MAT 3121: Complex Analysis I2016)Jan11.pdf · Complex Analysis I Introduction C Let us follow the same scheme as with the construction of the other number systems, i.e., let us enlarge

MAT 3121:Complex Analysis I

Introduction

MAT 3121: Complex Analysis I

Paul-Eugene ParentDepartment of Mathematics and Statistics

University of Ottawa

January 11th, 2014


Introduction

Outline

1 Introduction


Introduction

Introduction: The Complex Numbers

Recall that

• The natural numbers are N = {0, 1, 2, 3, ...}.Unfortunately the equation x + a = b doesn’t alwaysadmit a solution within N. For example: x + 2 = 0.Hence we need to expand the system of numbers thatwe are working with, i.e.,

• The integers Z = N× N/ ∼, where (n,m) ∼ (q, p) ifand only if n + p = m + q. Of course we have a naturalinclusion

N −→ Zn 7→ (n, 0).


IntroductionExercise: Show that this equivalence relation induces alegitimate abelian group structure.

Again the equation ax = b doesn’t always admit a solutionwithin Z. For example: 2x = 1. Hence again we need toexpand the system of numbers that we are working with, i.e.,

• The rational numbers

Q =

{p

q

∣∣∣ p ∈ Z, q ∈ Z− {0}}/

∼, wherep

q∼ a

bif

and only if bp = aq. Of course we have a naturalinclusion

Z −→ Qn 7→ n

1.


Introduction

Unfortunately again not all equations ax2 = b admit asolution within Q. For example: x2 = 2. Again we need toexpand the system of numbers that we are working with,

i.e.,

• The real numbers

R = {(xn) | (xn) is a Cauchy sequence in Q}/ ∼,

where(xn) ∼ (yn) ⇐⇒

∀ε > 0, ∃N ∈ N | n ≥ N ⇒ |xn − yn| < ε.


Introduction

We have a natural inclusion

Q −→ Rr 7→ (r),

i.e., the constant sequence at r ∈ Q.


Introduction

Some properties of R

Recall that (R,+, ∗) is a complete field, i.e., every Cauchysequence in R converges in R. Moreover the natural order onR is a total order compatible with the field structure, i.e.,

1 a ≤ a,

2 a ≤ b and b ≤ a implies a = b,

3 a ≤ b and b ≤ c implies a ≤ c ,

4 a, b ∈ R implies that either a ≤ b or b ≤ a,

5 a ≤ b implies a + x ≤ b + x for all x ∈ R, and

6 a, b ≥ 0 implies ab ≥ 0.


Introduction

But for all our cunning, scheming, plotting, ... the realnumbers are still not “strong” enough to solve the simpleequation

x2 + 1 = 0.

Is there any hope to construct a set of numbers large enough(and with “good” properties) such that within that systemthe equation

p(x) = 0

admits a solution where p(x) is any non-constant polynomialwith integer coefficients?


Introduction

C

Let us follow the same scheme as with the construction ofthe other number systems, i.e., let us enlarge R and then“kill” what we are missing. We start with the polynomialring R[x ] and we “kill” what needs to be zero, i.e., x2 + 1.We get

C = R[x ]/(x2 + 1),

where (x2 + 1) is the ideal generated by x2 + 1. Because ofthe Euclidean algorithm C ∼= R2 as real vector spaces. Thetwo operations on C can be described explicitly :

• (x , y) + (u, v) := (x + u, y + v), and

• (x , y) · (u, v) := (xu − yv , xv + yu).


IntroductionTheorem(C,+, ·) is a complete field.

Proof.The fact that C is complete comes from the fact that R2 iscomplete. The remaining field axioms are tedious but easyverifications. You can easily show that (0, 0) is “the” zero ofC and that (1, 0) is “the” 1 of C.Let’s show that every (x , y) 6= (0, 0) is invertible.

Here is the candidate for (x , y)−1, namely1

x2 + y2(x ,−y).

(x , y) · 1

x2 + y2(x ,−y) =

1

x2 + y2(x2 + y2, 0)

= (1, 0)


IntroductionThere is a natural inclusion

R −→ Cx 7→ (x , 0)

such that

• (x , 0) + (x ′, 0) = (x + x ′, 0), and

• (x , 0) · (x ′, 0) = (xx ′, 0).

Historically we write

1 := (1, 0) and i := (0, 1).

Hence a complex number z ∈ C is written as

z = x + iy for some unique x , y ∈ R.


Introduction

Let z = x + iy ∈ C.

Definition

• The real part of z is Re(z) := x , and

• the imaginary part of z is Im(z) := y .

Now notice that

i2 = (0, 1) · (0, 1) = (−1, 0) = −1.

Hence in C the equation z2 + 1 = 0 admits a solution,namely z = i .

Remark: z = −i is another such solution.


Introduction

The natural question now begs to be asked!!! i.e., given apolynomial

p(z) = anzn + an−1z

n−1 + ...+ a1z + a0

with all ai ∈ C, an 6= 0, and n ≥ 1, is there zo ∈ C such that

p(zo) = 0?

Giving the answer to this question is one the main objectivesof this course. The answer is yes and the statement isknown as the Fundamental Theorem of Algebra.


Introduction

The price we pay

Once we prove the Fundamental Theorem of Algebra we’llhave achieved our objective of finding a set of numbers largeenough such that every polynomial equation admits asolution.

Unfortunately we had to pay a price! C does not carry anatural total order compatible with its field structure.Asking a question like “is the complex number z big, small,positive, ...?” makes no sense!


Introduction

Order

Definition A field k is said to be ordered if there is a subsetP of k, called the positive elements of k, such that

1 if α, β ∈ P, then both α + β and αβ belong to P; and

2 for every α ∈ k, one and only one of the followingpossibilities happens

α ∈ P, α = 0, or − α ∈ P.


Introduction

In that case, we can define a total order on k by declaringx < y if and only if y − x ∈ P. Indeed, because of condition(2), for every x , y ∈ k one and only one of the threefollowing possibilities always happens

x < y , x = y or y < x .

Examples The fields Q and R are both ordered.


Introduction

Suppose C could be ordered, i.e., according to the definition,there would be a subset P of C called the “positive complexnumbers”. Then i 6= 0 would mean that either i ∈ P or−i ∈ P. In either case, it would mean that

i2 = (−i)2 = −1 ∈ P.

But then we would arrive at the contradiction

1 = (−1)2 ∈ P.


Introduction

Exercise

Find all solutions to the equation z2 = 1 + i .

Solution: We are searching for all z = x + iy such that

z2 = x2 − y2 + 2ixy = 1 + i ,

i.e., we need to solve simultaneously in R the system

x2 − y2 = 1 and 2xy = 1.

Remark: We deduce that xy > 0, i.e., both x and y musthave the same sign.


Introduction

Let’s expand (x2 + y2)2, i.e.,

(x2 + y2)2 = x4 + 2x2y2 + y4

= x4 − 2x2y2 + y4 + 4x2y2

= (x2 − y2)2 + (2xy)2

= 1 + 1 = 2.

Since in R, x2 + y2 ≥ 0, then x2 + y2 =√

2. Hence we needto solve the system

x2 + y2 =√

2 and x2 − y2 = 1,

under the requirement that xy > 0.


Introduction

Hence

x2 =1 +√

2

2and y2 =

√2− 1

2.

Since both x and y must be of the same sign, we get twosolutions, i.e.,

z1 =

√1 +√

2

2+ i

√√2− 1

2and z2 = −z1.

Question: Could there be other solutions?


Introduction

Exercise at home

Given a fix zo ∈ C and mimicking the preceding exercise,find a general scheme to solve

z2 = zo .


Introduction

Basic definitions

Recall that C ∼= R2 as a real vector space, i.e., we canrepresent complex numbers by points in the plane.


Introduction

Polar representation

We can talk about the length of a complex numberz = a + ib as the euclidian norm of the corresponding vectorin R2, i.e., the modulus of z is

|z | :=√

a2 + b2.

For every non zero z = a + ib, there is a unique θ ∈ [0, 2π)such that

cos(θ) =a

|z |and sin(θ) =

b

|z |,

hence the polar representation of z is

z = |z |(

cos(θ) + i sin(θ)).


Introduction

Argument

We can be more flexible with the values that θ can assume.Indeed, given a fixed y ∈ R, we can define a function, calledargument function,

argy : C− {0} −→ [y , y + 2π)

z 7→ argy (z).

Choosing such an interval [y , y + 2π) is known as choosing abranch of the argument.

• The θ of the preceding slide corresponds to arg0.

• The Principal Branch is arg−π corresponding to theinterval [−π, π).


Introduction

Complex conjugate

The following function

C −→ Cz = a + ib 7→ z := a− ib

is known as the complex conjugate of z . One can show

• zz = |z |2;

• z + z ′ = z + z ′;

• zz ′ = z · z ′;

• if z ′ 6= 0, then( z

z ′

)=

z

z ′;

• z = z if and only if z ∈ R; and z = z .


IntroductionNotice that

Re(z) =z + z

2and Im(z) =

z − z

2i.

Moreover, if z 6= 0, then z−1 =z

|z |2.

Proposition

• |zz ′| = |z | · |z ′|;

• if z ′ 6= 0, then∣∣∣ zz ′

∣∣∣ =|z ||z ′|

;

• |Re(z)| ≤ |z | and |Im(z)| ≤ |z |;• |z | = |z |;

• |z + z ′| ≤ |z |+ |z ′|; and |z − z ′| ≥∣∣∣|z | − |z ′|∣∣∣.

Documents

MAT 3121: Complex Analysis I2016)Jan11.pdf · Complex Analysis I Introduction C Let us follow the same scheme as with the construction of the other number systems, i.e., let us enlarge