1 Complex Numbers

1.1 Sums and Products

Definition: The complex plane, denoted C is the set of all ordered pairs (x, y) with x, y ∈ R,where Re z = x is called the real part and Imz = y is called the imaginary part.

• We call the x-axis the real axis.

• We call the y-axis the imaginary axis.

• Usually we write z = (x, y) or z = x+ iy, this is called rectangular form.

• For two complex numbers, z1 = (x1, y1) and z2 = (x2, y2) we have z1 = z2 if and only if

x1 = x2 and y1 = y2

For z1 = (x1, y1) and z2 = (x2, y2), we define addition as

z1 + z2 = (x1, y1) + (x2, y2) = (x1 + x2, y1 + y2)

and multiplication as

z1z2 = (x1, y1)(x2, y2) = (x1x2 − y1y2, y1x2 + x1y2)

How do the real numbers interact here?

• For any r ∈ R we have (r, 0) ∈ C, and

(r, 0) + (s, 0) = (r + s, 0)

and(r, 0)(s, 0) = (rs, 0)

• For i = (0, 1) ∈ C we havei2 = (0, 1)(0, 1) = (−1, 0)

Now we can see that multiplication is well-defined, since

z1z2 = (x1 + iy1)(x2 + iy2) = x1x2 + i2(y1y2) + i(y1x2 + x1y2) = (x1x2 − y1y2, y1x2 + x1y2)

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1.2 Basic Algebraic Properties

• Complex numbers are associative

z1 + (z2 + z3) = (z1 + z2) + z3

and commutativez1 + z2 = z2 + z1

so we can rearrange and drop parenthesis.

• We have an additive identity 0 = (0, 0) ∈ C, and mult. identity 1 = (1, 0) ∈ C. Checkthese!

• For any z = x+ iy we have an additive inverse −z = −x− iy since

x+ iy + (−x− iy) = 0.

For z = x + iy non-zero we have a multiplicative inverse z−1 = u + iv so that zz−1 = 1defined by

(x+ iy)(u+ iv) = 1

and soux− vy + i(uy + vx) = 1

therefore ux− vy = 1 and uy + vx = 0. Solving the linear equation, we get

z−1 =

(x

x2 + y2,−y

x2 + y2

)• If z1z2 = 0 then z1 = 0 or z2 = 0. To see this, suppose z1 6= 0, then

z2 = z2 · 1 = z2 · z1 · z−11 = 0 · z−11 = 0

• Some interesting things happen with quotients, for example

1

i=

1

i· ii

=i

−1= −i

and4 + i

2− 3i=

(4 + i)(2 + 3i)

(2− 3i)(2 + 3i)=

5 + 14i

13=

5

13+ i

14

13.

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1.3 Vectors and Moduli

We can consider z = x+ iy = (x, y) as a vector on the complex plane.

Then the sum can also be interpreted vectorially, as follows.

Definition: The modulus of a complex number z = x + iy gives the distance from z to theorigin, and is given by

| z |=√x2 + y2

extending the notion of absolute value for the reals, and

| z |2= (Re z)2 + (Im z)2

now gives a relation between real numbers. From here

• | z |=| −z |

• Re z ≤| Re z |≤| z | and Im z ≤| Im z |≤| z |

• Note that z1 < z2 is meaningless, but | z1 |<| z2 | has meaning.

Consider the difference

| z1 − z2 |=| (x1 − x2) + i(y1 − y2) |=√

(x1 − x2)2 + (y1 − y2)2

What does this mean geometrically?

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Examples:

1. Consider| −3 + 2i |=

√9 + 4 =

√13

and| 1 + 4i |=

√1 + 16 =

√17

What does this mean, geometrically?

2. The equation | z |= 1 is the unit circle centered at the origin.

3. The equation | z − 1 + 3i |= 2 is the circle with center z = (1,−3) and radius 2.

4. Exercise: Show that | z1 · z2 |=| z1 | · | z2 |.

5. Exercise: Show that | z2 |=| z |2.

1.4 Triangle Inequality

Definition: The triangle inequality is given by

| z1 + z2 |≤| z1 | + | z2 | .

Illustrated by figure 3. Some consequences:

• We have | z1 + z2 |≥|| z1 | − | z2 ||, to see this, observe

| z1 |=| (z1 + z2) + (−z2) |≤| (z1 + z2) | + | −z2 |

and hence| (z1 + z2) |≥| z1 | − | z2 | if | z1 |≥| z2 |

and| (z1 + z2) |≥ −(| z1 | − | z2 |) if | z1 |<| z2 | .

What does this mean geometrically? Length of one side is greater than difference of othertwo.

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Examples:

1. Suppose z is on the circle | z |= 1. Then

| z − 2 |=| z + (−2) |≤| z | + | 2 |= 1 + 2 = 3

and| z − 2 |=| z + (−2) |≥|| z | − | −2 ||=| 1− 2 |= 1.

This means the distance from z to 2 is between 1 and 3.

2. The triangle inequality can be extended inductively, that is

| z1 + z2 + ...+ zn |≤| z1 | + | z2 | +...+ | zn |

for any n. So for example, for z on the circle | z |= 2, we have

| 3 + z + z2 |≤ 3+ | z | + | z2 |= 3+ | z | + | z |2≤ 9.

3. Consider the degree n polynomial given by

P (z) = a0 + a1z + a2z2 + ...+ anz

n

where n ∈ Z+ and a0, ..., an ∈ C. We will show that for some positive R ∈ R,∣∣∣∣ 1

P (z)

∣∣∣∣ < 2

anRn

whenever | z |> R. Morally: Reciprocal 1/P (z) is bounded from above when z is outsideof the circle | z |= R. Recall,

• | z1 · z2 |=| z1 | · | z2 |• | z2 |=| 2 |2

and write

w =P (z)− anzn

zn=a0zn

+a1zn−1

+ ...+an−1z

.

ThenP (z) = (an + w)zn

Then the equation above becomes

wzn = a0 + a1z + ...+ an−1zn−1

and hence| w || zn |≤| a0 | + | a1z | +...+ | an−1zn−1 |

implying

| w |≤ | a0 || z |n

+| a1 || z |n−1

+ ...+| an−1 || z |

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Now pick R large enough so that

| ai || z |n−i

<| an |2n

when | z |> R. Then,

w < n · | an |2n

=| an |

2whenever | z |< R.

Consequently,

| an + w |≥|| an | − | w ||>| an |

2

and hence

| P (z) |=| an + w | · | zn |> | an |2·Rn.

1.5 Complex Conjugates

Definition: The complex conjugate of a number z = x+ iy is given by

z = x− iy

Some observations:

• | z |=| z |

• z = z.

• z1 + z2 = z1 + z2, since

z1 + z2 = x1 + iy1 + x2 + iy2 = xi + x2 − i(y1 + y2) = x1 − iy1 + x2 − iy2 = z1 + z2.

• z1 − z2 = z1 − z2

• z1 · z2 = z1 · z2

•(z1z2

)= z1

z2

• z + z = 2Re z.

• z − z = 2iIm z.

• zz = (x+ iy)(x− iy) = x2 + y2 =| z |2.

Examples:

1. −1+3i2−i = (−1+3i)(2+i)

(2−i)(2+i) = −5+5i|2−i|2 = −5+5i

5 = −1 + i

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2. We can use conjugates to establish modulus properties. We know zz =| z |2. Therefore,we obtain

| z1 · z2 |2 = (z1 · z2)(z1 · z2)= z1 · z2 · z1 · z2= z1 · z1 · z2 · z2= | z1 |2 · | x2 |2

= (| z1 | · | z2 |)2

from which we conclude that | z1 · z2 |=| z1 | · | z2 |, since modulus is always positive.

1.6 Exponential Form

Let r > 0 and θ be polar coordinates for z = (x, y), then

z = r(cos θ + i sin θ)

then| z |= r

√cos2 θ + sin2 θ = r

Definition: The value θ = arg(z) is an argument for z, and it is the angle z = (x, y) makeswith the real axis. Infinite arguments exist for z, namely,

θ + 2π, θ + 4π, ..., θ + 2kπ, ...

but we call Θ = Arg(z) with −π < Θ ≤ π the principal argument of z.

Examples:

1. The complex number −1− i has

Arg(−1− i) = −3π

4

but

arg(−1− i) =5π

4+ 2kπ

for n ∈ Z.

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Definition: The symbol eiθ is defined by Euler’s formula as

eiθ = cos θ + i sin θ

and hencez = reiθ = r(cosθ + i sin θ)

Examples:

1. Writing −1− i in exponential form we have

−1− i =√

2ei−3π4

or−1− i =

√2ei(

−3π4

+2πk)

2. The set of numbers z = eiθ are the set of numbers on the unit circle. Accordingly,

eiπ = −1 and eiπ/2 = i and ei4π = 1.

Also,z = Reiθ where 0 ≤ θ ≤ 2π

is a parametrization of the unit circle, moving counter-clockwise, and

z = z0 +Reiθ where 0 ≤ θ ≤ 2π

moves the circle to be centered at z0.

Some basic properties of exponential forms:

• eiθeiθ′ = ei(θ+θ′)

• reiθ · r′eiθ′ = (rr′) · ei(θ + θ′)

• reiθ

r′eiθ′= r

r′ ei(θ−θ′)

• If z = reiθ then

z−1 =1

reiθ=

1

re−iθ

andzn = rneinθ.

Examples:

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1. From the properties above we see that

arg(z1z2) = arg(z1) + arg(z2).

but this doesn’t necessarily hold when arg is replaced with Arg. For example, z1 = −1and z2 = i, then

Arg(−1) = π and Arg(i) =π

2so

Arg(−1) + Arg(i) = π +π

2=

3π

2

but

Arg(−1 · i) = Arg(−i) =−π2

2. Find the principal argument ofi

−1− i.

We already know that i = 1eiπ2 and −1− i =

√2e−i

3π4 and so

i

−1− i=

eiπ2

√2e−i

3π4

=1√2ei(π/2+3π/4)

Therefore

arg

(i

−1− i

)=π

2+

3π

4=

5π

4

and hence

Arg

(i

−1− i

)=

5π

4− 2π =

−3π

4

3. Let’s write the expression (−1 + i)7 in rectangular form. First, for z = (−1 + i) we have

| z |=√

1 + i =√

2

and

Arg(z) =3π

4

so −1 + i =√

2ei3π4 , and

(−1 + i)7 =(√

2eii3π4

)7= 8√

2ei21π4 = 8

√2ei5πe

iπ4 = −8

√2e

iπ4

But we can easily see that

(1 + i) =√

2eiπ4

and so(−1 + i)7 = −8(1 + i)

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4. When r = 1, for any value n ∈ Z, we have(eiθ)n

= einθ

but this means(cos θ + i sin θ)n = cosnθ + i sinnθ

this is known as De Moivre’s formula.

5. Using the above with n = 2, we reclaim known identities

cos2 θ − sin2 θ + 2i cos θ sin θ = (cos θ + i sin θ)2 = cos 2θ + i sin 2θ.

Now equating real and imaginary parts, we have

cos 2θ = cos2 θ − sin2 θ and sin 2θ = 2 cos θ sin θ

1.7 Roots of Complex Numbers

Two non-zero complex numbers can be arrived at in infinitely many ways, for example

z = reiθ and z′ = r′eiθ′

are equal precisely whenr = r′ and θ = θ′ + 2πk

for k ∈ Z. We also know that zn = rneinθ, so we should know something about nth roots.Definition: An nth root of z0 ∈ C is a complex number z = reiθ such that

zn = z0

that isr0e

iθ0 = rneinθ,

and hencer0 = rn and nθ = θ0 + 2πk.

• Since r0 is real, this just means r = n√r0.

• For k ∈ Z, we know that

θ =θ0n

+2πk

n

So the nth roots of z0 are given by the infinite family

z = n√r0exp

[i

(θ0n

+2πk

n

)]all lying on the circle | z |= n

√r0.

So n of them should be more special than the others, namely, those with

ck = n√r0exp

[i

(θ0n

+2πk

n

)]and k ∈ 0, 1, 2, ..., n− 1.

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• When z0 is a positive real then n√r0 means the familiar unique positive root.

• When θ0 = Arg(z0) then we call c0 the principal root.

• When z0 is a positive real, its principal root is n√r0.

Examples:

1. Let’s find all values of (−16)14 , that is, the 4th roots of −16. Since

−16 = 16ei(π+2kπ)

so the roots are just

ck = 2ei(π4+ kπ

2 )

where k = 0, 1, 2, 3. So we have

ck = 2ei(π4 )ei(

kπ2 ) = 2

(cos

π

4+ i sin

π

4

)ei(

kπ2 ) = 2

(1√2

+ i1√2

)ei(

kπ2 ) =

√2(1 + i)ei(

kπ2 )

and therefore

c0 =√

2(1 + i) · 1 =√

2(1 + i)

c1 =√

2(1 + i) · i =√

2(−1 + i)

c2 =√

2(1 + i) · −1 =√

2(−1− i)c3 =

√2(1 + i) · −i =

√2(1− i)

2. Let’s examine the nth roots of 1, called the nth roots of unity. We start with

1 = 1 · ei(0+2πk)

where k ∈ Z. Then,

ck =n√

1 · ei(0+2πkn ) = ei(

2πkn )

where k ∈ {0, 1, 2, ..., n − 1}. When n = 2 we get c0 = 1 and c1 = −1. For larger n, weget roots lying on the regular n-gon inscribed in the circle | z |= 1.

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1.8 Regions in the Complex Plane

Definition: We define an ε neighborhood around a given point z0 by

| z − z0 |< ε

we can also define the deleted neighborhood, that is the neighborhood minus the point z0 itself,by

0 <| z − z0 |< ε

For a given set S contained in C,

• A point z0 is called interior if there exists a neighborhood containing only z0 and pointsof S.

• A point z0 is called exterior if there exists a neighborhood containing z0 and no points inS.

• If z0 is neither of these, then it is a boundary point.

• A set is open if it contains none of its boundary points.

• A set is closed if it contains all of its boundary points.

• A set is a closure of S if it contains S plus its boundary points. For example, | z |≤ 1 isthe closure of | z |< 1.

• Some sets like 0 <| z |≤ 1 is neither open nor closed.

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• An open set S is connected if every pair of points in S can be joined by a polygonal line.

• A domain is an non-empty open connected set.

• A domain with some, none, or all of its boundary points is called a region.

• A set S is bounded if every point of S lies in some circle | z |< R.

Examples:

1. Consider the set Im(1z

)> 1 in relation to these properties, for z 6= 0 we have

1

z=

z

zz=

z

| z |2=

x− iyx2 + y2

so we want

Im

(1

z

)=

−yx2 + y2

> 1

and thereforex2 + y2 + y < 0.

Completing the square, we get

x2 + y2 + y +1

4<

1

4

which means

(x− 0)2 +

(y +

1

2

)2

<

(1

2

)2

.

This is the region whose boundary is the circle centered at x = 0 and y = −1/2 withradius 1/2.

• A point z0 is an accumulation point, or limit point, of a set S if each deleted neighborhoodof z0 contains at least one point of S.

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