integral of f along any polygonal path from p to z.

Then

• F is well-dened by the result on triangles above and the polygonally simply connected property of D (or Ω).

• for each z 6= z₀ ∈ D (or Ω), since F (z) − F (z₀)

z − z₀ = 1 z − z₀

Z z z0

f (ζ) dζ for z 6= z₀, thus we have

F (z) − F (z₀)

z − z₀ − f (z₀)    

=    

1 z − z₀

Z z z0

(f (ζ) − f (z₀)) dζ     ,

where we note that 1 z − z₀

Z z z0

f (z₀) dζ = f (z₀) z − z₀

Z z z0

dζ = f (z₀)

z − z₀ (z − z₀) = f (z₀).

For each ε > 0, since f is holomorphic at z₀, there exists a δ > 0 such that if 0 < |z − z₀| < δ, then z ∈ D (or Ω) and

|f (z) − f (z0)| < ε.

This implies that 

1 z − z₀

Z z z0

(f (ζ) − f (z₀)) dζ    

≤ 1

|z − z₀| Z z

z0

|f (ζ) − f (z₀)| |dζ|< ε

|z − z₀||z − z₀| = ε

and

F⁰(z₀) = f (z₀) ∀ z₀ ∈ D (or Ω).

Proof of Claim (3) Let F (z) be an antiderivative of f on D. Then F⁰(z) = f (z) for all z ∈ D and thus

Z

C

f (z) dz = F (final) − F (initial) = 0, for any closed contour C in D.

Corollary (Closed Curve Theorem for regions) If f (z) is holomorphic on an open disk D (or an open polygonally simply connected region Ω ⊆ C), then

Z

C

f (z) dz = 0 for any closed contour C in D (or Ω).

Proof Take F (z), the antiderivative of f (z), so that Z

C

f (z) dz = F (final) − f (initial) = 0.

Lemma If f (z) is holomorphic on an open disk D (or an open polygonally simply connected region Ω ⊆ C) and if w is a point in D (or Ω), then the function

g : D (or Ω) → C defined by

g(z) =







f (z) − f (w)

z − w if z 6= w f⁰(w) if z = w satisfies (the closed curve theorem)

Z

C

g(z) dz = 0 for any closed contour C in D (or Ω).

Proof Since

• lim

z→wg(z) = f⁰(w) = g(w), g(z) is continuous on D and

• M = sup{|g(z)| : z ∈ ¯T } < ∞.

The goal is

• to prove Z

Γ

g(z) dz = 0 along any trianglular path Γ in D;

• then it will follow ∃ G(z) such that G⁰(z) = g(z) for all z ∈ D.

• and hence the closed curve theorem holds for g(z).

If w /∈ ¯T is outside the triangle T, then g(z) is holomorphic in an open neighborhood of T and hence

Z

Γ

g(z) dz = 0 by the closed curve theorem for D.

If w ∈ Γ = ∂T, then for any ε > 0,

• we divide T into smaller triangles {T_i} with boundaries {Γ_i} such that w is an interior point of Γ₁ and

• the length of Γ₁ is ≤ ε

M + 1, where M = sup{|g(z)| : z ∈ ¯T } < ∞.

Thus we have

| Z

Γ

g(z) dz| = | Z

Γ1

g(z) dz| ≤ Z

Γ1

|g(z)| |dz| ≤ M ε

M + 1 < ε =⇒

Z

Γ

g(z) dz = 0.

If w ∈ Int T is in the interior, we can use the same trick with w being an interior point of T₁ to show that

Z

Γ

g(z) dz = 0.

Cauchy Integral Formula Let f (z) be holomorphic on an open disk D. For each w ∈ D, if C ⊂ D is a closed curve surrounding w, then we have the formula

f (w) = 1 2πi

Z

C

f (z) z − w dz.

Proof For each w ∈ D, let C ⊂ D be a circle centered at w that is parameterized by C : γ(t) = w + Re^it for 0 ≤ t ≤ 2π.

By letting

g(z) =







f (z) − f (w)

z − w if z 6= w f⁰(w) if z = w and by the above Lemma, we have

0 = 1 2πi

Z

C

g(z) dz = 1 2πi

Z

C

f (z) − f (w)

z − w dz for any closed contour C in D since w /∈ C. Thus we have

1 2πi

Z

C

f (z)

z − w dz = 1 2πi

Z

C

f (w) z − wdz

= f (w) 2πi

Z 2π 0

1

Re^it Rie^itdt

= f (w).

In general, if C = ∂Ω ⊂ D is a closed contour surrounding w, there exists a positive number

• r > min

z∈C dist(z, w) > 0 such that

• Γ = {z ∈ D | |z − w| = r} is a circle centered at w, we have

f (w) = 1 2πi

Z

C

f (z) z − w dz.

Furthermore, since f (z)

z − w is holomorphic on and inside Ω \ {z ∈ D | |z − w| < r

2}, we have Z

C

f (z) z − w dz =

Z

Γ

f (z) z − w dz.

Theorem Suppose f (z) isholomorphic on the open disk D = D(0, R)of radius R around 0 ∈ C.

Then there exists a convergent power series

∞

X

k=0

c_kz^k with lim

k→∞|c_k|^1/k ≤ 1

R such that f (z) =

∞

X

k=0

c_kz^k for each z ∈ D.

Proof For each z ∈ D = D(0, R), choose ˜R such that |z| < ˜R < R and let ˜C be the circle centered at 0 of radius ˜R. By the Cauchy’s integral formula, we have

f (z) = 1 2πi

Z

C˜

f (w) w − z dw.

Since    z w

  = |z|

R˜ < 1, so

∞

X

k=0

z^k

w^k+1 converges uniformly to 1

w − z for all w ∈ ˜C, i.e.

1

w − z = 1 w

1

1 − _w^z = 1

w 1 + z w+ z²

w² + · · ·
=

∞

X

k=0

z^k

w^k+1 for all w ∈ ˜C, and

f (z) = 1 2πi

Z

C˜

f (w) w − z dw

= 1

2πi Z

C˜

f (w) 1 w+ z

w² + z²

w³ + · · ·
dw

= 1

2πi Z

C˜

f (w)

w dw + 1 2πi

Z

C˜

f (w)

w² dw
z + 1 2πi

Z

C˜

f (w)

w³ dw
z² + · · ·

=

∞

X

k=0

ckz^k i.e. f has a power series representation, where

c_k = 1 2πi

Z

C˜

f (w)

w^k+1 dw for k = 0, 1, 2, . . . . Let M be a positive number such that

sup{|f (w)| : |w| ≤ ˜R} ≤ M.

Since

|c_k| =    

1 2πi

Z

C˜

f (w) w^k+1 dw

≤ M 2π

2π ˜R R˜^k+1 = M

R˜^k for k = 0, 1, 2, . . . . we have

lim

k→∞|c_k|^1/k ≤ lim

k→∞

M^1/k R˜ = 1

R˜ for all ˜R < R =⇒ lim

k→∞|c_k|^1/k ≤ lim

R→R˜

1 R˜ = 1

R, the raius of convergence of the power series

∞

X

k=0

c_kz^k is not less than R and the power series

∞

X

k=0

c_kz^k converges to f (z) for each z ∈ D.

Remarks

(a) Note that c_k = f^k(0) k! = 1

2πi Z

C

f (w)

w^k+1 dw for any circle C centered at 0.

(b) By the above theorem, we conclude that if f is holomorphic in an open disk, then f is analytic and infinitely differentiable there by the term-by-term differentiation theorem.

(c) Let f : R → R be defined by

f (x) =

(e^−1/x if x > 0

0 if x ≤ 0

Then f is infinitely differentiable on R, f^(k)(0) = 0 for all k = 0, 1, 2, . . . and f is not (real) analytic at 0 since it does not have Taylor’s expansion (a power series representation) at x = 0.

(d) Corollary Iff (z) is holomorphic in the disk D(α, r) thenthere exist constants cksuch that f (z) =

∞

X

k=0

ck(z − α)^k for all z ∈ D(α, r).

Proof For each z ∈ D = D(α, r), choose ˜R such that |z − α| < ˜r < r and let ˜C be the circle centered at α of radius ˜r. By the Cauchy’s integral formula, we have

f (z) = 1 2πi

Z

C˜

f (w) w − zdw.

Since    

z − α w − α    

= |z − α|

˜

r < 1, so

∞

X

k=0

(z − α)^k

(w − α)^k+1 converges uniformly to 1

w − z for all w ∈ ˜C, i.e.

1

w − z = 1

(w − α) − (z − α) = 1 w − α

1 1 −_w−α^z−α =

∞

X

k=0

(z − α)^k

(w − α)^k+1 for all w ∈ ˜C, and

f (z) = 1 2πi

Z

C˜

f (w) w − z dw

= 1

2πi Z

C˜

f (w) 1

w − α + z − α

(w − α)² + (z − α)²

(w − α)³ + · · ·
dw

= 1

2πi Z

C˜

f (w)

w − αdw + 1 2πi

Z

C˜

f (w)

(w − α)² dw
(z − α) + 1 2πi

Z

C˜

f (w)

(w − α)³dw
(z − α)²+ · · ·

=

∞

X

k=0

c_k(z − α)^k i.e. f has a power series representation, where

c_k = 1 2πi

Z

C˜

f (w)

(w − α)^k+1 dw = f^k(α)

k! for k = 0, 1, 2, . . . . with

k→∞lim |ck|^1/k ≤ 1

R =⇒ the radius of convergence is ≥ R, and the power series

∞

X

k=0

c_k(z − α)^k converges to f (z) for each z ∈ D = D(α, r).

(e) Corollary If f (z) is holomorphic in an arbitrary domain D, then for each α ∈ D, there exist constants c_k such that

f (z) =

∞

X

k=0

ck(z − α)^k

for all points z inside the largest disc centered at α and contained in D.

(f) Examples

(1) Find a power series representation for f (z) = 1

z − 1 near z = 2.

(2) Find a power series representation for f (z) = 1

z² near z = 3.

(3) Find the first three terms of the power series representation for f (z) = sin(1/z) around z = 1.

Theorems

(a) An entire function is infinitely differentiable everywhere in C.

(b) Taylor Expansion If f is entire and if a is any complex number, then f (z) = f (a) + f⁰(a)(z − a) + f⁰⁰(a)

2! (z − a)²+ · · · for all z ∈ C and f^(k)(a), for k = 0, 1, 2, . . . , satisfies

f^(k)(a) k! = 1

2πi Z

C

f (w)

(w − a)^k+1 dw,

where C = {a + Re^it| 0 ≤ t ≤ 2π} is a circle of radius R with center a.

(c) If f is entire and if

g(z) =







f (z) − f (a)

z − a if z 6= a f⁰(a) if z = a then g is entire.

Proof By (b),

g(z) =







f⁰(a) +f⁰⁰(a)

2! (z − a) + f⁽³⁾(a)

3! (z − a)²+ · · · for z 6= a,

f⁰(a) for z = a,

and since

∞

X

k=1

f^(k)(a)

k! (z − a)^k−1 is an everwhere convergent power series and since

g(z) =

∞

X

k=1

f^(k)(a)

k! (z − a)^k−1 for all z ∈ C, g is entire.

(d) Suppose f is entire with zeroes at a₁, a₂, . . . , a_N. Then if g is defined by

g(z) =











f (z)

(z − a₁)(z − a₂) · · · (z − a_N) for z 6= a₁, a₂, . . . , a_N,

z→alimk

f (z)

(z − a₁)(z − a₂) · · · (z − a_N) for k = 1, 2, . . . , N, then g is entire.

Proof Let f₀(z) = f (z).

For k ≥ 1 and for z 6= a_k, . . . , a₁, let f_k(z) = f_k−1(z) − f_k−1(a_k)

z − ak

= f_k−1(z) z − ak

= · · · = f₀(z)

(z − ak)(z − ak−1) · · · (z − a1). Assuming that f_k−1 is entire, it follows that f_k(a_k) = lim

z→ak

f_k(z) = f_k−1⁰ (a_k) exists and f_k is entire. Since f₀ is entire, the proof follows by induction.

Applications of Cauchy’s Integral Formula

(a) Liouville Theorem A bounded entire function is constant, i.e. If f is entire and if there exist a nonnegative constant M such that

|f (z)| ≤ M for all z ∈ C, then f is constant.

Proof Let a and b represent any two complex numbers and let C be any positively oriented circle centered at 0 and with radius R > max(|a|, |b|). Then according to the Cauchy Integral Formula, we have

|f (b) − f (a)| =    

1 2πi

Z

C

f (z)

z − bdz − 1 2πi

Z

C

f (z) z − adz

=    

1 2πi

Z

C

f (z)(b − a) (z − b)(z − a)dz

≤ 1

2π Z

C

|f (z)| |b − a|

(|z| − |b|) (|z| − |a|)|dz|

≤ M |b − a| R

(R − |b|) (R − |a|) for all R > max(|a|, |b|).

Since lim

R→∞

M |b − a| R

(R − |b|) (R − |a|) = 0, f (b) = f (a) and f is constant.

(b) Extended Liouville Theorem If f is entire and if, for some integer k ≥ 0, there exist positive constants A and B such that

|f (z)| ≤ A + B|z|^k for all z ∈ C, then f is a polynomial of degree at most k.

Proof Note that the case k = 0 is the original Liouville Theorem.

Suppose that the Extended Liouville Theorem holds for the case of k − 1, and g(z) is a function defined by

g(z) =







f (z) − f (0)

z if z 6= 0

f⁰(0) if z = 0.

Since f is an entire function such that |f (z)| ≤ A + B|z|^k for all z ∈ C, g is also entire and there exist positive constants C and D such that

|g(z)| ≤ C + D|z|^k−1 for all z ∈ C.

This implies that g is a polynomial of degree at most k − 1 and thus f is a polynomial of degree at most k.

(c) Fundamental Theorem of Algebra Every non-constant polynomial with complex coef- ficients has a zero in C.

Proof Let P (z) be any polynomial. If P (z) 6= 0 for all z ∈ C, then f (z) = 1/P (z) is an entire function. Furthermore if P is non-constant, then lim

z→∞P (z) = ∞ and f is bounded on C. But then, by Liouville’s Theorem, f is constant, and so is P, contrary to our assumption.

(d) Mean Value Theorem If f (z) is holomorphic on D = D(p, R), then for any 0 < r < R, we have

f (p) = 1 2π

Z 2π 0

f (p + re^iθ) dθ.

This is true as well for u = Re(f ), v = Im(f ) in place of f.

Proof Use the Cauchy integral formula.

Examples (1) Show that

Z ∞ 0

sin x

x dx = π 2.

Proof For R > r > 0, let γ₁ be the line from x = r to x = R, C_R be the upper half-circle defined by {Re^iθ | θ ∈ [0, π]}, γ3 be the line from x = −R to x = −r, and Cr be the upper half-circle defined by {re^iθ | θ ∈ [0, π]}. Suppose that C is the contour defined by

C = γ₁∪ C_R∪ γ₃∪ (−C_r).

Since e^iz

z is holomorphic on and inside C, we have 0 =

Z

C

e^iz z dz

= Z R

r

e^ix x dx +

Z π 0

e^iReiθ

Re^iθ iRe^iθdθ + Z −r

−R

e^ix x dx −

Z π 0

e^ireiθ

re^iθ ire^iθdθ

= Z R

r

e^ix x dx +

Z π 0

ie^iReiθdθ − Z R

r

e^−ix x dx −

Z π 0

ie^ireiθdθ

= Z R

r

2i sin x x dx +

Z π 0

ie^iReiθdθ − Z π

0

ie^ireiθdθ for all R > r > 0.

Since lim

R→∞

Z π 0

ie^iReiθdθ = 0 and lim

r→0

Z π 0

ie^ireiθdθ = πi, we obtain Z ∞

0

sin x

x dx = π 2.

(2) Show that Z ∞

0

sin(x²) dx = Z ∞

0

cos(x²) dx =

√2π 4 . Proof Recall that

Z ∞

−∞

e^−x2dx

2

= Z ∞

−∞

Z ∞

−∞

e^−(y2+z2)dydz = Z 2π

0

Z ∞ 0

re^−r2drdθ = π.

For R > 0, let γ₁ be the line from x = 0 to x = R, C_R be the eighths-circle defined by {Re^it | 0 < t < π

4} and γ₃ be the line from Re^iπ/4 to 0. Suppose that C is the contour defined by

C = γ₁∪ C_R∪ γ₃. Since e^−z2 is holomorphic on and inside C, we have

0 = Z

C

e^−z2dz

= Z R

0

e^−x2dx + Z π/4

0

e^−R2ei2tiRe^itdt + Z 0

R

e^{−(teiπ/4)2}e^iπ/4dt

Since lim

R→∞

Z π/4 0

e^−R2ei2tiRe^itdt = 0 and lim

R→∞

Z R 0

e^−x2dx =

√π

2 , we obtain

√π

2 =

Z ∞ 0

e^{−(teiπ/4)2}e^iπ/4dt

= e^iπ/4 Z ∞

0

e^−it2dt

=

√2 2 + i

√2 2

Z ∞

0

 cos(t²) − i sin(t²) dt

= Z ∞

0



√2

2 cos(t²) +

√2

2 sin(t²)
+ i

√2

2 cos(t²) −

√2

2 sin(t²)
 dt By equating the real and imaginary parts, we have

Z ∞ 0

√2

2 cos(t²) +

√2

2 sin(t²)
dt =

√π

2 and

Z ∞ 0

√2

2 cos(t²) −

√2

2 sin(t²)
 dt = 0 which implies that

Z ∞ 0

sin(t²) dt = Z ∞

0

cos(t²) dt =

√2π 4 .

Proposition (Uniqueness Theorem) If f and g are holomorphic on a region D ⊆ C with f (z) = g(z) on a collection of points accumulating somewhere, then f (z) = g(z) on D.

Proof Let

A = {z ∈ D | ∃ infinitely many points w in any disk containing z s.t. f (w) = g(w)}

and let B = D \ A.

Claim : both A and B are open.

If z ∈ B, then ∃ a disk containing z with only finitely many such w’s. Let δ < lim

such w, w6=z|w − z|.

Since every point in the open disk of radius δ is in B, i.e. D(z, δ) ⊂ B, B is open.

If z ∈ A, since f and g are holomorphic on D, ∃ a disk D(z, r) ⊂ D such that f and g have convergent power series expansions centered at z of radius of convergence ≥ r. So by uniqueness for power series, f = g on D(z, r) which implies that D(z, r) ⊂ A and A is open.

Since A and B are both open, A ∪ B = D, and A ∩ B = ∅, this implies that either A = ∅ or B = ∅. Since A 6= ∅, this implies that A = D, or equivalently f (z) = g(z) on D.

Remark It turns out that polynomials are the only entire functions that go to infinity with z in the limit. Geometrically, this means that if we consider the Riemannian sphere ˆC with f : ˆC →C being entire, then limˆ

z→∞f (z) = a for some finite a ∈ C implies that f (z) is bounded, i.e. a constant. If lim

z→∞f (z) = ∞, then f (z) is a polynomial.

Proposition If f (z) is entire and lim

z→∞f (z) = ∞, then f (z) is a polynomial.

Proof Since lim

z→∞f (z) = ∞, there exists R > 0 such that the zeroes of f all lie in a disk D(0, R).

There are only finitely many zeroes, else they accumulate somewhere and then f = 0. Suppose that the zeros of f are at z₁, z₂, . . . , z_k of order m₁, m₂, . . . , m_k, respectively.

Consider the function g : C → C defiend by

g(z) = f (z)

(z − z₁)^m1(z − z₂)^m2· · · (z − z_k)^mk for z ∈ C Then g(z) is an entire function and g has no zeros in C.

Since lim

z→∞f (z) = ∞ =⇒ lim

z→∞|g(z)z^m| = ∞, where m =

k

X

j=1

m_j, there exists R₁ ≥ R such that

|g(z)| ≥ 1

|z|^m for all |z| ≥ R₁. Let A = max

|z|≤R₁

1

|g(z)|. Then 1

g(z) is entire and satisfies that 1

|g(z)| ≤ A + |z|^m for all z ∈ C.

So, by the (Extended) Liouville’s theorem and the Fundamental Theorem of Algebra, 1

g(z) must be a constant. Thus f (z) is a polynomial.

Maximum Modulus Principle A non-constant analytic function in a region D does not have any interior maximum points, i.e. for each z ∈ D and for each δ > 0, there exists some w ∈ D(z; δ) ⊂ D such that

|f (w)| > |f (z)|.

Proof Suppose, on the contrary, that there exists z ∈ D and r > 0 such that

|f (w)| ≤ |f (z)| ∀ w ∈ D(z; r) ∪ C(z; r) ⊂ D.

For each 0 ≤ ρ ≤ r, since

f (z) = 1 2π

Z 2π 0

f (z + ρe^iθ) dθ, it follows that

|f (z)| ≤ 1 2π

Z 2π 0

|f (z + ρe^iθ)| dθ ≤ max

θ∈[0,2π]|f (z + ρe^iθ)| ≤ |f (z)| ∀ 0 ≤ ρ ≤ r.

This implies that

|f (z + ρe^iθ)| = |f (z)| ∀ θ ∈ [0, 2π], ∀ 0 ≤ ρ ≤ r,

i.e. |f | is constant throughout D(z; r). But then by Theorem 3.7, f is constant in D(z; r), and by the Uniqueness Theorem, f is in the region D. This contradicts to the hypothesis on f.

Minimum Modulus Principle If f is a non-constant analytic function in a region D, then no point z ∈ D can be a relative minimum of f unless f (z) = 0, i.e. for each z ∈ D such that f (z) 6= 0 and for each δ > 0, there exists some w ∈ D(z; δ) ⊂ D such that

|f (w)| < |f (z)|.

Proof Suppose that f (z) 6= 0 and consider g = 1/f. If z were a minimum point for f, it would be a maximum point for g. Hence g would be constant in D, contrary to our hypothesis on f.

Corollary If f is continuous on a closed bounded set E and holomorphic in its interior E \ ∂E, then the maximum of |f | is attained on ∂E.

Proof (Ahlfors) Since E is compact, |f (z)| has a maximum on E. Suppose that it is attained at z₀; if z₀ ∈ ∂E, we’re done. Else if z₀ is an interior point, then |f (z₀)| is also the maximum of

|f (z)| in a disk |z − z₀| < δ contained in E. But this is not possible unless f (z) is constant in the component of the interior of E which contains z₀. It follows by continuity that |f (z)| is equal to its maximum on the whole boundary of that component. This boundary is not empty and is contained in the boundary of E, so the maximum is always attained at a boundary point.

Proposition Suppose f is non-constant and analytic on the closed disk D, and assumes its maximum modulus at the boundary point z₀. Then f⁰(z₀) 6= 0.

Proof Since f is non-constant and analytic on the closed disk D, there exists z₀ ∈ ∂D such that maxz∈D |f (z)| = |f (z₀)| > 0. Suppose that f⁰(z₀) = 0, since f is analytic at z₀, there exists a disk D(z₀; r) such that

f (z₀+ w) = f (z₀) + f^(k)(z₀)

k! w^k+ f^(k+1)(z₀)

(k + 1)! w^k+1+ · · · for each |w| < r,

where k is the least integer with f^(k)(z₀) 6= 0. Multiplying the above expression by its conjugate shows

|f (z₀+ w)|² = f (z₀+ w)f (z₀+ w)

= |f (z₀)|²+ 2

k!Re f (z₀)f^(k)(z₀)w^k
+ · · ·

= |f (z₀)|²+2A

k! |w|^k cos(kθ + α) + · · · ,

where A = |f (z₀)f^(k)(z₀)| > 0, 0 ≤ α < 2π satisfying f (z₀)f^(k)(z₀) = Ae^iα and e^iθ = w

|w| for 0 < |w| < r.

For each j = 0, 1, . . . , k − 1, since cos(kθ+α) > 0 for θ ∈ −π + 4πj − 2α

2k ,π + 4πj − 2α 2k



=⇒ |f (z₀+w)| > |f (z₀)| for 0 < |w| < r,

and if k ≥ 2, there exists at least one 0 ≤ j ≤ k − 1 such that



z₀+ |w|e^iθ | θ ∈ −π + 4πj − 2α

2k ,π + 4πj − 2α 2k



, 0 < |w| < r



∩ D 6= ∅,

|f (z0)| < max

z∈D |f (z)| contrary to the assumption |f (z0)| = max

z∈D |f (z)|. Hence |f⁰(z0)| 6= 0, i.e.

k = 1.

Remark Similarly, if min

z∈D|f (z)| = |f (z₀)| > 0, then for each j = 0, 1, . . . , k − 1, since cos(kθ+α) < 0 for θ ∈ π + 4πj − 2α

2k ,3π + 4πj − 2α 2k



=⇒ |f (z0+w)| < |f (z0)| for 0 < |w| < r,

and if k ≥ 2, there exists at least one 0 ≤ j ≤ k − 1 such that



z₀+ |w|e^iθ | θ ∈ −π + 4πj − 2α

2k ,π + 4πj − 2α 2k



, 0 < |w| < r



∩ D 6= ∅,

|f (z₀)| > min

z∈D|f (z)| contrary to the assumption |f (z₀)| = min

z∈D|f (z)|. Hence |f⁰(z₀)| 6= 0.