Download Notes on logic, sets and complex numbers

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Propositional logic, sets, cardinality and
complex numbers
Mohamed Khaled
I. L OGIC
Logic is the branch of knowledge that studies and uses
the valid reasoning.
Reasoning:
Every horse is beautiful
Mohamed is not beautiful
———————————————Mohamed is not a horse!
Here is another example:
Every mozza is tatli
(1)
Salma is not tatli
(2)
———————————————Salma is not a mozza!
(3)
To recognize the fact that (3) is indeed a consequence of
(1) and (2) we do not need to know whether or not any
of these three statements is true. Moreover, we even do
not need to know what the words “mozza” and “tatli”
mean and who is the person named Salma.
Propositional Logic: (also known as sentential logic)
It studies the reasoning of sentences; we call them
propositions.
Definition 1.1: A proposition is a statement either true
(T) or false (F), but not both.
Example 1.2:
• “1 + 1 > 3” is false proposition.
• “5 > 3” is true proposition.
• “What a wonderful lecture!” is not proposition.
• “Singapore is in Europe” is false proposition.
• “Is Singapore in Asia?” is not a proposition.
Definition 1.3: A paradox is a statement that can not
be assigned a truth value; neither T nor F.
Here is a paradox called the liar paradox: “This statement is false”. Suppose that “This statement is false”
takes the value T, then it must be that the statement is
false, but then if “this statement is false” is false, then
the statement is true.
Truth values: To formalize statements, we shall use
symbols. These symbols will have the same truth values as the statements. For example, if we consider
the proposition “1 + 1 > 5”, we will denote it by p:
p = “1 + 1 > 500 . Since “1 + 1 > 5” is false, p will take
the truth value false (F).
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Sometimes, the truth value of some particular proposition may vary from special circumstance to another. For
example the proposition p = “Bob is single” might be
true if Bob has been married yet, but it can be also false
if he is married already.
Definition 1.4: An assignment ν is a particular choice
for the truth values of the proposition symbols. In other
words, ν shall decide for us whether p is T or F.
Logical operators: These operators are use to combine
statements and to form more complex statements. The
propositional operators are ¬, ∧, ∨, → and ↔.
1) Negation ¬ (Not):
¬p
F
T
p
T
F
So ¬ p takes the truth value T if p takes the value F,
and, ¬ p takes the value F if p takes the value T.
Example 1.5: if p = “you shall pass” then ¬ p =
“you shall not pass”.
2) Disjunction ∨ (Or):
p
T
T
F
F
q
T
F
T
F
p∨q
T
T
T
F
q∨p
T
T
T
F
Note that p ∨ q takes the value F if and only if both
p and q have the value F.
3) Conjunction ∧ (And):
p
T
T
F
F
q
T
F
T
F
p∧q
T
F
F
F
q∧p
T
F
F
F
Note that p ∨ q takes the value T if and only if both
p and q have the value T.
4) Implication → (If · · · then · · · ):
p
T
T
F
F
q
T
F
T
F
p→q
T
F
T
T
Thus, p → q is false if and only if p is true but q is
false. There are alternative ways to read p → q:
2
If p then q.
p only if q.
• p is a sufficient condition for q.
• q is a necessary condition for p.
Well-formed formulas: The well-formed formulas of
propositional logic are obtained by using the rules below:
• A proposition p is a well-formed formula.
• If ϕ is a well-formed formula, then so is ¬ ϕ.
• If ϕ and ψ are well-formed formulas, then so are
ϕ ∧ ψ, ϕ ∨ ψ and ϕ → ψ.
• If ϕ is a well-formed formula, then so is ( ϕ ).
For instance, p ∧ q¬( is not a well-formed formula, while
p → ((¬q ∧ r ) → s) is a well-formed formula. One also
can compute the truth tables for any of the well-formed
formulas by using the tables given above. For example,
let us compute the truth table of the formula ( p ∧ q) ∨
(¬ p ∧ q).
•
•
p
T
T
F
F
q
T
F
T
F
¬p
F
F
T
T
p∧q
T
F
F
F
¬p ∧ q
F
F
T
F
( p ∧ q) ∨ (¬ p ∧ q)
T
F
T
F
Each row of the truth table represent the value of the
formula with respect to some certain assignment. For
instance, the third row in the above table represent that
assignment that gives F to p and T to q, then the formula
( p ∧ q) ∨ (¬ p ∧ q) takes the value T with respect to this
assignment.
Consequence and equivalence: Note that both p ∨ q and
q ∨ p have the same truth tables. In this case, we will say
that p ∨ q and q ∨ p are equivalent.
Definition 1.6: We say that ϕ and ψ are equivalent, in
symbols ϕ ≡ ψ, if both have the same truth table. We
say that ϕ is a consequence of ψ, in symbols ψ |= ϕ, if
ϕ gets the value T whenever ψ has the value T.
Example 1.7: p ∨ q is a consequence of p, indeed,
p
T
T
F
F
q
T
F
T
F
p |= p ∨ q
p∨q
T
T
T
F
In the following theorem we shall see examples on the
equivalence. These laws are important, one may need to
remember them in some occasions.
Theorem 1.8:
• De Morgan laws:
1) ¬( p ∧ q) ≡ ¬ p ∨ ¬q.
2) ¬( p ∨ q) ≡ ¬ p ∧ ¬q.
• Commutativity of ∧ and ∨:
1) p ∧ q ≡ q ∧ p.
2) p ∨ q ≡ q ∨ p.
• Double negation: ¬(¬ p ) ≡ p.
• Absorption laws:
1) p ∨ ( p ∧ q) ≡ p.
2) p ∧ ( p ∨ q) ≡ p.
Proof: It is proved in the class.
Example 1.9: The following three statements are equivalent:
• Alice is not married but Bob is not single ¬ a ∧ ¬ b.
• Bob is not single and Alice is not married ¬ b ∧ ¬ a.
• Neither Bob is single nor Alice is married ¬( b ∨ a ).
Definition 1.10:
• The converse of p → q is q → p.
• The inverse of p → q is ¬ p → ¬ q.
• The contrapositive of p → q is ¬ q → ¬ p.
Theorem 1.11:
1) p → q ≡ ¬ p ∨ q ≡ ¬( p ∧ ¬q).
2) ¬q → ¬ p ≡ p → q.
Proof: It is proved in the class.
Example 1.12: The following statements are equivalent:
• Peter pays taxes only if his income is more than 100
000 HUF.
• If Peter’s income is less than 100 000 HUF, then he
does not need to pay taxes.
Definition 1.13: A contradiction is a statement that is
always false. A tautology is a statement that always gives
a true value.
Example 1.14:
• p ∧ ¬ p is a contradiction.
p
T
F
•
¬p
F
T
p ∧ ¬p
F
F
p ∨ ¬ p is a tautology.
p
T
F
¬p
F
T
p ∨ ¬p
T
T
Puzzles:
There is a wide variety of puzzles about an island in
which certain inhabitants called “Knights” always tell
the truth, and others called “Knaves” always lie. It is
assumed that every inhabitant of the island is either a
knight or a Knave.
• You meet two inhabitants A and B. A says: “I am a
knave but he is not”. What are A and B?
To determine A and B, we use logic. Let us call p the
statement “A is a knight”, and q the statement “B
is a knight”. With that, we rephrase the statement
of A. So, A says “¬ p ∧ q”, the truth table of this
formula is as follows.
p
T
T
F
F
¬p
F
F
T
T
q
T
F
T
F
¬p ∧ q
F
F
T
F
3
•
Now either A is a knight, or A is a knave. If A is
a knight, then the statement p is true, and A must
always tell the truth, thus ¬ p ∧ q must be true as
well. This cannot be from the truth table. If A is a
knave, then the statement p is not true, and A must
always lie, thus ¬ p ∧ q must be false as well. This
happens in the last row of the table, where p is false,
q is false, and ¬ p ∧ q is false, thus we conclude that
A is a knave, and B is a knave.
You meet two inhabitants A and B. A says: “If I am
a knight then so is he”. What are A and B?
Let p be “A is a knight”, and q be “B is a knight”.
Thus, A says “p → q”, the truth table of this formula
is as follows.
p
T
T
F
F
q
T
F
T
F
p→q
T
F
T
T
If A is a knight, then the statement p is true, and A
must always tell the truth, thus p → q must be true
as well. This is possible with the first row. If A is a
knave, then the statement p is not true, and A must
always lie, thus p → q must be false as well. This
cannot happen. We conclude that A is a knight, and
B is a knight.
II. S ETS
A set is a collection of objects, and it is considered as
an object in itself. Suppose that A is a set, to say that
a is an object from this collection that is contained in
A we write a ∈ A. Let us see some example, suppose
that you are living in the same house with Anna, Pál
and Chang. One can consider the set, that describes your
house, that contains you together with your housemates;
H = {You, Anna, Pál, Chang}. Another person might be
interested in the furniture of your house, but not in you
and your neighbors. So, this person would define his set
H 0 = {Closet, Bed, Fridge, Stove, Washingmachine}. Note
that, however intuitively both H and H 0 are meant to
describe the same house, they describe your house from
completely different point of views, and also H 6= H 0 .
Notations 1:
• Sets are usually denoted by capital letters
A, B, C, . . .. The elements of the set are usually
denoted by small letters a, b, c, . . ..
• If X is a set and x is an element of X, we write
x ∈ X. We also say that x belongs to X.
• If X is a set and y is not an element of X, we write
y 6∈ X. We also say that y does not belong to X.
• There are two ways to describe a set:
1) S = { a, b, c, . . .} where the elements are listed
between braces.
2) S = { x ∈ U : p( x ) is true} where U is a universe
and p( x ) is a property that defines the set S.
Example 2.1:
1) N = {0, 1, 2, 3, . . .} is the set of natural numbers.
2) Z = {0, 1, −1, 2, −2, . . .} is the set of all integers.
p
3) Q = { q : p ∈ Z, q ∈ N and q 6= 0} the set of
rational numbers.
4) R is the set of all real numbers.
5) C is the set of all complex numbers.
Set-theoretic notions
1) Equality between sets: Two sets A and B are equal if
and only if they have exactly the same elements.
A = B ⇐⇒ for any x ( x ∈ A ↔ x ∈ B).
2) Subsets and Proper subsets: We say that A is a subset
of B and we write A ⊆ B or B ⊇ A if every element of A
is also an element of B. We say that A is a proper subset
of B and we write A ⊂ B if A ⊆ B and A 6= B.
A ⊆ B ⇐⇒ B ⊇ A ⇐⇒ for any x ( x ∈ A → x ∈ B),
A ⊂ B ⇐⇒ A ⊆ B and there is x ( x ∈ B but x 6∈ A).
3) The empty set: The set which has no element is
called the empty set and is denoted by ∅.
4) Power sets: Let X be any set. The set of all subsets of
X is called the power set of X and is denoted by P ( X ).
That is, P ( X ) = { A : A ⊆ X }.
Remark 2.2: Let A and B be any sets. Then:
• A = B if and only if A ⊆ B and B ⊆ A.
• A ⊆ A and ∅ ⊆ A.
• A ∈ P ( A ) and ∅ ∈ P ( A ).
• P ( ∅ ) = { ∅ }, that is P ( ∅ ) is not empty and it
contains exactly one element.
5) Universal Set: There is no universal set, that is a set
which contains every thing. To show this, let us assume
toward a contradiction that X is a set of everything.
Construct the following set:
A = { x ∈ X : x 6∈ x } ∈ X.
The set A contains all of these things that do not contain
themselves as elements. Since X is the collection of
everything, then A ∈ X. Now, by the definition of A,
we have
A ∈ A if and only if A 6∈ A,
which gives a contradiction. Therefore, there is no such
set X of everything. Instead, we consider a (local) universe set, usually denoted by U. This set represents a
local part of the universe where we live. So U is the
set of everything that we can speak about it in the time
being (or in this specific problem).
6) Union of two sets: Let A and B be any two sets. The
union A ∪ B is the a set that contains any element that
belongs to A or B. In other words,
For any x [ x ∈ A ∪ B ⇐⇒ x ∈ A or x ∈ B].
7) Intersection of two sets: Let A and B be any two
sets. The intersection A ∩ B is the a set that contains any
element that belongs to both A and B. In other words,
For any x [ x ∈ A ∩ B ⇐⇒ x ∈ A and x ∈ B].
4
8) Difference of sets: Let A and B be any two sets.
The set A \ B is defined to be the set that contains any
element of A that is not an element of B. That is,
A \ B = { x ∈ A : x 6 ∈ B }.
Suppose that the universe is U, then we can define the
complement of the set A to be the set of any thing (that
should be in the universe U) that is not in A, i.e., Ā =
U \ A.
A
E
B
D
C
Theorem 2.3:
The union and the intersection are commutative and
associative operations:
1) A ∩ B = B ∩ A and A ∩ ( B ∩ C ) = ( A ∩ B) ∩ C.
2) A ∪ B = B ∪ A and A ∪ ( B ∪ C ) = ( A ∪ B) ∪ C.
• The union is distributive with respect to the intersection and the intersection is distributive with respect
to the union:
1) A ∪ ( B ∩ C ) = ( A ∪ B) ∩ ( A ∪ C ).
2) A ∩ ( B ∪ C ) = ( A ∩ B) ∪ ( A ∩ C ).
• De Morgan’s law:
¯ B = Ā ∪ B̄.
1) A ∩
¯ B = Ā ∩ B̄.
2) A ∪
Proof: It is proved in the class.
9) Cartesian products: Let A and B be two sets. The
Cartesian product of A and B is defined by
•
A × B = {( a, b) : a ∈ A and b ∈ B},
i.e. the Cartesian product A × B is the set of all ordered
pairs ( a, b) with a ∈ A and b ∈ B.
Remark 2.4:
•
•
( x, y) = (u, v) if and only if x = u and y = v.
We write A2 for A × A and, more generally, we write
An for A × · · · × A.
n−times
a1
b1
a2
b2
a3
b3
a4
b4
The domain of the function f , denoted by D ( f ), is the
set of all possible inputs, so the domain of the function
f above is D ( f ) = A = { a1 , a2 , a3 , a4 }. The range of the
function f , denoted by R( f ), is the set of all outputs
of the function f , that is R( f ) = {b1 , b2 , b4 }. Note that
R( f ) 6= B, so B is not usually the range of f and we call
it the co-domain of f .
Definition 2.5: Let f : A → B be a function.
• We say that f is one-one or injective if and only if
every output has a unique input. That is,
For any a1 , a2 ∈ A, if f ( a1 ) = f ( a2 ) then a1 = a2 .
• We say that f is onto or surjective if and only if
every element of the co-domain is an output. That
is,
For any b ∈ B, there is a ∈ A such that f ( a) = b.
• We say that f is correspondence or bijective if it is
both injective and surjective.
For example, the following function is bijective.
a1
b1
a2
b2
a3
b3
a4
b4
Example 2.6:
• The function ι A : A → A that maps any a ∈ A
to itself, i.e., ι A ( a) = a is bijective. This function is
called the identity function on A.
x
• The function f : (0, ∞ ) → (0, 1) given by f ( x ) = 1+
x
is bijective.
• The function f : Q → N given by
p
f ( ) = 2 p · 3q · 5sign( p)+1 ,
q
p
Functions or maps
A map f : A → B from the set A to the set B is a
machine that maps A to B, i.e., f sends each element
a ∈ A to a unique element b ∈ B. You can think about
a ∈ A as an input and f ( a) = b as the output of the
machine f .
•
•
where q is in the irreducible form, is injective but
not surjective.
The function f : A × B → A given by f ( a, b) = a
is surjective but not injective. This function is called
the projection of A × B onto B.
The function f : R → R given by f ( x ) =| x | is
neither injective nor surjective.
5
Definition 2.7: Let f : A → B be a bijective function.
The inverse of f , denoted by f −1 : B → A, is the function
that takes any element b ∈ B to the unique element a ∈ A
for which f ( a) = b. That is,
For any a ∈ A and b ∈ B [ f −1 (b) = a ⇐⇒ f ( a) = b].
Note that the inverse f −1 is defined only if f is bijective.
The inverse of the bijective function in the above figure
can be illustrated as follows:
the sum of a real number together with an imaginary
number. In other words, if a, b ∈ R then a + ib is a
complex number.
Notations 2: Let z = a + ib be a complex number.
•
•
•
a is called the real part of z, and denoted by Re(z).
b is called the imaginary part of z, and denoted by
Im(z).
The set of all complex numbers is denoted by C.
4i
a1
b1
a2
b2
a3
b3
2i
1i
−4
a4
2 + 3i
3i
−3
−2
−1
1
2
3
4
b4
−1i
Definition 2.8: Let f : A → B and g : B → C be two
functions. The composition g ◦ f : A → C is defined as
g ◦ f ( a) = g( f ( a)). In other words, g ◦ f ( a) = c if and
only if there is b ∈ B such that f ( a) = b and g(b) = c.
a1
a2
a3
a4
b1
b2
b3
b4
−2i
−3i
−4i
c1
c2
c3
c4
Remark 2.9: Let f : A → B be a bijection. Then f −1 is
also a bijection, and f ◦ f −1 = ι B and f −1 ◦ f = ι A .
Example 2.10: The function f : (0, ∞) → (0, 1) given by
f ( x ) = 1+x x is bijective. The inverse f −1 : (0, 1) → (0, ∞)
y
is defined by f −1 (y) = 1−y . One can check that f −1 is
also a bijection. Moreover, f −1 ◦ f is the identity function
on (0, ∞) while f ◦ f −1 is the identity function on (0, 1).
III. C OMPLEX NUMBERS
We all know that one can not take the square root of a
negative number because there is no real number whose
square is negative. But if we define i to be a new kind of
numbers (of course not a real number) such that i2 = −1,
then we might be able to find the square
√
√ negative
√ roots of
−
5
=
−1 5 =
numbers
in
terms
of
i.
For
example
√
i 5. This new number is called the imaginary unit, and
it follows immediately from its definition that
i2 = −1, i3 = −i, i4 = 1, i5 = i, · · ·
√
√
Recall that −5 = i 5, we call this number an imaginary number. In a general an imaginary number is of the
form ia, where a ∈ R. A complex number is defined as
Argand plane
In R, one could compare between the real numbers, for
example one could say that a < b, a > b or a = b. That
means the set of real numbers R is ordered by <. This
property fails for C, we cannot compare between two
complex numbers unless if they are equal. In particular,
we can not tell whether a complex number is positive or
not.
Equality of complex numbers: We say that two complex
numbers z and w are equal, and we write z = w, if and
only if Re(z) = Re(w) and Im(z) = Im(w).
Arithmetic of complex numbers
As usual, one can define addition, subtraction, multiplication, division, powers, etc, on the complex numbers.
Let z1 = a1 + ib1 and z2 = a2 + b2 be two complex
numbers.
1) Addition: The sum of z1 and z2 is given by
z1 + z2 = ( a1 + a2 ) + i (b1 + b2 ).
2) Subtraction: The difference of z1 and z2 in that order
is given as follows:
z1 − z2 = ( a1 − a2 ) + i (b1 − b2 ).
3) Multiplication with a real number: Let λ be a real
number, then
λz1 = (λa1 ) + i (λb1 ).
6
4) Multiplication: The multiplication of complex numbers follows the rules of ordinary algebra. Thus, the
product of z1 and z2 can be calculated as follows:
z1 z2
= ( a1 + ib1 )( a2 + ib2 )
= a1 a2 + (ib1 )(ib2 ) + a1 (ib2 ) + (ib1 ) a2
= a1 a2 − b1 b2 + i ( a1 b2 + b1 a2 ).
Example 3.1:
• (3 + 2i ) + (−1 + 5i ) = 2 + 7i.
• (4 − 3i ) − (2 + 6i ) = 2 − 9i.
• (2 + 3i )(1 + 4i ) = −10 + 11i.
• (1 + i )(1 − i ) = 1 − i2 − i + i = 2.
In general ( a + ib)( a − ib) = a2 + b2 which is a real
number. These numbers z = a + ib and w = a − ib are
called the conjugate of each others and we write
z∗ = w and w∗ = z.
Definition 3.2: The multiplication of z = a + ib with its
complex conjugate z∗ = a − ib gives a2 + b2 . This real
number is called modulus squared (or, radius squared)
of z, denoted by | z |2 , and its positive square root is
known as the modulus (or, radius) of z, denoted by | z |.
Thus,
p
| z |2 = zz∗ = a2 + b2 and | z |= a2 + b2 .
Example 3.3:
• The complex conjugate of (6 + 5i ) is (6 − 5i ).
• The complex conjugate of (32 − 9i ) is (32 + 9i ).
• The complex conjugate of any real number is itself.
• The modulus
√ squared of (1 − 7i ) is 50 and its
modulus is 25.
5) Division: The division of complex numbers does
not require a new definition, it can be reduced to the
rule of multiplication. If we want to compute z1 ÷ z2 ,
where z2 6= 0, we multiply both numbers by z2∗ . So, we
get the following:
z1 z2∗
z1 z2∗
z1
( a + ib1 )( a2 − ib2 )
=
=
= 1
.
∗
2
z2
z2 z2
| z2 |
a22 + b22
Example 3.4:
3i
7
22
• 24+
−5i = − 41 + 41 i.
1+ i
• 1−i = i.
6) Reciprocals: Let z 6= 0 be a given complex number.
We want to find the reciprocal 1z , i.e., we want to find
the complex number w for which zw = 1. Suppose that
z = a + ib and w = u + iv, then
zw = ( a + ib)(u + iv) = au − bv + ( av + bu)i = 1.
Thus the real part should be equal to 1, while the
complex number should be 0. That means,
au − bv = 1 and av + bu = 0.
Solving these equations together gives
u=
a2
a
−b
and v = 2
.
2
+b
a + b2
Therefore,
−b
1
a
+i 2
.
=w= 2
2
z
a +b
a + b2
(Note that if z = 0 then a2 + b2 = 0 and so we cannot
compute the reciprocal of z.)
Lemma 3.5 (Properties of complex conjugate):
• ( z1 + z2 )∗ = z1∗ + z2∗ .
• (− z )∗ = −( z∗ ).
• ( z1 − z2 )∗ = z1∗ − z2∗ .
• ( z1 z2 )∗ = z1∗ z2∗ .
• ( 1z )∗ = z1∗ .
z∗
z
• ( z1 )∗ = z1∗ .
2
2
• z is real if and only if z = z∗ .
∗
z∗
• Re (z ) = z+2z and Im ( z ) = z−
2i .
• If z = a + ib then zz∗ = a2 + b2 =| z |2 .
Proof: Easy to verify.
Lemma 3.6 (Properties of modulus):
• | z1 z2 |=| z1 | | z2 |.
• | 1z |= 1 .
|z|
| zz12 |= ||zz1 || .
2
Proof: Easy to verify.
√
Example 3.7: Find the square roots of −2 + 2 3i. √
We need to solve the equation w2 = −2 + 2 3i.
2
2
2
Suppose
√ that w = a + ib, then w = a − b + 2iab =
−2 + 2 3i. By equating the real and the imaginary parts,
we get
√
a2 − b2 = −2 and 2ab = 2 3.
•
√
Then b = a3 . So, a2 − a32 = −2 and a4 + 2a2 − 3 = 0.
Thus, ( a2 + 3)( a2 − 1) = 0. But a is√real, then a2 −√1 = 0.
Hence, a ∈ {−1, 1}. Therefore,
√ 1 + 3i and −1 − 3i are
the square roots of −2 + 2 3i.
√
Example 3.8: Solve the equation z2 + ( 3 + i )z + 1 = 0.
Because every complex number has a square root, the
familiar formula
√
−b ± b2 − 4ac
z=
2a
for the solution of the general quadratic equation az2 +
bz + c = 0 can be used, where now a, b, c ∈ C (and of
course a 6= 0). Hence,
q √
√
−( 3 + i ) ± ( 3 + i )2 − 4
z =
2
q
√
√
−( 3 + i ) ± (3 + 2 3i − 1) − 4
=
p 2
√
√
−( 3 + i ) ± −2 + 2 3i
=
2
√
2
Now we have to solve w = −2 +
the example
√2 3i. By√
above, we know that w ∈ {1 + 3i, −1 − 3i }. Therefore,
√
√
− 3 − i ± (1 + 3i )
z =
2
√
√
√
√
1 − 3 + (1 + 3) i
−1 − 3 − (1 + 3) i
z =
or
.
2
2
7
Polar representation of complex numbers
Any point P in the plan can be described by the
distance r from the origin O and the angle θ which
the vector OP makes with the positive real axis. As
usual the the angles measured in anticlockwise direction
are positive while the angles measured in clockwise
direction are negative.
a = rcosθ
r
a + bi
Lemma 3.9 (Properties of exp):
• e z1 e z2 = e z1 + z2 .
• ez 6 = 0.
• e1z = e−z .
z1
• eez2 = ez1 −z2 .
∗
• ( ez )∗ = e(z ) .
Proof: Here, we only prove the first item. The rest
is similar and easy to verify. Suppose that z1 = a1 + ib1
and z2 = a2 + ib2 be complex numbers, then:
e z1 e z2
b = rsinθ
= e a1 e a2 [(cos b1 + i sin b1 )(cos b2 + i sin b2 )]
= e a1 e a2 [(cos b1 cos b2 − sin b1 sin b2 )
+ i (cos b1 sin b2 + sin b1 cos b2 )]
a1 + a2
(cos (b1 + b2 ) + i sin (b1 + b2 ))
= e
= e(a1 +a2 )+i(b1 +b2 ) = ez1 +z2
θ
By definition of the exponential, the trigonometric representation of a complex number z = r (cosθ + i sinθ ) is
equivalent to the following exponent form z = reiθ . This
exponential representation of complex numbers is very
useful for the multiplication and division of complex
numbers. Consider the complex numbers z1 = r1 eiθ1 and
z2 = r2 eiθ2 , then
z1 z2 = (r1 eiθ1 )(r2 eiθ2 ) = r1 r2 ei(θ1 +θ2 ) .
With this definition we can immediately get the relation
between the polar and the algebraic representations of
complex numbers. Consider the complex number z =
a + ib and denote its polar coordinates by r and θ. Thus,
a = r cosθ and b = r sinθ,
and hence
z = a + ib = r (cosθ + isinθ ).
Note that r is the modulus of z, indeed
q
p
| z |= a2 + b2 = (r cosθ )2 + (r sinθ )2 = r
because cos2 θ + sin2 θ = 1. The angle θ is called the
argument of z, and it is easy to see that

Im(z)
arctan( Re(z) )




Im(z)


arctan( Re(z) ) + π




Im(z)
θ = arg(z) = arctan( Re(z) ) − π

π


2




− π2



undefined
if x > 0
if x < 0 and y ≥ 0
if x < 0 and y < 0
if x = 0 and y > 0
if x = 0 and y < 0
if x = 0 and y = 0.
Exponential:
For any real number t, the exponential et is welldefined. We need to extend the exponential to cover the
complex numbers, too. Define
r
z1
r eiθ1
= 1 iθ = 1 ei(θ1 −θ2 ) .
z2
r2
r2 e 2
A special case of multiplication is raising to a power.
Consider raising the complex number z = reiθ to the nth
power, then
zn = (reiθ )n = r n einθ .
That means, the modulus of z is raised to the nth power
and its argument is multiplies by n.
Theorem 3.10 (de Moiver’s theorem):
(cosθ + i sinθ )n = cos nθ + i sin nθ.
Proof: Consider z = eiθ = cosθ + i sinθ. Now, we
raise z to the power n and we get:
(cosθ + i sinθ )n = zn = einθ = cos nθ + i sin nθ,
because of the equivalence between the trigonometric
and the exponential representations of the complex numbers.
de Moiver’s theorem can be used to get some expressions for the cosines and sines of multiples of θ in terms
of powers of cosθ and sinθ.
Example 3.11: We will show the following well known
formulas of elementary trigonometry:
cos 2θ = cos2 θ − sin2 θ and cos 2θ = 2 sinθ cosθ.
By de Moiver’s theorem, we have the following
(cos θ + i sin θ )2 = cos 2θ + i sin 2θ.
e a+ib = e a (cos b + i sin b).
The exponential have some interesting properties, we list
some of them in the following lemma.
Then,
cos2 θ − sin2 θ + 2 i (cosθ sinθ ) = cos 2θ + i sin 2θ.
8
After equating the real and the imaginary parts, we get
cos 2θ = cos2 θ − sin2 θ and cos 2θ = 2 sinθ cosθ.
Note that the sin and cos of an angle do not change if
we add a multiple of 2π to that angle:
cos(θ + 2kπ ) = cos θ and sin(θ + 2kπ ) = sin θ,
where k is any arbitrary integer in Z.
Lemma 3.12 (Properties of argument):
• arg ( z1 z2 ) = arg ( z1 ) + arg ( z2 ) + 2kπ.
• arg ( 1z ) = − arg ( z ) + 2kπ.
z
• arg ( z1 ) = arg ( z1 ) − arg ( z2 ) + 2kπ.
2
• arg ( zn ) = n arg ( z ) + 2kπ.
Where k is any integer.
Proof: Easy to verify.
The roots of the complex numbers: The fact that sines and
cosines are periodical with period 2π is important when
we take the roots of a complex number. Let z = reiθ =
rei(θ +2kπ ) , then
q
√
√
1
θ + 2kπ
n
z = z n = n r exp(iθ + 2kπ ) = n r exp(i
),
n
and we get n-many different complex numbers (exactly
n-many roots of z) for k = 0, 1, . . . , (n − 1). The complex
number corresponding to k = n coincides with that for
k = 0, so this is not new root.
Example 3.13: Let us find all the cube roots of 1. For
that, we write 1 and its cube roots in the exponential
form as follows:
√
2kπi
3
).
1 = 1 exp(i 0) = exp(2kπi ) and 1 = exp(
3
We know that there are exactly three cube roots corre2
4
spond to k = 0, 1, 2. Then 1, w = e 3 πi and w2 = e 3 πi are
all the cube roots of 1.
w
2
3π
1
1
3π
w2
Theorem 3.14: If f (z) = an zn + an−1 zn−1 + · · · + a1 z +
a0 , where an , . . . , a0 ∈ C, an 6= 0 and n ≥ 1, then
f (z) = an (z − z1 )(z − z2 ) · · · (z − zn )
and f (z1 ) = f (z2 ) = · · · = f (zn ) = 0.
Theorem 3.15: If f (z) = an zn + an−1 zn−1 + · · · + a1 z +
a0 and f (w) = 0, then f (w∗ ) = 0.
Theorem 3.16: If f (z) is a nonconstant polynomial with
real coefficients, then f (z) can be factorized as a product
of real linear factors and real quadratic factors.
Example 3.17: Find a factorization of z4 + 1 into real
linear and quadratic factors.
We have z4 + 1 = (z2 − i )(z2 + i ). Also, one can easily
(1+ i )
show that ± √ are the roots of z2 − i. If z2 + i = 0 then
2
−z2 = −i. Multiplying both sides by i2 gives (iz)2 = i.
(1+ i )
Thus, the roots of z2 + i must be ± √
= ± (1√−i) . In
2i
2
other words the roots are w = 1√+i , w∗ , −w, (−w)∗ .
2
Grouping conjugatecomplex terms gives the factorization
z4 + 1
= (z − w)(z − w∗ )(z + w)(z + w∗ )
= (z2 − 2zRe(w) + ww∗ )(z2 + 2zRe(w) + ww∗ )
√
√
= (z2 − 2z + 1)(z2 + 2z + 1)/