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Solutions to Ahlfors’ Complex Analysis

By:

Dustin Smith

1 Complex Numbers

1.1 The Algebra of Complex Numbers

1.1.1 Arithmetic Operations

1. Find the values of

( 1 + 2i)^3 ,

− 3 + 4i

( (^2) + i 3 − 2i

, ( 1 + i)n^ + ( 1 − i)n

For the first problem, we have ( 1 + 2i)^3 = (− 3 + 4i)( 1 + 2i) = − 11 − 2i. For the second problem, we

should multiple by the conjugate ¯z = − 3 − 4i.

− 3 + 4i

− 3 − 4i − 3 − 4i

− 15 − 20i 25

i

For the third problem, we should first multiple by ¯z = 3 + 2i.

2 + i 3 − 2i

3 + 2i 3 + 2i

8 + i 13

Now we need to just square the result.

( 8 + i)^2 = 63 + 16i 169

For the last problem, we will need to find the polar form of the complex numbers. Let z 1 = 1 + i

and z 2 = 1 − i. Then the modulus of z 1 =

2 = z 2. Let φ 1 and φ 2 be the angles associated with z 1

and z 2 , respectively. Then φ 1 = arctan( 1 ) = π 4 and φ 2 = arctan(− 1 ) = − 4 π. Then z 1 =

2eπi/4^ and

z 2 =

2e−πi/4.

zn 1 + zn 2 = 2 n/

[

enπi/4^ + e−nπi/

]

= 2 n/2+^1

[

enπi/4^ + e−nπi/ 2

]

= 2 n/2+^1 cos

( (^) nπ 4

2. If z = x + iy (x and y real), find the real and imaginary parts of

z^4 ,

z

z − 1 z + 1

z^2

For z^4 , we can use the binomial theorem since (a + b)n^ =

∑n k= 0

(n k

anbn−k. Therefore,

(x + iy)^4 =

(iy)^4 +

x(iy)^3 +

x^2 (iy)^2 +

x^3 (iy) +

x^4 = y^4 − 4xy^3 i − 6x^2 y^2 + 4x^3 yi + x^4

Then the real and imaginary parts are

u(x, y) = x^4 + y^4 − 6x^2 y^2

v(x, y) = 4x^3 y − 4xy^3

For second problem, we need to multiple by the conjugate ¯z.

x + iy

x − iy x − iy

x − iy x^2 + y^2

From equation ( 1. 1 ), we see that x^2 = y^2 or ±x = ±y. Also, note that i is the upper half plane (UHP).

That is, the angle is positive so x = y and 2x^2 = 1 from equation ( 1. 1 ). Therefore,

i = √^12 ( 1 + i). We

also could have done this problem using the polar form of z. Let z = i. Then z = eiπ/2^ so

z = eiπ/

which is exactly what we obtained. For

√ −i, let^ z^ = −i.^ Then^ z^ in polar form is^ z^ =^ e−iπ/2^ so

z = e−iπ/4^ = √^12 ( 1 − i). For

1 + i, let z = 1 + i. Then z =

2eiπ/4^ so

z = 2 1/4eiπ/8. Finally, for

1 −i√ 3

2 , let^ z^ =^

1 −i√ 3

2. Then^ z^ =^ e

−iπ/3 so √z = e−iπ/6 = 1

3 − i).

2. Find the four values of 4

Let z = 4

− 1 so z^4 = − 1. Let z = reiθ^ so r^4 e4iθ^ = − 1 = eiπ(^1 +2k).

r^4 = 1 θ =

π 4 ( 1 + 2k)

where k = 0 , 1 , 2 , 3. Since when k = 4 , we have k = 0. Then θ = π 4 , 3π 4 , 5π 4 , and 7π 4.

z = eiπ/4, e3iπ/4, e5iπ/4, e7iπ/

3. Compute 4

i and 4

−i.

Let z = 4

i and z = reiθ. Then r^4 e4iθ^ = i = eiπ/2.

r^4 = 1 θ = π 8

so z = eiπ/8. Now, let z = 4

−i. Then r^4 e4iθ^ = e−iπ/2^ so z = e−iπ/8.

4. Solve the quadratic equation

z^2 + (α + iβ)z + γ + iδ = 0.

The quadratic equation is x = −b±

√b (^2) −ac

2. For the complex polynomial, we have

z = −α − βi ±

α^2 − β^2 − 4γ + i(2αβ − 4δ) 2

Let a + bi =

α^2 − β^2 − 4γ + i(2αβ − 4δ). Then

z = −α − β ± (a + bi) 2

1.1.3 Justification

1. Show that the system of all matrices of the special form

α β −β α

combined by matrix addition and matrix multiplication, is isomorphic to the field of complex numbers.

2. Show that the complex number system can be thought of as the field of all polynomials with real

coefficients modulo the irreducible polynomial x^2 + 1.

1.1.4 Conjugation, Absolute Value

1. Verify by calculation the values of

z z^2 + 1

for z = x + iy and ¯z = x − iy are conjugate.

For z, we have that z^2 = x^2 − y^2 + 2xyi.

z z^2 + 1

x + iy x^2 − y^2 + 1 + 2xyi

= x + iy x^2 − y^2 + 1 + 2xyi

x^2 − y^2 + 1 − 2xyi x^2 − y^2 + 1 − 2xyi

= x(x^2 − y^2 + 1 ) + 2xy^2 + iy(x^2 − y^2 + 1 − 2x^2 ) (x^2 − y^2 + 1 )^2 + 4x^2 y^2

For ¯z, we have that ¯z^2 = x^2 − y^2 − 2xyi.

z¯

z ¯^2 + 1

x − iy x^2 − y^2 + 1 − 2xyi

= x − iy x^2 − y^2 + 1 − 2xyi

x^2 − y^2 + 1 + 2xyi x^2 − y^2 + 1 + 2xyi

= x(x^2 − y^2 + 1 ) + 2xy^2 − iy(x^2 − y^2 + 1 − 2x^2 ) (x^2 − y^2 + 1 )^2 + 4x^2 y^2

Therefore, we have that equations ( 1. 3 ) and ( 1. 4 ) are conjugates.

2. Find the absolute value (modulus) of

−2i( 3 + i)( 2 + 4i)( 1 + i) and

( 3 + 4i)(− 1 + 2i) (− 1 − i)( 3 − i)

When we expand the first problem, we have that

z 1 = −2i( 3 + i)( 2 + 4i)( 1 + i) = 32 + 24i

so

|z 1 | =

For the second problem, we have that

z 2 = ( 3 + 4i)(− 1 + 2i) (− 1 − i)( 3 − i)

i

so

|z 2 | =

3. Prove that ∣

∣∣ a^ −^ b

1 − ab¯

if either |a| = 1 or |b| = 1. What exception must be made if |a| = |b| = 1?

Recall that |z|^2 = zz¯.

a − b

1 − ab¯

2

( (^) a − b

1 − ab¯

)( (^) a − b

1 − ab¯

( (^) a − b

1 − ab¯

)( a¯ − b¯

1 − a b¯

a a¯ − a b¯ − ab¯ + b b¯

1 − ab¯ − a b¯ + a ab¯ b¯

If |a| = 1 , then |a|^2 = a a¯ = 1 and similarly for |b|^2 = 1. Then equation ( 1. 5 ) becomes

1 − a b¯ − ab¯ + b b¯

1 − ab¯ − a b¯ + b b¯

and

1 − a b¯ − ab¯ + a a¯

1 − ab¯ − a b¯ + a a¯

resepctively which is one. If |a| = |b| = 1 , then |a|^2 = |b|^2 = 1 so equation ( 1. 5 ) can be written as

2 − a b¯ − ab¯

2 − ab¯ − a b¯

Therefore, we must have that a b¯ + ab¯ 6 = 2.

1.1.5 Inequalities

1. Prove that ∣

∣∣ a^ −^ b

1 − ab¯

if |a| < 1 and |b| < 1.

From the properties of the modulus, we have that

∣∣ a^ −^ b

1 − ab¯

∣∣ = |a^ −^ b|

| 1 − ab¯ |

|a − b|^2

| 1 − ab¯ |^2

(a − b)( a¯ − b¯)

( 1 − ab¯ )( 1 − a b¯)

|a|^2 + |b|^2 − a b¯ − ab¯

1 + |a|^2 |b|^2 − ab¯ − a b¯

2 − a b¯ − ab¯

2 − ab¯ − a b¯

From equations ( 1. 8 ) and ( 1. 9 ), we have

|a − b|^2

| 1 − ab¯ |^2

|a − b|

| 1 − ab¯ |

2. Prove Cauchy’s inequality by induction.

Cauchy’s inequality is

|a 1 b 1 + · · · + anbn|^2

|a 1 |^2 + · · · + |an|^2

|b 1 |^2 + · · · + |bn|^2

which can be written more compactly as

∑^ n

i= 1

aibi

2 6

∑^ n

i= 1

|ai|^2

∑n

i= 1

|bi|^2.

For the base case, i = 1 , we have

|a 1 b 1 |^2 = (a 1 b 1 )( a¯ 1 b¯ 1 ) = a 1 a¯ 1 b 1 b¯ 1 = |a 1 |^2 |b 1 |^2

so the base case is true. Now let the equality hold for all k − 1 ∈ Z where k − 1 6 n. That is, we assume

that

k∑− 1

i= 1

aibi

∣∣^26

k∑− 1

i= 1

|ai|^2

k∑− 1

i= 1

|bi|^2

to be true.

k∑− 1

i= 1

aibi

∣∣^2 + |a kbk|^2

k∑− 1

i= 1

|ai|^2

k∑− 1

i= 1

|bi|^2 + |akbk|^2

∣ ∣∣ ∑^ k

i= 1

aibi

∣∣^26

k∑− 1

i= 1

|ai|^2

k∑− 1

i= 1

|bi|^2 + (akbk)( a¯k b¯k)

k∑− 1

i= 1

|ai|^2

k∑− 1

i= 1

|bi|^2 + |ak|^2 |bk|^2

∑^ k

i= 1

|ai|^2

∑k

i= 1

|bi|^2

Therefore, by the principal of mathematical induction, Cauchy’s inequality is true for all n > 1 for

n ∈ Z +.

3. If |ai| < 1, λi > 0 for i = 1 ,... , n and λ 1 + λ 2 + · · · + λn = 1 , show that

|λ 1 a 1 + λ 2 a 2 + · · · + λnan| < 1.

Since

∑n

i= 1 λi^ =^1 and^ λi^ >^0 ,^0 6 λi^ < 1. By the triangle inequality,

|λ 1 a 1 + λ 2 a 2 + · · · + λnan| 6 |λ 1 ||a 1 | + · · · + |an||λn|

<

∑^ n

i= 1

λi

= 1

4. Show that there are complex numbers z satisfying

|z − a| + |z + a| = 2 |c|

if and only if |a| 6 |c|. If this condition is fulfilled, what are the smallest and largest values |z|?

By the triangle inequality,

|z − a| + |z + a| > |(z − a) − (z + a)| = 2 |a|

so

2 |c| = |z − a| + |z + a|

|(z − a) − (z + a)| = 2 |a|

Thus, |c| > |a|. For the second implication, if a = 0 , the result follow. Suppose a 6 = 0. Then let z = |c| |aa|.

2 |c| = |a|(|c|/|a| − 1 ) + |a|(|c|/|a| + 1 ) = |z − a| + |z + a|

The smallest and largest values of z can be found below.

2 |c| = |z + a| + |z − a| 4 |c|^2 =

|z + a| + |z − a|

= 2 (|z|^2 + |a|^2 ) 6 4 (|z|^2 + |a|^2 ) |c|^2 6 |z|^2 + |a|^2 √ |c|^2 − |a|^2 6 |z|

1.2 The Geometric Representation of Complex Numbers

1.2.1 Geometric Addition and Multiplication

1. Find the symmetric points of a with respect to the lines which bisect the angles between the coordinate

axes.

2. Prove that the points a 1 , a 2 , a 3 are vertices of an equilateral triangle if and only if a^21 + a^22 + a^23 =

a 1 a 2 + a 2 a 3 + a 1 a 3.

3. Suppose that a and b are two vertices of a square. Find the two other vertices in all possible cases.

4. Find the center and the radius of the circle which circumscribes the triangle with vertices a 1 , a 2 , a 3.

Express the result in symmetric form.

3. Express the fifth and tenth roots of unity in algebraic form.

To find the roots of unity, we are looking to solve zn^ = 1. Let z = eiθ^ and 1 = e2ikπ. Then θ = 2kπn. For

the fifth roots of unity, n = 5 and k = 0 , 1 ,... , 4 so we have

ω 0 = e^0 = cos( 0 ) + i sin( 0 )

ω 1 = e2π/5^ = cos

( (^) 2π 5

+ i sin

( (^) 2π 5

ω 2 = e4π/5^ = cos

( (^) 4π 5

+ i sin

( (^) 4π 5

ω 3 = e6π/5^ = cos

( (^) 6π 5

+ i sin

( (^) 6π 5

ω 4 = e8π/5^ = cos

( (^) 8π 5

+ i sin

( (^) 8π 5

Now we can plot the roots of unity on the unit circle.

Figure 1. 1 : The fifth roots of unity.

For the tenth roots of unity, n = 10 and k = 0 , 1 ,... , 9 so we have

ω 0 = e^0 = cos( 0 ) + i sin( 0 )

ω 1 = e2π/10^ = cos

( (^) π 5

+ i sin

( (^) π 5

ω 2 = e4π/10^ = cos

( (^) 2π 5

+ i sin

( (^) 2π 5

ω 3 = e6π/10^ = cos

( (^) 3π 5

+ i sin

( (^) 3π 5

ω 4 = e8π/10^ = cos

( (^) 4π 5

+ i sin

( (^) 4π 5

ω 5 = e10π/10^ = cos(π) + i sin(π)

ω 6 = e12π/10^ = cos

( (^) 6π 5

+ i sin

( (^) 6π 5

ω 7 = e14π/10^ = cos

( (^) 7π 5

+ i sin

( (^) 7π 5

ω 8 = e16π/10^ = cos

( (^) 8π 5

+ i sin

( (^) 8π 5

ω 9 = e18π/10^ = cos

( (^) 9π 5

+ i sin

( (^) 9π 5

Now we can plot the roots of unity on the unit circle.

Figure 1. 2 : The tenth roots of unity.

4. If ω is given by ω = cos

( (^) 2π n

+ i sin

( (^) 2π n

, prove that

1 + ωh^ + ω2h^ + · · · + ω(n−^1 )h^ = 0

for any integer h which is not a multiple of n.

Let ω = cos

( (^) 2π n

+ i sin

( (^) 2π n

be written in exonential form as ω = e2πi/n. Then the series can be written

as

n∑− 1

k= 0

e2πih/n

)k

e2ihπ^ − 1 e2hiπ/n^ − 1

Since h is an integer, e2ihπ^ = 1 ; therefore, the series zero.

5. What is the value of

1 − ωh^ + ω2h^ − · · · + (− 1 )n−^1 ω(n−^1 )h?

We can represent this series similarly as

n∑− 1

k= 0

−e2πih/n

)k

(− 1 )ne2ihπ^ − 1 −e2hiπ/n^ − 1

1 + (− 1 )n+^1 e2ihπ 1 + e2hiπ/n^

Again, since h is an intger, we have that e2ihπ^ = 1 which leaves us with

1 + (− 1 )n+^1 1 + e2hiπ/n^

0 , if n is even

2

1 +e2hiπ/n^ ,^ if^ n^ is odd

1.2.3 Analytic Geometry

1. When does az + bz¯ + c = 0 represent a line?

2. Write the equation of an ellipse, hyperbola, parabola in complex form.

For x, y, h, k, a, b ∈ R such that a, b 6 = 0 , we define a real ellipse as

(x − h)^2 a^2

(y − k)^2 b^2

Let z = xa + i yb and z 0 = ha + i bk. If we expand the equation for an ellipse, we have

x^2 a^2

y^2 b^2

h^2 a^2

k^2 b^2

2xh a^2

2yk b^2

Notice that |z|^2 = x

2 a^2 +^

y^2

b^2 and^ |z^0 |

(^2) = h^2 a^2 +^

k^2

b^2. Now, let’s write the ellipse as

|z|^2 + |z 0 |^2 − 2xh a^2

2yk b^2

yh ab i − yh ab i + xk ab i − xk ab

i = |z|^2 + |z 0 |^2 − zz¯ 0 − z¯z 0 = 1.

Thus, the equation of an ellipse in the complex plane is

(z − z 0 )(z¯ − ¯z 0 ) = |z − z 0 |^2 = 1 ⇒ |z − z 0 | = 1

2 Complex Functions

2.1 Introduction to the Concept of Analytical Function

2.1.1 Limits and Continuity

2.1.2 Analytic Functions

1. If g(w) and f(z) are analytic functions, show that g(f(z)) is also analytic.

Let g(w) = h(x, y) + it(x, y) and f(z) = u(x, y) + iv(x, y) where z = w = x + iy for x, y ∈ R. Then

(g ◦ f)(z) = h(u(x, y), v(x, y)) + it(u(x, y), v(x, y)).

Since f and g satisfy the Cauchy-Riemann equations,

∂u ∂x

∂v ∂y

∂u ∂y

∂v ∂x ∂h ∂x

∂t ∂y

∂h ∂u

∂t ∂x

The partial deivatives of (g ◦ f)(z) are

∂h ∂x

∂h ∂u

∂u ∂x

∂t ∂v

∂v ∂x

∂t ∂y

∂t ∂u

∂u ∂y

∂t ∂v

∂v ∂y ∂h ∂y

∂h ∂u

∂u ∂y

∂t ∂v

∂v ∂y

∂t ∂x

∂t ∂u

∂u ∂x

∂t ∂v

∂v ∂x

In order for g(f(z)) to be analytic, ∂h∂x = ∂y∂t and ∂h∂y = − ∂t∂x. We can then write

∂h ∂x

∂t ∂y

∂h ∂u

∂u ∂x

∂h ∂v

∂v ∂x

∂t ∂u

∂u ∂y

∂t ∂v

∂v ∂y = ∂h ∂u

∂u ∂x

∂t ∂u

∂u ︸ ︷︷ ∂y︸ term 1

∂h ∂v

∂v ∂x

∂t ∂v

∂v ︸ ︷︷ ∂y︸ term 2

In order for the right hand side of equation ( 2. 1 ) to be zero, we need both terms to be zero.

∂h ∂u

∂u ∂x

∂t ∂u

∂u ∂y

∂h ∂u

∂u ∂x

∂t ∂y

∂u ∂u = ∂h ∂u

∂u ∂x

∂h ∂u

∂u ∂x

Equation ( 2. 2 ) occurs since g is analytic and satisfies the Cauchy-Riemann equations.

For the second term in equation ( 2. 1 ), we again use the analyticity of g.

∂h ∂v

∂v ∂x

∂t ∂v

∂v ∂y

∂h ∂v

∂v ∂x

∂h ∂v

∂v ∂x = 0

Therefore, from equation ( 2. 1 ), we have

∂h ∂x

∂t ∂y

∂h ∂x

∂t ∂y

By similar analysis, we are able to conclude that ∂h∂y = − ∂t∂x. Therefore, g(f(z)) satisfies the Cauchy-

Riemann so it is analytic.

uuy + vux = 0 ( 2. 3 b)

Let’s write equations ( 2. 3 a) and ( 2. 3 b) in matrix form. Then we have

[

u −v v u

] [

∂u ∂x∂u ∂x

]

[

]

Suppose the matrix is not invertible. Then u^2 + v^2 = 0. Since u^2 , v^2 ∈ R , u^2 , v^2 > 0. Therefore, u = v = 0

so f(z) = 0. Now, suppose that the matrix is invertible. Then we have

[

∂u ∂u∂x ∂x

]

[

]

so f′(z) = 0 and f(z) = c for some constant c.

5. Prove rigorously that the functions f(z) and f(z¯) are simultaneously analytic.

Let g(z) = f(z¯) and suppose f is analytic. Then g′(z) is

g′(z) = lim

∆z→ 0

g(z + ∆z) − g(z) ∆z

= lim

∆z→ 0

f(z¯ + ∆z) − f(z¯)

∆z

= lim

∆z→ 0

[

f(z¯ + ∆z) − f(z¯)

∆z

]

Since conjugation is continuous, we can move the limit inside the conjugation.

= lim

∆z→ 0

f(z¯ + ∆z) − f(z¯)

∆z

= f′(z¯)

Thus, g is differentiable with derivative f′(z¯). Suppose f(z¯) is analytic and let g(z¯) = f(z). Then by the

same argument, f is differentiable with derivative g′(z¯). Therefore, f(z) and f(z¯) are simultaneously

analytic.

We could also use the Cauchy-Riemann equations. Let f(z) = u(x, y) + iv(x, y) where z = x + iy so

z¯ = x − iy. Then f(z¯) = α(x, y) − iβ(x, y) where α(x, y) = u(x, −y) and β(x, y) = v(x, −y). In order for

both to be analytic, they both need to satisfy the Cauchy-Riemann equations. That is, ux = vy, uy = −vx,

αx = βy and αy = −βx.

ux(x, y) = vy(x, y)

uy(x, y) = −vx(x, y)

αx(x, y) = ux(x, −y)

αy(x, y) = −uy(x, −y)

−βx(x, y) = vx(x, −y)

βy(x, y) = vy(x, −y)

Suppose that f(z¯) satisfies the Cauchy-Riemann equations. Then αx = ux(x, −y) = vy(x, −y) = βy and

αy = −uy(x, −y) = vx(x, −y) = −βx. Therefore,

ux(x, −y) = vy(x, −y)

uy(x, −y) = −vx(x, −y)

which means f(z¯) satisfies the Cauchy-Riemann equations. Now, recall that |z| = |z¯|. Since f(z¯) satisfies

the Cauchy-Riemann equations, for an  > 0 there exists a δ > 0 such that when 0 < |∆z| < δ,

|f(z¯) − z¯ 0 | = |f(z) − z 0 | < . Thus, lim∆z→ 0 f(z) = z 0 so f(z) is analytic if f(z¯) is analytic.

6. Prove that the functions u(z) and u(z¯) are simultaneously harmonic.

Since u is the real part of f(z), u(z) = u(x, y) where z = x + iy. Suppose u(z) is harmonic. Then u(z)

satisfies Laplace equation.

∇^2 u(z) = uxx + uyy = 0

Now, u(z¯) = u(x, −y) where ∂

2

∂x^2 u(z¯) =^ uxx^ and^

∂^2

∂y^2 u(z¯) =^ uyy^ so

∇^2 u(z¯) = uxx + uyy = 0.

Since u(z) is harmonic, uxx + uyy = 0 so it follows that u(z¯) is harmonic as well.

7. Show that a harmonic function satisfies the formal differential equation

∂^2 u

∂z∂z¯

Let u be a harmonic. Then ∇^2 u = 0.

∂z¯

∂x

• i

∂y

( 2. 4 a)

∂z

∂x − i

∂y

( 2. 4 b)

From equation ( 2. 4 a), we have

∂x

• i

∂y

u =

(ux + iuy).

Then we have

∂^2 u

∂z∂z¯

∂x − i

∂y

(ux + iuy) =

[

uxx + uyy + i(uyx − uxy)

]

Since u is a solution to the Laplace equation, u has continuous first and second derivatives. That is,

u ∈ C^2 at a minimum. By Schwarz’s theorem, uxy = uyx so

∂^2 u

∂z∂z¯

Schwarz’s theorem states that if f is a function of two variables such that fxy and fyx both exist and are

continuous at some point (x 0 , y 0 ), then fxy = fyx.

2.1.3 Polynomials

2.1.4 Rational Functions

1. Use the method of the text to develop

z^4 z^3 − 1

and

z(z + 1 )^2 (z + 2 )^3

in partial fractions.

Let R(z) = z 3 z−^41 = z + z 3 z− 1. The poles of R(z) occur when z^3 = 1. Then the distinct poles are

z = 1 , e2iπ/3, e4iπ/3. Let H(z) = z 3 z− 1 , z 7 → βi + 1/w, and βi ∈ { 1 , e2iπ/3, e4iπ/3}.

H( 1 + 1/w) = w 3

w 3 (3w^2 + 3w + 1 ) H(e2iπ/3^ + 1/w) = w 3e2iπ/^

w 3e2iπ/3(3e2iπ/3w^2 + 3e4iπ/3w + 1 ) H(e4iπ/3^ + 1/w) = w 3e4iπ/^

w 3e4iπ/3(3e4iπ/3w^2 + 3e2iπ/3w + 1 ) H(βi + 1/w) = w 3βi − Q(w)