MATH101 LECTURE NOTE.pdf

GOMBE STATE UNIVERSITY FACULTY OF SCIENCE DEPARTMENT OF MATHEMATICS MATH101:GENERAL MATHEMATICS I LECTURE NOTE BY MR. AJIYA YAHAYA COORDINATOR 2018/2019 SESSION MR. AYUBA SANDA MR. AMINU AUDU MR. ABUBAKAR MUH...

GOMBE STATE UNIVERSITY FACULTY OF SCIENCE DEPARTMENT OF MATHEMATICS MATH101:GENERAL MATHEMATICS I LECTURE NOTE BY MR. AJIYA YAHAYA COORDINATOR 2018/2019 SESSION MR. AYUBA SANDA MR. AMINU AUDU MR. ABUBAKAR MUHAMMAD BAKOJI MR. BITRUS SAMBO November, 2018 Contents Title Page...................................... 1 1 SET THEORY 1 1.1 Introduction................................. 1 1.1.1 Finite and infinite set......................... 2 1.1.2 Subset................................. 2 1.1.3 Empty sets.............................. 2 1.1.4 Singleton set............................. 2 1.1.5 Equality of a Set........................... 2 1.1.6 Universal set............................. 2 1.1.7 Union of a Set............................. 3 1.1.8 Intersection of a set.......................... 3 1.1.9 Complement of a Set......................... 3 1.1.10 Power set............................... 3 1.1.11 Properties of union and intersection set............... 3 1.2 Symmetric difference............................. 4 1.2.1 Properties of symmetric difference.................. 4 1.3 Venn Diagram................................. 5 1.3.1 Inclusion-exclusion principle..................... 5 1.4 Elements of relations and functions..................... 6 1.4.1 Reflexive................................ 7 1.4.2 Symmetric............................... 7 1.4.3 Anti-symmetric............................ 7 1.4.4 Transitive............................... 7 1.4.5 Equivalence relation......................... 8 1.5 Functions................................... 8 1.5.1 Composition of a Function (Product of Function)......... 10 i ii MR. AJIYA YAHAYA (Coordinator 2018/2019 session) 2 COMPLEX NUMBER 11 2.1 Introduction.................................. 11 2.1.1 Equality of a Complex Number................... 11 2.1.2 Addition, Subtraction, and Multiplication of Complex Number.. 12 2.1.3 Conjugate of a complex number................... 13 2.1.4 Division of Complex Number.................... 13 2.1.5 Complex or Argand Plane...................... 14 2.1.6 Polar representation of a complex number............. 15 2.1.7 Exponential form of complex number................ 16 2.1.8 Power of complex numbers...................... 17 3 TRIGONOMETRY 19 3.1 Introduction.................................. 19 3.2 Additional formulae.............................. 20 3.3 Multiple and submultiple angle formulae.................. 22 3.4 Factor formulae................................ 25 Chapter 1 SET THEORY 1.1 Introduction The idea of sets is perhaps the most fundamental aspects of mathematics and cannot be defined in terms of simplex concept. It is common among people to talk about collection of objects or things. Collection in mathematics means selection of countable objects or members. Example, collection of students in GSU. Definition: A set is simply defined as a collection of well-defined, distinct and distin- guishable objects or things. Sets are generally denoted by capital letters such as A, B, C, · · · , while elements (mem- bers) by small letters a, b, c, · · ·. The symbols are enclosed the elements of the sets in a pair of brace or curly bracket {}, e.g A = {a, b, c}. There are two ways of specifying a set one way is by listening the elements in the set such as A = {1, 2, 3} and the second way is by set builder notation for instance B = {x|2 < x < 5}. Note the stroke (|) or colon (:) can be used to interchangeable, with each read as “such that” the set B = {x : 2 < x < 5} is read as follows, B is a set consisting of elements x, such that 2 is less than x and x is less than 5. Exercise 1.1: 1. Write in tabular form the set of first ten prime numbers 2. Write in set of the builder form, the set of odd numbers which lie between 10 and 20 3. List the elements of the given set Q = {p : 2 < p < 20}, where p is prime. 1 2 MR. AJIYA YAHAYA (Coordinator 2018/2019 session) 1.1.1 Finite and infinite set Definition: A set is said to be finite if it is elements can be listed or counted. For example the set of the Student in GSU Definition A set is said to be infinite if it is elements cannot be counted. For example (1) Set of real numbers (2) Q = {x : x is a Rational number} The main difference between finite and infinite set is that, a finite set has a definite beginning and end, while infinite set may have a beginning and no end or vice versa or may not have both beginning and end. 1.1.2 Subset A subset is a set written in another set. Suppose P = {1, 2, 3, 4, 5} and Q = {2, 3, 4}, then we say Q is contained in P , and we use the symbol ⊂ to denote the statement “is contained in” or “is subset of”. Thus we say that Q ⊆ P Q is called a proper subset of P if there is at least one member of P which is not a member of Q. 1.1.3 Empty sets A set is said to be empty if it does not contain any element or member. It is also called a Null set, and it is denoted as {}. 1.1.4 Singleton set Any set which has only one member or element is called singleton. Example A = {a}. 1.1.5 Equality of a Set Two sets A and B are said to be equal if A is a subset of B and B is a subset of A, implies A = B if and only if A ⊆ B and B ⊆ A. Example Suppose X = {1, 2, 3} and Y = {3, 2, 1} then X = Y since X ⊆ Y and Y ⊆ X. 1.1.6 Universal set In any given context, the total collection of elements under discussion is called the uni- versal set. It is denoted by symbol µ or E MATH101: GENERAL MATHEMATICS I LECTURE NOTE 3 1.1.7 Union of a Set The union of sets A and B is the set of all elements which belongs to A or B or to both A and B. This is usually written as A ∪ B. For Example A = {1, 2, 3} and B = {2, 3, 4} =⇒ A ∪ B = {1, 2, 3, 4} 1.1.8 Intersection of a set The intersection of sets A and B is the set of elements which are common to both A and B. Simply, A intersection B, consist of elements which are common to both A and B Example A = {1, 2, 3, 4} and B = {2, 4, 5}, then A ∩ B = {2, 4} 1.1.9 Complement of a Set The complement of a set A is the set of element which do not belong to A but belong to 0 the universal set. The complement of a set A is usually denoted by A or Ac 1.1.10 Power set The collection of all the subsets of any set A is called the power set of A. If a set has n member, where n is finite, then the total number of subsets of A is 2n. For example Let A = {1, 2, 3}. Then P (A) = 23 = 8, which shows that set A has 8 subsets. 1.1.11 Properties of union and intersection set 1. A ∪ A = A and A ∩ A = A 2. A ∪ B = B ∪ A and A ∩ B = B ∩ A 3. A ∪ (B ∪ C) = (A ∪ B) ∪ C and A ∩ (B ∩ C) = (A ∩ B) ∩ C 4. A ⊆ (A ∪ B), B ⊆ (A ∪ B), (A ∩ B) ⊆ A, (A ∩ B) ⊆ B 5. A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) 6. A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) Example 1.1: If A = {1, 2, 3, 4, 5, 6} and B = {9, 6, 4, 8, 10, 12}. 4 MR. AJIYA YAHAYA (Coordinator 2018/2019 session) Find (1) A ∪ B, (2) A ∩ B, Solution Given A = {1, 2, 3, 4, 5, 6} and B = {9, 6, 4, 8, 10, 12} 1. A ∪ B = {1, 2, 3, 4, 5, 6, 8, 9, 10, 12} 2. A ∩ B = {4, 6} Exercises 1.2: 1. Let the universal set µ be the set integer, µ = {x : 0 < x < 10}. Find the comple- ment of the set P = {x : x ∈ µ, x is not divisible by 4} 2. Given that µ = {x : x ∈ N, x < 25} A = {even number} B = {perfect squares} and C = {numbers divisible by4} List the element of sets A, B, C and hence fine, 0 0 (1) (A ∪ B ) ∪ C (2) (A ∪ C) ∩ C 0 (3) (A ∩ B) 0 0 (4) B ∩ C 0 (5) (B ∪ C) ∩ A 1.2 Symmetric difference The symmetric difference of two sets A and B denoted as A∆B is defined as A∆B = {x : x ∈ A ∪ B and x ∈ / A ∩ B} = {(A ∪ B)\(A ∩ B)} = (A\B) ∪ (B\A) 1.2.1 Properties of symmetric difference 1. A∆φ = A 2. A∆A = φ 3. A∆(B∆C) = (A∆B)\(A∆C) 4. A∆B = B∆A 0 NB: A\B = A ∩ B. MATH101: GENERAL MATHEMATICS I LECTURE NOTE 5 1.3 Venn Diagram The Venn diagram is geometric interpretation of sets using diagrams (or is a pictorial representation of a set) which shows different relationship between sets. The diagram was invented by John Venn(1834-1923), in Venn diagram, the rectangle is the universal set while the ovals shape inside represent the subsets. If the oval shaped overlap, it shows the intersection, if does not overlap it shows that there is no intersection between the subsets. Complements are written only inside the rectangle and never in any of the circle contained in the rectangle. Exercises 1.3: 1. Use Venn diagram to illustrate the information below µ = {1, 2, 3, 4, · · · , 10} P = {2, 4, 5, 7} Q = {1, 2, 5, 6, 8} and R = {1, 2, 4, 6, 9} Find (a) P ∩ Q ∩ R (b) P ∩ Q (c) P ∩ R (d) Q ∩ R (e) (P ∪ Q ∪ R) 2. In a class containing 32 students, a student can either do Chemistry or Biology or both. If 16 students do Chemistry, 18 do Biology and 3 do none of the subjects. Find how many do both? 3. In a class of 40 students, 25 speak Hausa, 16 speak Igbo, and 21 speak Yoruba. Assuming each of the student speak at least one these three languages. If 8 Speaks Hausa and Igbo; 11 speaks Hausa and Yoruba; and 6 speak Igbo and Yoruba. (a) Draw a Venn diagram to illustrate the information. (b) Find the number of student that speaks all the three subject. (c) How many speaks only two languages? (d) How many speaks only one language? 1.3.1 Inclusion-exclusion principle Let A and B be a finite set, then the inclusion and exclusion state that n(A ∪ B) = n(A) + n(B) − n(A ∩ B) n(A ∪ B ∪ C) = n(A) + n(B) + n(C) − n(A ∩ B) − n(A ∩ C) − n(B ∩ C) + n(A ∩ B ∩ C) 6 MR. AJIYA YAHAYA (Coordinator 2018/2019 session) Exercises 1.4: 1. In a class of 50 students, 25 offer Mathematics, 22 offer Physics, 30 offer Chem- istry and all the student take at least one these subjects; 10 offer Physics and Mathematics, 8 offer Chemistry and Mathematics, 16 offer Physics and Chemistry. a. Draw a Venn diagram to illustrate the information. b. Find the number of students who offer all the subject. 2. A platoon of 32 soldiers went for a night raid in an army location. On their return it was found that 12 were shot in head, 16 in hand and 10 in the leg; 8 had wounds on head and hand, 4 were wounded in the hand and leg and 6 in the leg and head. Assuming all soldiers were wounded as above, describe, how many were shot in all the three part of the body? 1.4 Elements of relations and functions Let x and y be element of a set X. As before {x, y} denotes the subsets of X where elements are precisely x and y. This subsets could of course have been written as {x, y}; There is no preference given to x over y or y over x. We may called {x, y} an unordered pair. However, it is often useful to consider ordered pair (x, y), say, where the order in which the element are written (x, y) = (u, v), if and only if x = u and y = v. Thus (2, 3) is not equal to (3, 2). Let X and Y be sets, then the Cartesian product X × Y of X and Y is the set of all ordered pair (x, y) such that x ∈ X and y ∈ Y. Definitions 1. Cartesian product or direct product The Cartesian or ordered pair pair is the sets A and B is defined as A × B = {(a, b) : a ∈ A, b ∈ B} 2. Ordered pair An ordered pair is the set (a, b) where a ∈ A and b ∈ B such that preference is given to a or b. 3. Relation Let A be a sets, then a relation R exist on A if xRy have meaning for any ordered pair (x, y) ∈ A. Otherwise, no relation between x and y, for all x, y ∈ A. MATH101: GENERAL MATHEMATICS I LECTURE NOTE 7 A relation is part of an open sentence of the type “is twice as”,“is greater than”,“is less than”, “is a husband of”, “is a girl friend of”, etc. A relation is a proportion which may be true or false. If R is a symbol representing the relation determined by Pxy , then R = {(x, y) : Pxy } Probably, if the relation is recast, it may make the relation to be reasonable. In connection with a relation in a set , the word Reflexive, Symmetry, Anti-symmetry and Transitive occur. 1.4.1 Reflexive A relation R in a set is said to be a reflexive iff aRa where a ∈ A which =⇒ (a, a) ∈ R. In other word, A is reflexive iff all the elements are related to each one another. Example: A = {1, 2, 3} then R = {(1, 1), (2, 2), (3, 3)}. 1.4.2 Symmetric A relation in a set A is said to be symmetric iff aRb and bRa but a 6= b, e.g, let A be a set of handsets and R be relation on A defined by x can call y, then R is symmetric since if phone x can call y , then y can also call x. 1.4.3 Anti-symmetric Let A be a set and R be a relation, but in this case a = b. Example, let N be the set of natural numbers and R be a relation in N defined by “x divide y”, then R is anti- symmetric since for any ordered pair (a, b) ∈ A, if a divides b and a = b, implies b divides a. 1.4.4 Transitive A relation in a set A is called transitive iff aRb and bRc, implies aRc, i.e ∀ {a, b, c} ∈ A, the ordered pair (a, b) ∈ R, (b, c) ∈ R, implies (a, c) ∈ R. Example, let R be a relation 8 MR. AJIYA YAHAYA (Coordinator 2018/2019 session) in the set of real numbers defined by “is less than” i.e x < y, then for all {a, b, c} ∈ R, if a < b and b < c, then a < c. 1.4.5 Equivalence relation A relation R in a set A is said to be an equivalent relation iff it is refllexive, symmetric and transitive. We often used the symbol 0 ≈0 to represent an equivalence relation, i.e if x is equivalent to y, then (a) R is reflexive since every ∆a is similar to itself (b) R is symmetric, since if (a, b) ∈ R, it implies (b, a) ∈ R i.e if ∆a is similar to ∆b, then ∆b is also similar to ∆a. (c) R is transitive, since if (a, b) ∈ R and (b, c) ∈ R → (a, c) ∈ R. Example 1.2: 1. Let a relation R be defined on the set R of real number by xRy if and only if kxk = kyk then determine if R is an equivalence relation. 2. Show that “is equivalent to” on the set T of all triangles in a plane is an equivalence relation. 3. Show that “ < ” on Z is not an equivalence relation. 4. If the relation R on the set Z, is defined by a − b = 5k, where k, a, b ∈ Z. Show that R is an equivalence relation. 5. Let R be a relation on a set A defined by “ In the same blood group with”, show that R is an equivalence relation. How many blood group do we have? Mention them. 1.5 Functions Let A and B be two sets. A function f from A to B is a rule which associate to each element a ∈ A a unique element f (a) ∈ B. Then notation is f : A → B or f : a → f (a), ∀a ∈ A, meaning that f maps the element a to the element f (a).We call A the domain of the function f and B the co-domain (range) of A. Then f (a) = {b ∈ B : b = f (a)} , ∀ {a ∈ A}. Example: if f : R → R by saying f : x → x2 , we can see that the function f maps each x ∈ R to its square which is also in R MATH101: GENERAL MATHEMATICS I LECTURE NOTE 9 Type of Functions (1) One to one (Injective) function Let f : A → B; if (x, y) ∈ F and (x, y) ∈ F implying x1 = x2 , then f is called a one to one function or if A and B are two sets, then if the different elements in A always map to different elements in B, then f is 1-1 or injective that is f (x1 ) = f (x2 ) → x1 = x2 , ∀x1 , x2 ∈ A. But if x1 6= x2 → f (x1 ) 6= f (x2 ), then f is called many to one function. Let f : R → R be defined by f : x → (x + 1) then f is 1-1, since for different values of x in the domain, there are different images in the codomain. (2) Onto (Surjective) function A function f is said to be onto, if every element of B is an image of some elements in A. That is ∀y ∈ B, there exist some x ∈ A such that f (x) = y. Example, f : R → R defined by f : x → x2 , then f is onto, since if we consider x and −x they have the same image under f but x 6= −x (3) Bijective function We say that f is bijective if it is both one to one and onto. i.e; f is bijective because different elements of A are into different elements of B is an image. Let f be defined as f : Z → Z by f : x → x + 2, you see that no two or more different values of x give the same image. (4) Equality of two functions Two functions f and g are said to be equal iff f and g are defined on a set A, implies f (a) = g(a) ∀a ∈ A. In other words, two functions are said to equal if both functions (say f and g) have some output using same input. Example: If f (x) = 2x + 1, g(x) = (1 + 2x), then ∀x, g(x) = f (x) (5) Identity function Let Ie : A −→ A be a function which maps every element of A onto itself i.e Ie (x) = x. (6) Constant function A function f : A −→ B is called a constant function if f (x) = b, ∀ x ∈ A and b ∈ B is a fixed element. E.g, let f : R −→ R such that f : x −→ x0 , we can see that ∀x ∈ R, x0 = 1. Hence f is constant. (7) Inverse function Let f be a one-to-one function from A onto B. The inverse of f written as f −1 is the set {(x, y)|(x, y) ∈ f }. That is f −1 : B −→ A is a one-to-one function. In general, the inverse function f −1 of a function f : A −→ B need not be a function. The inverse can 10 MR. AJIYA YAHAYA (Coordinator 2018/2019 session) only be a function iff it is both one to one and onto. Example: (1) Let A = {1, 2, 3} and B = {p, q, r}. Suppose f : A → B is defined by (1, r), (2, p), (3, q) Then, the inverse function f −1 consist the ordered pair (r, 1), (p, 2), (q, 3). (2) Let f : R → R defined by f (x) = 3x − 1∀x ∈ R find f −1. SOLUTION: let y such that f (x) = y =⇒ y = 3x − 1 y+1 =⇒ y + 1 = 3x =⇒ x = 3 but y = f (x) =⇒ x = f −1 (y) hence, f −1 (x) = x+1 3. 1.5.1 Composition of a Function (Product of Function) Suppose f and g are function defined by f : A −→ B and g : B −→ C i.e, ∀x ∈ A =⇒ f (x) ∈ B =⇒ g(f (x)) ∈ C. In other words for some a ∈ A, its image f (a) is in B which is the domain of g. suppose that for some f (a) ∈ B its image g(f (a)) ∈ C. Therefore, the composition of f and g denoted by f og is given by (f og)(a) for some a ∈ A, and (f og)(a) = f (g(a)). Examples 1.3: 1. Let f : R −→ R be defined by f (x) = 3x+2 3−x for x 6= 3 and x ∈ R. Find f −1 and values of x for which f −1 is not defined. Hence find f −1 ( 31 ). 2. Let f : R −→ R and g : R −→ R be defined by f (x) = 2x2 and g(x) = 3x + 1 (a) Compute (f og)(x), (gof )(x) (b) Find (f og)(2), (gof )(2). 3. Let f (x) = (x2 + 3x + 1), g(x) = (3x + 1) and h(x) = 3x x2. Find (f ogoh)(x) Chapter 2 COMPLEX NUMBER 2.1 Introduction Given the equation, x2 = 2, the solution is x = ±1.414 · · · , but given x2 = −2 or x2 = −1, we face a complication since no real number times itself will yield a negative real number. To cope with this problem, larger system of numbers “the Complex system” is usually presented. This system will yield solution not only to equation like x2 = −2 and x2 = −1 but also to complicated polynomial equation of the form n X ai z i = an z n + an−1 z n−1 + · · · + a2 z 2 + a1 z1 + a0 = 0 i=0 where a0 , a1 , · · · , an are complex numbers, n is a positive integer and z is unknown vari- able. Definition A complex number is a number that is written in the form, z = a + ib where a and b represent the real numbers. We say that a is a real part of z and b is the imaginary part of z, that is a = Re(z), b = im(z). The number −2 + 3i is a complex number with real part −2 and imaginary part 3. Note 3i is not a imaginary part 2.1.1 Equality of a Complex Number Two complex numbers z1 = (a + ib) and z2 = (c + id) are said to be equal if and only if a = c and b = d. Using real numbers, we can say for example 5 > 3 but it makes no sense to assert that either (1 + i) > (2 + 3i) or (2 + 3i) > (1 + i), an equality like a > b, will implies that both a and b are real numbers 11 12 MR. AJIYA YAHAYA (Coordinator 2018/2019 session) 2.1.2 Addition, Subtraction, and Multiplication of Complex Num- ber Let z1 = (a + ib) and z2 = (c + id), then z1 + z2 = (a + ib) + (c + id) = (a + c) + i(b + d) z1 − z2 = (a + ib) − (c + id) = (a − c) + i(b − d) z1 × z2 = (a + ib)(c + id) = (ac − bd) + i(ab + bc) The result in the product is obtained through use of ordinary rules of algebra and noting the crucial fact that i×i = i2 = −1. Observe that complex number obey the commutative, associative, and distributive laws. Now, consider two complex numbers z1 and z2 , whose imaginary part are zero. Let z1 = a + 0i and z2 = c + 0i, then, z1 + z2 ∈ R and z1 z2 ∈ R. We speak of complex numbers of the form of a + 0i as purely real and those of the form of 0 + ib a purely imaginary. The additive identity of complex numbers is given as 0 + 0i. A multiplication involving a real and a complex number is treated, by definition, as if the real number were complex but with zero imaginary part. For example if k ∈ R, then k(a + ib) = (k + oi)(a + ib) = ka + ikb. Example 2.1: Perform the following operation and express the result in the form a = ib, where a, b ∈ R. (1) (−1 + 3i) + (5 − 7i) (2) (−1 + 3i)(5 − 7i) Exercises 2.1: Perform each of the indicated operation and express the result in the form a + ib, where a, b ∈ R. 1. (3 + 2i) + (−7 − i) 2. (4 + 2i)(2 − 3i) 3. [Re [(5 + 3i) + (−1 + 2i)]]2 4. [Im (1 + i)]2 5. (3 − 2i)(4 + 3i)(3 + 2i) 6. (x + iy)(u − iv)(x − iy)(u + iv) MATH101: GENERAL MATHEMATICS I LECTURE NOTE 13 2.1.3 Conjugate of a complex number A pair of complex number are said to be conjugate of each other if they have identical real parts and imaginary parts that are identical except for being opposite in sign. If z = a + ib, then the conjugate of z, denoted by z or z ∗ = a − ib e.g −2 + i4 = −2 − 4i Note that (z) = z Other important identification for complex numbers 1. z = a + ib and z = a − ib 2. z + z = 2Rez = 2Rez 3. z − z = 2iImZ 4. zz = a2 + b2   k 5. (z) = z k for every k integer 2.1.4 Division of Complex Number Let z1 = a + ib and z2 = c + id, then z1 a + ib (a + ib) (c − id) (ac + bd) (bc − ad) = = = 2 2 +i 2 z2 c + id (c + id) (c − id) c +d c + d2 Example 2.2: compute the numerical values of the following expression.  2 3 − 4i 3 + 4i 1 1 (1) + (2) i+ (3) 1 + 2i 1 − 2i 1 − 2i 1 + 2i Definition: Modulus of a complex number The modulus of a complex number is a positive square root of the sum of the squares of its real and its imaginary parts. The term absolute value and magnitude are used to mean the modulus. If the complex number is z, then its modulus is written kzk. If p z = x + iy, we have, from the definition, |z| = x2 + y 2 Properties of modulus of complex numbers 1. |z1 z2 | = |z1 | |z2 | or |z1 z2 · · · zk | = |z1 | |z2 | · · · |zk | z1 |z1 | 2. z2 = |z2 | for z2 6= 0 3. |z|2 = zz 14 MR. AJIYA YAHAYA (Coordinator 2018/2019 session) 4. |Re(z)| ≤ |z| 5. |Im(z)| ≤ |z| 6. |z1 + z2 | ≤ |z1 | + |z2 | triangular inequality 7. |z1 − z2 | ≥ |z1 | − |z2 | 8. z k = |z|k for k ∈ N Example 2.3: Find the modulus of the following complex number (1) − 4 + 2i (2) − 4i + 3 Exercises 2.2: Find the modulus of the following complex number. (i) (2 − i) (−3 + 2i) (5 − 4i) (3+i) (ii) −i + 1−i √ 3 (iii) If z1 = 2 + i, z2 = 3 − 2i and z3 = − 12 + 2 i, evaluate each of the following (a) |3z1 − 4z2 | (b) |z13 − 3z12 + 4z1 − 8| (c) z3 − 12 z2 + 2z1 (z) (z3 )4 2.1.5 Complex or Argand Plane If the complex number z = (x + iy) were written as an ordered pair z = (x, y), we would perhaps be reminded of the notation for the coordinates of a point on the xy−plane. The p expression |z| = x2 + y 2 , also recalls the Pythagoras expression for the distance of that point from the origin. The xy−plane in complex number is what we called the argand plane, the z−plane or the complex plane. Here, the x−axis is called the real axis and y−axis is called imaginary axis. Note: When representing the addition or subtraction of two or more complex number on an argand plane or diagram, we employ the familiar parallelogram rule, which is used is adding vectors. Example 2.4: Represent the following complex numbers on a complex plane. (i) z1 = 3 + 4i (ii) z2 = −4 − 6i (iii) z3 = (3 + 4i) + (5 + 2i) MATH101: GENERAL MATHEMATICS I LECTURE NOTE 15 (iv) z4 = (6 − 2i) − (2 − 5i) (v) z5 = (−1 − 3i) + (1 + 4i) Exercises 2.3: Show that the following complex numbers on a complex plane (a) (−1 − 3i) + (1 + 4i) (b) (−1 − 3i) − (1 + 4i) 2.1.6 Polar representation of a complex number p Let x and y be real numbers and x + iy a complex number, and let r = x2 + y 2. If (r, θ) is the polar coordinate of the point (x, y) in the plane then, x = rcosθ and y = rsinθ, hence, z = x + iy = rcosθ + irsinθ. The number θ is sometimes called the angle or argument of z and we write θ = argz. At a glance, we see that tanθ = xy. Important features of θ 1. θ is positive in counter clockwise sense and negative otherwise. 2. Since both cosine and sine functions are periodic with period 2πrad or 360◦ , then θ is given by θ = θp + 2πk, k = 0, ±1, ±2, · · · 3. θ is multivalued. Example 2.5: Find the argument of the following complex numbers 1 (1) 1 + i (2) 1 − √ i (3) −i (4) −3 (5) −1−i 3 Properties of the argument of complex numbers 1. arg (z1 z2 ) = arg (z1 ) + arg (z2 )   z1 2. arg z2 = arg (z1 ) − arg (z2 ) 3. arg (z) = −arg (z) 1
4. arg z = argz 16 MR. AJIYA YAHAYA (Coordinator 2018/2019 session) Principal Argument The principal value of the argument or principal argument of a complex number z is that value of arg(z) that is greater than −π and less than or equal to π. Thus, the principal value of θ satisfies −π < θ ≤ π. NB: In computing the principal argument, you must not cross the negative real axis. Examples 2.6: 1. Transform the following complex numbers in to their polar form √ √ (i) − 1 + i 3 (ii) − 1 + i (iii) 1 + i 3 2. Transform the following complex numbers in to their Cartesian form (i) 2 cos π3 + isin π3 (ii) cos 2π + isin 2π
5 5 3. Express each of the following complex numbers in polar form using the principal argument √ √ √ (i) 2 − 2i (ii) −1+ 3i (iii) 2 2 + 2 2i (iv) −4 (1+i)(3+i)(−2−i) 4. Express and √1+i in form a + ib. i(3+4i)(5+i) 3+i 5. Find the argument of each of the following complex numbers and write each in polar form 1 √ (i) − 2 (ii) − 3 + 3i (iii) − πi (iv) − 2 3 − 2i 2.1.7 Exponential form of complex number There is still another way of expressing a complex number which we must deal with, for it has its uses. Many functions can be expressed as series, for example ∞ θ X θn θ2 θ3 θ4 e = =1+θ+ + + + ··· i=0 n! 2! 3! 4! ∞ X (−1)n θ2n+1 θ3 θ5 θ7 sinθ = =θ− + − + ··· i=0 (2n + 1)! 3! 5! 7! ∞ X (−1)n θ2n θ2 θ4 θ6 cosθ = =1− + − + ··· i=0 (2n)! 2! 4! 6! MATH101: GENERAL MATHEMATICS I LECTURE NOTE 17 Now, replacing θ by iθ in the series eθ , we have (iθ)2 (iθ)3 (iθ)4 (iθ)5 (iθ)6 eiθ = 1 + (iθ) + + + + + + ··· 2! 3! 4! 5! 6! i2 θ2 i3 θ3 i4 θ4 i5 θ5 i6 θ6 = 1 + iθ + + + + + + ··· 2! 3! 4! 5! 6! θ2 iθ3 θ4 iθ5 θ6 = 1 + iθ − − + + − + ··· 2!  3!  4! 5! 6! θ2 θ4 θ6 θ3 θ5   = 1− + − + ··· + i θ − + + ··· 2! 4! 6! 3! 5! = cosθ + isinθ Therefore, r (cosθ + isinθ) can be written as reiθ. This is called the exponential form of the complex number, the angle must be in radians. Exercises 2.4: 1. Write the following complex numbers into their exponential and polar forms √ (i) z = 1 − 3i (ii) z = −1 2. Write each of the given complex number in the form a + ib π e1+3πi π (i) z = ei 4 (ii) z = −1+ iπ (iii) 2e3+ 6 i e 2 3. Write each of the given complex number in the form reiθ (i) 1−i 3 (ii) 2+2i √ − 3+i (iii) (1 + i)6 4. Show that eiθ − e−iθ eiθ + e−iθ eiθ − e−iθ (i) sinθ = (ii) cosθ = (iii) tanθ = 2i 2 i (eiθ + e−iθ ) 2.1.8 Power of complex numbers De Moivre’s theorem: State that if z = r (cosθ + isinθ) and n is a natural numbers, then z n = rn (cosnθ + isinnθ) This theorem was named after the French Mathematician Abraham de Moivre (1667- 1754). Prof Let z = r (cosθ + isinθ), and z = reiθ. Then
n z n = reiθ = rn eiθn = rn (cosnθ + isinnθ) 18 MR. AJIYA YAHAYA (Coordinator 2018/2019 session) where n ∈ N 1. De Moivre’s theorem also allows us to calculate the nth roots of any complex number, where n ∈ N 2. The nth roots of complex numbers w are the values of z such that z n = w. There are exactly n such values of z. 3. To solve the equation z n = w for some w ∈ C n∈N (i) Express w in polar form as w = r (cosθ + isinθ) (ii) since the angle θ is equivalent to the angle (θ + 2πk) for any k ∈ Z, we know that r (cosθ + isinθ) = r [cos (θ + 2πk) + isin (θ + 2πk)], where k ∈ Z; 1 (iii) Thus z = (r [cos (θ + 2πk) + isin (θ + 2πk)]) n and by the extension of De Moivre’s theorem      1 θ + 2πk θ + 2πk z = r cos n + isin n n Exercises 2.5: 1. Evaluate the following √
6 1 1 1 (i) = 1 − 3i (ii) z = (−16) 4 (iii) z = 1 5 (iv) z = i 4 2. Find the fourth roots of unity 3. Find the fourth roots of complex number 4. Find all the four roots of the equation z 4 + 1 = 0 √ √ 5. Find all the cube roots of 2+i 2 6. Find the square roots of the following (i) 8 + 6i (ii) 5 + 12i (iii) 3 − 4i (iv) − 8 + 6i √
7 √ 7. Using De Movre’s formula, show that 3 − i = −64 3 + 64i Chapter 3 TRIGONOMETRY 3.1 Introduction When a line OP rotates from a position OX to some other position OP , the angle P OX is produced and is said to be positive if the sense of rotation is anticlockwise, and negative if the sense of rotation is clockwise. Angles are generally measured in degree or radians. The trigonometric rations for an acute angle is Considering the figure above, we have the following relationships: y x y sinθ sin θ = , cosθ = , tanθ = = r r x cosθ 1 1 x cosθ cosecθ = , secθ = , cotθ = = sinθ cosθ y sinθ By Pythagoras theorem, we get
2
2 x2 + y 2 = r 2 =⇒ cos2 θ + sin2 θ = r2 19 20 MR. AJIYA YAHAYA (Coordinator 2018/2019 session) cos2 θ + sin2 θ = 1 (1) Dividing (1) by sin2 θ and cos2 θ, we get 1 + cot2 θ = cosec2 θ (2) 1 + tan2 θ = sec2 θ (3) Examples 3.1 1. If sinθ = √1 , 3 and 0◦ ≤ θ ≤ 90◦. Find the values of the other trigonometric ratios of the angle θ. 2. Show that sec2 θsin2 θ − sec2 θ = −1 3. Verify the following identities a. sinθsecθ = tanθ b. cosec2 θ (1 − cos2 θ) = 1 1−2cos2 x c. sinxcosx = tanx − cotx d. tan2 xcosec2 xcot2 xsin2 x = 1 e. 1+sinx
1+secx
1+cosx 1+cosecx = tanx 3.2 Additional formulae Consider the figure below We have by cosine rule that |P Q|2 = r2 + r2 − 2r2 cos(A − B) = 2r2 − 2r2 cos(A − B) (4) MATH101: GENERAL MATHEMATICS I LECTURE NOTE 21 Observe that |P N | = rsinA − rsinB, |QN | = rcosA − rcosB and < P N Q = 90◦. Therefore, by Pythagoras theorem, |P Q|2 = |P N |2 + |QN |2 = (rsinA − rsinB)2 + (rcosA − rcosB)2 |P Q|2 = 2r2 − 2r2 (cosAcosB + sinAsinB) (5) equating (4) and (5), we get cos (A − B) = cosAcosB + sinAsinB (6) replacing B with −B in (6), we get cos (A + B) = cosAcosB − sinAsinB (7) replacing A by 90◦ − A in (6), we get cos [90 − (A + B)] = cos(90 − A)cosB + sin(90 − A)sinB sin (A + B) = sinAcosB + cosAsinB (8) replacing B with −B in (8), we get sin (A − B) = sinAcosB − cosAsinB (9) Dividing (8) by (7), we have sinAcosB + cosAsinB tan (A + B) = cosAcosB − sinAsinB dividing both numerator and de numerator by cosAcosB, we have tanA + tanB tan (A + B) = (10) 1 − tanAtanB replacing B with −B in (10), we obtain tanA − tanB tan (A − B) = (11) 1 + tanAtanB Summary 1. sin (A + B) = sinAcosB + cosAsinB 2. sin (A − B) = sinAcosB − cosAsinB 3. cos (A + B) = cosAcosB − sinAsinB 4. cos (A − B) = cosAcosB + sinAsinB 22 MR. AJIYA YAHAYA (Coordinator 2018/2019 session) tanA+tanB 5. tan (A + B) = 1−tanAtanB tanA−tanB 6. tan (A − B) = 1+tanAtanB Exercises 3.1: 1. Find the value of sine, cosine, and tangent of 15◦ 2. Simplify the following (a) sin (A + B) + sin (A − B) (b) sin (A + B) − sin (A − B) (c) cos (A + B) + cos (A − B) (d) cos (A + B) − cos (A − B) tan(α+β)−tanα 3. Show that 1+tan(α+β)tanα = tanβ 4. If A, B and C are angles of a triangle, show that (a) cosA + cos (B − C) = 2sinBsinC (b) cos 21 C + sin 21 (A − B) = 2sin 12 Acos 12 B 5. Show that: (a) cos (90 + A) = −sinA (b) sin (90 + A) = cosA (c) cos (180 − A) = sinA (d) sin (180 − A) = −cosA 3.3 Multiple and submultiple angle formulae Putting B = A in equations (7), (8) and (10), we obtain  cos2A = cos2 A − sin2 A       = 2cosA − 1 (since Sin2 A = 1 − cos2 A) (12)     = 1 − 2sin2 A (since cos2 A = 1 − Sin2 A)   sin2A = SinAcosA + cosAsinA = 2SinAcosA (13) MATH101: GENERAL MATHEMATICS I LECTURE NOTE 23 tanA + tanA 2tanA tan2A = = (14) 1 − tanAtanA 1 − tan2 A These results can be used to obtain the expression for sin3A, cos3A and tan3A as follows:      sin3A = sin(2A + A)         = sin2AcosA + cos2AsinA    = 2sinAcos2 A + (1 − 2sin2 A)sinA   (15) = 2sinA(1 − sin2 A) + sinA − 2sin3 A        = 2sinA − 2sin3 A + sinA − 2sin3 A        = 3sinA − 4sin3 A         cos3A = sin(2A + A)         = cos2AcosA − sin2AsinA    = (2cos2 A − 1)cosA − 2cosAsin2 A   (16) 3 2 = 2cos A − cosA − 2cos(1 − cos A)        = 2cos3 A − cosA − 2cosA + 2cos3 A        = 4cos3 A − 3cosA    If we replace A by 21 θ in (12),(13) and (14), we have θ θ sinθ = 2sin cos (17) 2 2  θ cosθ = cos 2 − sin2 2θ       = 1 − 2sin2 2θ (18)    = 2cos2 2θ − 1    2tan 2θ tanθ = (19) 1 − tan2 2θ Now, (17) − (19) can be expressed in terms of tan 2θ as follows: Let tan 2θ = t, then  2sin θ2 cos θ2 sinθ =      1  2sin θ2 cos θ2      = cos2 θ +sin 2θ  2 2 sin θ     2 2 cos θ = sin22 θ (20)   1+ 2 θ2   cos 2  θ 2tan       = 2 1+tan2 θ2     2t  = 1+t 2 24 MR. AJIYA YAHAYA (Coordinator 2018/2019 session)  2θ 2θ cosθ = cos 2 −sin 2     1  cos2 θ −sin2 θ      = cos2 θ2 +sin2 θ2  2 2 sin2 θ   1− 2 θ2   cos = sin2 2θ (21)   1+ 2 θ2   cos 2  2θ 1−tan       = 2 1+tan2 θ2   2  = 1−t   1+t2 2t 1 − t2 2t tanθ = 2 ÷ 2 = (22) 1+t 1+t 1 − t2 Summary 1. sin2A = 2sinAcosA 2. cos2A = cos2 A − sin2 A 2tanA 3. tan2A = 1−tan2 A 4. sinA = 2sin A2 cos A2 5. cosA = cos2 A2 − sin2 A2 2tan A 6. tanA = 2 1−tan2 A2 7. sin3A = 3sinA − 4sin3 A 8. cos3A = 4cos3 A − 3cosA 3tanA−tan3 A 9. tan3A = 1−3tan2 A 2t 10. siA = 1+t2 1−t2 11. cosA = 1+t2 2t 12. tanA = 1−t2 where t = tan A2 Exercises 3.2: 1. If cosA = 45 , find without table or calculator the values of (a) sin2A (b) cos A2 and tan A2 2. Using the half angle formula, find the exact values of (a) sin15◦ (b) cos135◦ MATH101: GENERAL MATHEMATICS I LECTURE NOTE 25 3.4 Factor formulae The sum and difference of sines and cosines may be expressed as product of sines and cosines and vice-versa. Now, recall that; sin(A + B) = sinAcosB + cosAsinB, sin(A − B) = sinAcosB − cosAsinB cos(A + B) = cosAcosB − sinAsinB and cos(A − B) = cosAcosB + sinAsinB  sin(A + B) + sin(A − B) = 2sinAcosB  (23) sin(A + B) − sin(A − B) = 2cosAsinB   cos(A + B) + cos(A − B) = 2cosAcosB  (24) cos(A + B) − cos(A − B) = −2sinAsinB  α+β α−β Let A + B = α and A − B = β, so that A = 2 and B = 2 , substituting these in (23) and (24), we get α+β α−β

1. sinα + sinβ = 2sin 2 cos 2 α+β α−β

2. sinα − sinβ = 2cos 2 sin 2 α+β α+β

3. cosα + cosβ = 2cos 2 cos 2 α+β α+β

4. cosα − cosβ = −2sin 2 sin 2 Exercises 3.3: 1. Show that sin2θ (a) 1+cos2θ = tanθ √ (b) sin50◦ + sin40◦ = 2cos5◦ sinA+sinB A+B
(c) cosA+cosB = tan 2 sinx−siny x+y
(d) cosx+cosy = tan 2 sinθ+sin5θ (e) cosθ+cos5θ = tan3θ (f) sin7x + sinx − 2sin2xcos3x = 4cos2 3xsinx 2. Express each of the following as a product (a) sin50◦ + sin40◦ (b) sin70◦ − sin20◦ (c) cos55◦ + cos25◦ 3. Express each of the following as a sum or difference (a) sin40◦ sin30◦ (b) cos50◦ cos35◦ (c) cos110◦ sin55◦ (d) sin55◦ sin46◦