discrete mathematics i - complex numberscompalg.inf.elte.hu/~vatai/dm1en-org/part1_complex/... ·...

Discrete mathematics I - Complex numbers

Discrete mathematics I - Complex numbers

Emil Vatai <[email protected]>(based on hungarian slides by László Mérai)1

January 31, 2018

1Financed from the financial support ELTE won from the Higher EducationRestructuring Fund of the Hungarian Government.

Discrete mathematics I - Complex numbersComplex numbers

Introduction and historical background

Cubic equation

Solving the cubic equationFor a 6= 0, we are looking for the solution ofax3 + bx2 + cx + d = 0 By dividing with a we obtain a monicpolynomial on the lhs:

x3 + b′x2 + c ′x + d ′ = 0 (1)

Reminder: Solving x2 + px + q = 0With the substitution x = y − p

2 we get y2 + q′ = 0. From this wecan easily obtain the solution by rearranging and taking the squareroots. By substituting x = y − b/3 into (1)

y3 + py + q = 0 (2)



Solving y 3 + py + q = 0

Idea: try to find the solution in a y = u + v form!By rearranging (u + v)3 = u3 + 3u2v + 3uv2 + v3 we obtain(u + v)3 −3uv(u + v) −(u3 + v3) = 0

y3 +py +q = 0Our goal: find u, v for which −3uv = p, −(u3 + v3) = q. Theny = u + v will be a solution!

Finding u, vu3v3 = (−p

3 )3 and u3 + v3 = −q, u3, v3 will be the roots of thequadratic equation z2 + qz + (−p

3 )3 = 0. Taking the cubic root ofthe solutions of the previous equation:

y = 3

√√√√−q2 +

√(q2

)2+(p3

)3+ 3

√√√√−q2 −

√(q2

)2+(p3

)3(3)



Example: solve y 3 − 21y + 20 = 0

It is pretty obvious that y = 1 is a solution.

Using the formulaSubstitution p = −21, q = 20 into (3)

x = 3

√√√√−q2 +

√(q2

)2+(p3

)3+ 3

√√√√−q2 −

√(q2

)2+(p3

)3

x = 3√−10 +

√−243 + 3

√−10−

√−243

We have negative −243 under the square root! Can we continueusing the formula like this?



Calculating x = 3√−10 +

√−243 + 3

√−10−

√−243

Let’s (formally) allow (√−3)2 = −3:

(2 +√−3)3 = 23 + 3 · 22 ·

√−3 + 3 · 2(

√−3)2 + (

√−3)3

= −10 + 9√−3 = −10 +

√−243

So −10 +√−243 = (2 +

√−3)3 and similarly

−10−√−243 = (2−

√−3)3

The solution is: x = (2 +√−3) + (2−

√−3) = 4.

I Is this formal calculation with√−3 allowed?

I Why do we calculate the value of −10 +√−243 like this?

I There was a solution y = 1, what happened to it?I Is there a third root? (There should be!)



Number sets

Natural numbers: N = {0, 1, 2, . . . }There is no x ∈ N such that: x + 2 = 1 i.e. x + 2 = 1 has nosolution because subtraction is not (always) defined in N!

Integers: Z = {. . . ,−2,−1, 0, 1, 2, . . . }Now we have subtraction (x = −1), but x · 2 = 1 has no solutionbecause division is not (always) defined in Z!

Rational numbers: Q ={

pq : p, q ∈ Z, q 6= 0

}Now we have division (x = 1

2), but x2 = 2 has no solution because

square root is not (always) defined in Q!

Real numbers: RNow we have square root (and a irrational numbers), but x2 = −1has no solution, because the square root of negative numbers is(never) defined!



x 2 = −1 can be solved in the set of complex numbers

Application of complex numbersI CS: Computer graphics, geometryI Math: geometry, solving equationsI Physics: quantum mechanics, relativity theory etc.

Introduction of complex numbers (informal definition)Let i be the solution of x2 = −1. We do our calculations as if iwas a variable, substituting i2 = −1. For example:

(1 + i)2 = 1 + 2i + i2 = 1 + 2i + (−1) = 2i .

More generally:

(a + bi)(c + di) = ac − bd + (ad + bc)i (4)


Complex numbers

Definition of complex numbers

Definition (The set of complex numbers)The C is the set of complex numbers, where if z ∈ C, then z is anexpression a + bi with a, b ∈ R. This is the rectangular (orCartesian or algebraic) form of z . Re(z) = a is the real part andIm(z) = b is the imaginary part of z . Important warning:Im(z) 6= bi!

Graphical representation

|z |a = Re(z)

b = Im(z)

z = a + bi


Complex numbers

Opertaions on complex numbers

Definition (Operations on C)

I Addition: (a + bi) + (c + di) = a + c + (b + d)i .I Multiplication: (a + bi)(c + di) = ac − bd + (ad + bc)i .I Equality: complex numbers are equal if both their real and

imaginary parts are equal i.e. a + bi = c + di if a = c andb = d!

Remarks: let z = a + bi (with a, b ∈ R).

I If b = 0 then z is a real number.I If a = 0 then z is a pure imaginary number.


Complex numbers

Alternative (more formal definition) of C

Definition (Complex numbers as ordered pairs)Let C = R× R (Cartesian product), (a, b), (c, d) ∈ C

I Addition: (a, b) + (c, d) = (a + c, d + b);I Multiplication: (a, b) · (c, d) = (ac − bd , ad + bc).

The two definitions are equivalent with i = 0 + 1 · i ≡ (0, 1).a + bi is more convenient when calculating with pencil and paper,(a, b) is used in computer programs.

Theorem (Fundamental theorem of algebra NP)If 0 < n ∈ N,a0, . . . , an ∈ C, where an 6= 0, then for everyexpression a0 + a1x + a2x2 + . . .+ anxn there is a complex numberz ∈ C, such that a0 + a1z + a2z2 + . . .+ anzn = 0.


Complex numbers

Calculation with complex numbers: absolute value

Definition (Additive inverse)In general, y is the additive inverse of x if x + y = 0. For r ∈ Rthe additive inverse is −r .

Theorem (Additive inverse)The additive inverse of z = a + bi ∈ C is −z = −a − bi ∈ C.

Definition (Absolute value)The absolute value of the z = a + bi ∈ C complex number inrectangular form is |z | = |a + bi | =

√a2 + b2

For real numbers: |a| = |a + 0i | =√a2 + 02 =

√a2

Theorem (Statement)If z ∈ C, then |z | ≥ 0, and |z | = 0⇔ z = 0


Complex numbers

Calculation with complex numbers: reciprocal

Definition (Reciprocal, multiplicative inverse)y is the reciprocal (or multiplicative inverse) of x if x · y = 1. Forr ∈ R \ {0} the reciprocal is 1/r sometimes denoted as r−1.

Reciprocal of complex numbers: rationalisationWhat is the rectangular form of 1

1+i ? Idea: rationalisation, i.e.multiply both numerator and denominator with the conjugate:

11 +√2

= 11 +√2· 1−

√2

1−√2

= 1−√2

(1 +√2)(1−

√2)

= 1−√2

12 − (√2)2 = 1−

√2

1− 2 = −1 +√2

Similarly:

11 + i

1− i1− i = 1− i

(1 + i)(1− i) = 1− i12 − i2 = 1− i

1− (−1) = 1− i2 = 1

2−12 i


Complex numbers

Calculation with complex numbers: quotient

ConjugateLet z = a + bi be the rectangular form of a complex number. Theconjugate of z is z = a + bi = a − bi szám.

Theorem (Statement)The reciprocal of z 6= 0 is 1

z = zz·z .

RemarkThe previous statement is correct, since the denominator isz · z = (a + bi)(a − bi) = a2 − (bi)2 = a2 + b2 = |z |2 ∈ R.

Theorem (No zero divisors)z · w = 0⇒ z = 0 vagy w = 0.

Definition (Quotient)The quotient of two complex numbers z

w = z · 1w .


Complex numbers

Calculation with complex numbers: rules

Theorem (Prove them for homework)

1. z = z;2. z + w = z + w;3. z · w = z · w;4. z + z = 2Re(z);5. z − z = 2 Im(z) · i ;6. z · z = |z |2;7. If z 6= 0, then z−1 = z

|z|2 ;8. |0| = 0 and if z 6= 0, then |z | > 0;9. |z | = |z |;10. |z · w | = |z | · |w |;11. |z + w | ≤ |z |+ |w | (triangle inequality theorem).


Complex numbers

Calculation with complex numbers: proof

Theorem (Prove them at homework)...10. |z · w | = |z | · |w |;...

Proof.

|z · w |2 = z · w · z · w = z · w · z · w= z · z · w · w = |z |2 · |w |2

= (|z | · |w |)2

Discrete mathematics I - Complex numbersPolar form of complex numbers

Polar form

Representation of complex numbersOn the complex plane

ϕ

r =|z |

r cosϕ

rsinϕ

z = r(cosϕ+ i sinϕ)

If z = a + bi ∈ C, then Re(z) = a, Im(z) = b.The length of the vector (Re(z), Im(z)) is r =

√a2 + b2 =

√|z |2.

The argument of z 6= 0 complex number is ϕ = arg(z) ∈ [0, 2π)Using trigonometric functions, we can express the coordinates as:

Re(z) = a = r · cosϕ, Im(z) = b = r · sinϕ


Polar form

Polar form of complex numbers

Definition (Polar form)The polar (or tigonometric) form of the non-zero z ∈ C isz = r(cosϕ+ i sinϕ), where r > 0 is the absolute value (or length)of z .

RemarksI 0 does not have a polar form (because it has no argument).I The polar form is not unique:

r(cosϕ+ i sinϕ) = r(cos(ϕ+ 2π) + i sin(ϕ+ 2π))

I z = r(cosϕ+ i sinϕ) is the polar form;I z = a + bi is the rectangular form (a = r cosϕ, b = r sinϕ)

Definition (Argument)The argument of a non-zero z ∈ C is ϕ = arg(z) ∈ [0, 2π), suchthat z = |z |(cosϕ+ i sinϕ).


Polar form

Converting rectangular to polar form

Using the inverse of tanLet a + bi = r(cosϕ+ i sinϕ), which implies a = r cosϕ andb = r sinϕ.If a 6= 0, then tanϕ = b

a , so

ϕ ={

arctan ba if a > 0;

arctan ba + π if a < 0

If a = 0 then we have a pure imaginary number, so

ϕ = sign(b)π2

Note: ϕ is not the argument, because ϕ ∈ [−π2 ,

3π2 ) is possible.

But this can be fixed by adding 2π.


Polar form

Multiplication with complex in polar formLet z ,w ∈ C non-zero complex numbers: z = |z |(cosϕ+ i sinϕ),and w = |w |(cosψ + i sinψ) then their product iszw = |z |(cosϕ+ i sinϕ) · |w |(cosψ + i sinψ) =

= |z ||w |(cosϕ cosψ − sinϕ sinψ + i(cosϕ sinψ + sinϕ cosψ)) == |z ||w |(cos(ϕ+ ψ) + i sin(ϕ+ ψ))

The last equality comes from the trigonometric addition formulas:I cos(ϕ+ ψ) = cosϕ cosψ − sinϕ sinψI sin(ϕ+ ψ) = cosϕ sinψ + sinϕ cosψ

The absolute value of the product is: |zw | = |z ||w |.The argument of the product is:

arg(zw) ={

arg(z) + arg(w) if arg(z) + arg(w) < 2πarg(z) + arg(w)− 2π if 2π ≤ arg(z) + arg(w) < 4π

i.e. the arguments are summed and then reduced by 2π.


Polar form

Multiplication in polar form

Theorem (De Moivre’s formula)Let z ,w ∈ C non-zero numbers z = |z |(cosϕ+ i sinϕ),w = |w |(cosψ + i sinψ), and let n ∈ N. Then

zw = |z ||w |(cos(ϕ+ ψ) + i(sin(ϕ+ ψ));zw = |z |

|w |(cos(ϕ− ψ) + i sin(ϕ− ψ)), if w 6= 0;

zn = |z |n(cos nϕ+ i sin nϕ)

The angles are added/subtracted/multiplied. To get the argument,we have to reduce the new angle by 2π

Geometric interpretationMultiplication with z ∈ C is a combined stretching and rotatingtransformation. The plane is “stretched” by |z |, and rotated byarg(z) around 0.


Polar form

Roots of complex numbers

Example (8-th root of 1)

(1 + i√2

)8=( 1√

2+ i 1√

2

)8=(

cos π4 + i sin π4

)8=

= cos(8 · π4

)+ i sin

(8 · π4

)= cos 2π + i sin 2π = 1

Other number which for which z8 = 1I 1, −1;I i : i8 = (i2)4 = (−1)4 = 1, and similarly −i ;I 1+i√

2 and −1+i√2 ;

I ±i · 1+i√2 also:

(i · 1+i√

2

)8= i8 ·

(1+i√

2

)8= 1 · 1 = 1


Polar form

Calculating the n-th root

Theorem (Equality in polar form)Two complex numbers z = |z |(cosϕ+ i sinϕ) andw = |w |(cosψ + i sinψ) in polar form are equal

|z |(cosϕ+ i sinϕ) = |w |(cosψ + i sinψ)

if |w | = |z | and ψ = ϕ+ k · 2π for some k ∈ Z

Taking the n-th root: Let wn = z (w is unknown):If wn = |w |n(cos nψ + i sin nψ) = |z |(cosϕ+ i sinϕ)then:

I |w |n = |z | ⇒ |w | = n√|z | and

I nψ = ϕ+ k · 2π for some k ∈ Z, i.e.I ψ = ϕ

n + k · 2πn for some k ∈ Z.

If k ∈ {0, 1, . . . , n − 1}, then these are all different numbers.


Polar form

Taking the n-th root

Theorem (Taking the n-th root)Let z = |z |(cosϕ+ i sinϕ), n ∈ N. The values of w ∈ C are then-th root of z if wn = z:

w = n√|z |(

cos(ϕ

n + 2πkn

)+ i sin

(ϕ

n + 2πkn

))k = 0, 1, . . . , n − 1.


Polar form

Example for: taking the n-th root

w = n√|z |(

cos(ϕ

n + 2kπn

)+ i sin

(ϕ

n + 2kπn

)): k = 0, 1, . . . , n−1

Example (Calculation)Let’s calculate the value 6

√1−i√3+i :

1− i =√2(√

22 − i

√2

2

)=√2(cos 7π

4 + i sin 7π4

)√3 + i = 2

(√3

2 + i 12

)= 2

(cos π6 + i sin π

6)

Since 7π4 −

π6 = 19π

12 :

6√

1−i√3+i = 6

√1√2

(cos 19π

12 + i sin 19π12

)=

= 112√2

(cos 19π+24kπ

72 + i sin 19π+24kπ72

): k = 0, 1, . . . , 5


Polar form

Complex roots of unity

n-th roots of unityThe complex numbers ε ∈ C which for which εn = 1 are the n-throots of unity

εk = ε(n)k =

(cos 2kπn + i sin 2kπ

n

): k = 0, 1, . . . , n − 1

Example (8-th roots of unity)

Re

Im

Oε0

ε1ε2

ε3

ε4

ε5ε6

ε7

α


Polar form

Taking the n-th root

The square roots a of a positive real number r is are the solutionsof x2 = r i.e. ±

√r .

Theorem (All n-th roots)Let z ∈ C be a non-zero complex number, n ∈ N and w ∈ C suchthat wn = z. Then the n-th roots of z can be written as wεk fork = 0, 1, . . . n − 1.

Proof.The values of wεk are n-th roots: (wεk)n = wnεn

k = z · 1 = z .This results in n different values, so we have all the n-th roots ofz .


Order and primitive roots of unity

OrderThe integer powers of some complex numbers are periodic:

I 1, 1, 1, . . .I −1, 1,−1, 1, . . .I i ,−1,−i , 1, i ,−1, . . .I 1+i√

2 , i ,−1+i√

2 ,−1, −1−i√2 ,−i , 1−i√

2 , 1,1+i√

2 , i , . . .

In general cos(2πn ) + i sin(2π

n ) has (only) n different powers.Definition (Order)The number of different powers of a complex number z is theorder of z , sometimes denoted by o(z).

Example (Order)

I The order of 1 is 1;I The order of 2 is ∞ : 2, 4, 8, 16, . . .;I The order of −1 is 2: 1,−1;I The order of i is 4: 1, i ,−1,−i .



Properties of order

Theorem (Properties of order)A complex number z

I either has all pairs of integer powers different, then o(z) =∞,I or there are two integer powers equal, and then the powers of

z are periodic, with the period equal to o(z).Also, o(z) = min{d ∈ N+ : zd = 1}, i.e. the order is the smallestpositive integer d such that zd = 1.Furthermore, zk = z` ⇔ o(z) | k − `. Specially: zk = 1⇔ o(z) | k

Proof.Let o(z) <∞. Then there are k, ` integers, such that zk = z`.Suppose k > `. Then zk−` = 1.Let d be the smallest positive integer, such that zd = 1. Let’sdivide n by d (with a remainder), so n = q · d + r , where0 ≤ r < d . Now 1 = zn = zq·d+r = (zd)qz r = 1qz r = z r . Since dis the smallest such integer, r = 0 implying d | n. In the otherdirection: d | n i.e. dq = n⇒ zn = 1. So: d | n⇔ zn = 1.



Primitive roots of unityNot all n-th roots of unity have order equal to n:4-th roots of unity: 1, i ,−1,−i .

I o(1) = 1;I o(−1) = 2I o(i) = 4

Primitive n-th roots of unityThe n-th roots of unity which have order equal to n are theprimitive n-th roots of unity

Theorem (Corollary (proof for homework))

I The powers of a primitive n-th root of unity are exactly then-th roots of unity.

I A primitive n-th root of unity is a k-th root of unity if, andonly if n | k.



Primitive roots of unity

Example (Primitive roots of unity)

I Primitive 1-st root of unity: 1;I Primitive 2-nd root of unity: −1;I Primitive 3-rd roots of unity: −1±i

√3

2 ;I Primitive 4-th roots of unity: ±i ;I Primitive 5-th roots of unity: . . . (HW)I Primitive 6-th roots of unity: 1±i

√3

2 .

Theorem (Statement)An n-th root of unity cos

(2kπ

n

)+ i sin

(2kπ

n

)is a primitive n-th

root of unity, if and only if gcd(n, k) = 1.

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