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1 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). [email protected] 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)/