Chapter 8 - Rational, real and complex numbers

Using the formal construction of complex numbers define addition and multiplication on complex numbers

(a,b)+(c,d) = (a+c, b+d) (a,b)×(c,d) = (ac−bd, ad+bc) (think of this as just expansion of the brackets (a +bi) * (x + yi))

(z^n)* =

(z*)^n

(z/n)* =

(z*/n*)

Determine e ^ 2+3i to 3s.f

-First split the complex number power up into powers of a real number and powers of an imaginary number -so e^x * e^yi e^x forms the value of r (modulus) and e^yi forms the argument part -Turn into x + yi to 3s.f

Use the F axioms in your lecture notes (slides) to prove: a x 0 = 0

0 = 0 + 0 by F5 a x 0 = a x (0 + 0) a x 0 = a x 0 + a x 0 by F4 (take away a x 0 from each side) a x 0 - a x 0 = a x 0 + a x 0 - a x 0 0 = a x 0

How do you turn a recurring decimal into a rational number in fraction form?

1) Check that the only part of the decimal is the recurring part if it isn't then multiply the decimal by a power of 10 which ensures only the recurring part is behind the decimal point 2) Let the decimal = x or 10^n x if you had to multiply you decimal in step one 3) Now multiply your decimal by a power of 10 to get one lot of the repeated part before the decimal point and don't forget to multiply your x by the same power of ten 4) Do equation in 3) - equation in 2) and solve for x to get the fraction

What is the method for finding the roots of a complex number using de Moivre's theorem? i.e solving z^n = w

1) Write the z as the general modulus argument form or exponential form and ensure to use de Moivre's theorem to expand power 2) Write w in modulus argument form or exponential form depending on what you used for z 3) Equate moduli to find r 4) Equate arguments to find θ 5) Add 2π/n "n-1" times to find all the arguments of the other roots.

Turn 1.54˙ into a fraction

10x = 15.4˙ 100x = 154. 4˙ So 90x = 139 so x = 139 / 90

z + z* =

2Re(z)

z - z* =

2iIm(z)

Which of the real numbers are rational?

A real number a0.a1a2a3 . . . is rational if and only if it is repeating decimals i.e. ∃k , t ∈ N : ∀n ≥ k we have ak +t = ak e.g 3/7 is rational and 3/7 = 0.428571428571428571... so 428571 repeats hence it is rational

Define a field?

A structure which follows all the F axioms such as the rational number or the prime numbers

state the important identity in exponential form when r =1 and θ = π

As e^ iπ = cos π + isin π = -1 so e^ iπ = -1 so e^ iπ + 1 = 0

Formally construct the complex numbers

C = {(x,y) | x ∈ R, y ∈ R} Note where x is the real part and y is the imaginary part so (x,0) is just real

Define a more common construction of complex numbers

C = {x+iy | x,y∈R} So a complex number is in the form z = x + yi

Use the F axioms in your lecture notes (slides) to prove: If ab = 0 then a = 0 or b = 0

CASE 1: b = 0 -then we have proven it CASE 2: b ≠ 0 -so by F6 we can say b^-1 exists - abb^-1 = 0b^-1 which is a = 0 so proven

Define an equivalence relation (that does not use division) to represent the rational numbers

Define a relation Q on Z × (Z \ {0}) by (a, b)Q (c, d) ⇔ ad = bc (this is basically saying a/b = c/d)

How do you divide complex numbers in polar form

Divide the moduli Subtract the arguments

What does the fundamental theorem of algebra tell you?

For a polynomial of degree n we have n roots now as now we have defined complexed roots we can find them.

State *Euler's Formula*

For a real number θ we have e^iθ =cosθ+isinθ.

Prove the real numbers are not countable:

Here we have shown there exists a number in the reals which is not mapped from a natural number hence |N| ≠ |R|

Give the relationship between the arguments of the roots and the modulus of the roots and what does this look like on an argand diagram?

If you are finding n roots, the arguments will always differ by 2π/n and modulus will always be the same. If you plot this on an argand diagram you will form a regular polynomial with n vertices which lie on a circle with radius of the modulus of roots e.g for the previous example with 3 roots (z^3 = 4√ 3 + 4i)

State De Moivre's theorem

If z = R (cosθ + i sinθ), then for any natural number n we have z^n = R^n [cos(nθ) + i sin(nθ)]

What does addition of complex numbers look like geometrically? e.g (1+3i)+(2+i)

It looks like the addition of two vectors e.g = (3 + 4i)

How would you derive a formula for cos(4θ) from cosθ?

Let z = cosθ + isinθso z^4 = (cosθ + isinθ)^4 Using the binomial expansion expand this Using de Moivre's theorem also expand Making the two expansions equal to each other and equate real and imaginary parts to find cos 4θ you only care about the real parts If it was find sin 4θ then you would care about the imaginary parts

Prove the set of all subsets of real numbers is bigger than the number of real numbers

Lets look at another structure called S which has similar properties to R and assume for contradiction that there exists f: S--> ρ(S)

How do you multiply in polar form?

Multiply the moduli Add the arguments

How do you perform division of complex numbers? e.g z/w?

Multiply top and bottom by the complex conjugate of denominator z/w = 1/ww* x z x w*

How can complex numbers be represented geometrically? e.g 3 + 2i

On an argand diagram -the horizontal axis represents the real parts; -the vertical axis represents the imaginary parts

If a polynomial has real coefficients and a complex root, what can you deduce about another root of the polynomial?

One other root will be the complex conjugate of the complex root

Prove the relation you just defined for the rational numbers is an equivalence relation

Reflexive xQx? (a, b)Q (a, b) and ab = ab so true Symmetric xQy = yQx? (a, b)Q (c, d) is ad = bc and (c, d)Q (a, b) is cb = da These are the same so true Transitive assume xQy and yQz is true, is xQz true? -Assume (a, b)Q (c, d) is true and (c, d)Q (e, f) is also true (Our aim is to show (a, b)Q (e, f) so af = be) - So ad = bc and cf = de - Multiply these equations together adcf = bcde - We know d ≠ 0 so we can say acf = bce - Now look at c CASE 1: c ≠ 0 - so we can say af = be which is our aim so transitivity proven CASE 2: c = 0 -then ad = b x 0 so ad = 0 and we know d ≠ 0 so a = 0 -also de = f x 0 so de = 0 and we know d ≠ 0 so e = 0 -So a = 0, e = 0 so af = 0 = be hence af = be so transitivity proven for case 2

Define roots of unity

The n solutions to the equation Z^n = 1 where Z is a complex number

In the model of real numbers we said that 0 = -0, why is it important we define this?

The real numbers are a field, for it to be a field the only unique additive identity is 0 hence if -0 works in the same way as 0 then 0 = -0

3) Define a relation R on Z by: xRy if 31|(x2 − y2). Show that Ris an equivalence relation.Find the equivalence classes [0]R and [1]R containing 0 and 1respectively. How many equivalence classes are there?

There are 16 equivalence classes because if you go through every element 0 to 31 for [0] there is a elements 0 for [1] there is elements 1 and -1 for [2] there is elements 2 and -2 ... So you get 15 classes looking at elements 1-31 and then 1 class for 0 so 16 classes

What are the real numbers?

These are the set or rational and irrational a0 . a1a2a3 ..., where a0 ∈ N ∪ {0} and a1,a2,··· ∈ {0,...,9}is a non-negative real number To get negative real numbers, we put a − in front. We say −0.000... = 0.000....

What can you say about the arguments for the roots of unity? So what is the pattern for the n roots of unity.

They are all multiples of 2π/n so each argument is some constant, k, multiplied by 2π/n Modulus is always 1 and argument will be 0 , 2π/n , 4π/n.... so roots 1 , e^(2π/n), e^(4π/n)i, e^(6π/n)i , e^((n-1)π/n)i

How do you represent complex numbers in polar form (modulus argument form)? e.g find the polar form of the general complex number a + bi

They exist in the form z = R· (cosθ+isinθ) The length R = a^2 + b^2 is called the modulus of z, and denoted by |z| θ is the angle (argument) the complex number makes with the real axis and its range is −π < θ ≤ π (when the argument is in this range it is called the principle argument of z)

Define addition on the equivalence relation for rational numbers which you have created

Think that a/b + c/d = ad + bc / bd So [(a, b)] ⊕ [(c, d)] = [(ad + bc, bd)]

Define multiplication on the equivalence relation for rational numbers which you have created

Think that a/b x c/d = ac / bd So [(a, b)] ⊗ [(c, d)] = [(ac, bd)]

Prove the size of the rationals is the same size as the natural numbers (i.e you can count the rational numbers)

This is showing that given we can count Z x (Z\{0}) we can find the different classes of Z x (Z\{0}) which are all the different rational numbers and count them. (i.e skipping any [(a,b)]Q which is the same). So if there is a bijection between the natural numbers and Z x (Z\{0}), then there is a bijection between the natural numbers and the rationals Note go to page 149 to explain what diagonal counting is (the reason we count diagonally is because if we counted in lines we would go on forever)

Prove the following theorem: Suppose x and y are any two real numbers with x < y. Then there is a rational number q such that x < q < y

This proof is based on the idea that if x and y are more than two integers apart from one another then there is definitely an integer in between them which will be rational

How do raise a complex number to a real number? e.g 3^(2 + 3i)

Turn the base into an exponential and then treat as e^(a + bi)

Prove that if A and B are countable infinite sets then `A ∪ B is also a countable set

Ways of adjusting the function h: -Instead of using set B turn B into B\A (B without A0 or -Define a rule that says "skip over counted elements" so if I have already counted a P from A then don't count again in set B

Prove the cartesian product of the natural numbers in countable i.e there is a bijection function f such that f: N --> N X N

We can count N x N using diagonal counting and we know this is definitely injective and is surjective because if you were to ask where does f(n) = (p,q) come up it is at the (p + q - 1 ) diagonal so definitely surjective

What does the inverse of a complex number do?

When multiplied to the complex number it equals the multiplicative identity which is 1 so z x z^-1 = 1

When does z = z*

When z is real (so no imaginary part)

Show the size of the infinite natural set is equal to the size of the infinite integer set (i.e you can count the integers)

You need to show bijection between the natural numbers and integers for them to have the same size Create a function where all odd natural numbers go to the non-negative integers and all the even natural numbers go to the negative integers. This is clearly surjective and injective so is bijective hence |N| = |Z|

If a ≠ 0 how would you define [(a, b)]^-1?

[(b, a)]

How would you define -[(a, b)]?

[(−a, b)]

What is a rational number?

a/b where a ∈ Z and b ∈ Z \ {0}.

(cos θ + i sin θ)^n =

cos(nθ) + i sin(nθ)

4) Show that, for any set of 20 integers a1, . . . , a20, there is some pair (ai , aj ) such that 31|(ai2 − aj2 )

map f: {a1... a20} --> equivalence class of R so ai ---> [ai] by pigeon hole Principle there are 20 elements and only 16 classes so there will be ai and aj where i ≠ j such that [ai] = [aj]

State the multiplicative inverse of w i.e what is w^-1?

w*/ww*

zz* = Where z = x + yi

x^2 + y^2 (real) = |z|^2

(z*)* =

z

In polar form what is the general inverse for a complex number? ie. find the inverse of z = R· (cosθ+isinθ)

z = 1/R· (cos-θ+isin-θ)

Given z = R· (cosθ+isinθ) write down the complex conjugate

z = R· (cos-θ+isin-θ)

Give the general solution to z = -a^2

z = i√a or z = i-√a

(z + w)* =

z* + w*

z = x + yi state the complex conjugate

z* = x - yi

(z x w)* =

z* x w*