THE UNIVERSITY OF NEW SOUTH WALES SCHOOL OF MATHEMATICS AND STATISTICS MATH 1141 HIGHER MATHEMATICS 1A ALGEBRA. Section 1: - Complex Numbers. 1. The Number Systems. Let us begin by trying to solve various algebraic equations. Suppose we only know about the set of natural numbers (written as N). Then we can solve the equation x 3 = 2 and obtain the solution x = 5. On the other hand, if we try to solve the equation x + 3 = 2 then there is no solution! To solve this equation we need a larger set of numbers which includes the negative whole numbers as well as the positive ones. This set is called the set of integers and is denoted by Z. Continuing this idea: In Z: x + 3 = 2 3x = 2 x = 1 No solution In Q: 3x = 2 x 2 = 2 In R: x = 2 No solution 3 x 2 = 2 x 2 + 1 = 0 x = ± 2 No solution i 2 C 2 i π 2 3 1 Z 1 0 N Q R 2 1
At each stage in the above we are able to solve each new type of equation by extending the set of numbers in which we are working. Hence, to solve the equation x 2 = 1 we introduce a new symbol i (much as we introduced the symbol 2 to solve x 2 = 2.) We define i to be the (complex) number whose square is 1. i.e. i 2 = 1. Using this new symbol, we can now solve x 2 = 1 to obtain solutions x = ±i. Furthermore, we can define the complex numbers by: Definition: The set of all numbers of the form a + bi where a, b are real numbers and i 2 = 1 is called the set of all complex numbers and denoted by C. Summary of Basic Rules and Notation: Let z = a + ib, w = c + id be complex numbers. Then (i) z ± w = (a ± c) + i(b ± d) (ii) zw = (ac bd) + i(ad + bc). (iii) z = a+ib c id = (ac+bd)+i(bc ad) w c+id c id c 2 +d 2 (iv) Re(z) = a, Im(z) = b. Example: z = 2 4i, w = 3 + i, find z + w, z w, zw, z. w Ex: Simplify (1 + i) 8. CP: Let n be an integer. By considering cases, simplify i n + i n+1 + i n+2. 2
As with the set of real numbers and the set of rational numbers, if we add, subtract, multiply, or divide (with the exception of division by 0) any two complex numbers, we again obtain a complex number. This property is called closure. The complex numbers are closed under addition, subtraction, multiplication and division (not by 0). A set of objects which has these (and a number of other properties) is called a field in mathematics. The real numbers and the rational numbers also form fields, but the integers do not, since they are not closed under division. The natural numbers are not closed under subtraction. The above examples are all infinite fields. In addition to these, there are finite fields, which you will study in more detail in higher years. Here are some simple examples. Definition: Given two integers a, b, we can write a = bq + r, with 0 r < b. We will write a r mod b, which we read as a is congruent to r modulo b. In words, r is the remainder when we divide a by b. Hence 7 2 mod 5, and 18 2 mod 4. We can now form the set Z 5 of the possible remainders when we divide by 5. So Z 5 = {0, 1, 2, 3, 4}. We can define the operations of addition and multiplication to be the same as in ordinary arithmetic, but the answers are calculated modulo 5. Thus, addition and multiplication tables can be drawn up as follows: + 0 1 2 3 4 0 0 1 2 3 4 1 1 2 3 4 0 2 2 3 4 0 1 3 3 4 0 1 2 4 4 0 1 2 3 0 1 2 3 4 0 0 0 0 0 0 1 0 1 2 3 4 2 0 2 4 1 3 3 0 3 1 4 2 4 0 4 3 2 1 From the tables, we can see that subtraction and division (except by 0) are defined. For example 2 4 = 3 since 4 + 3 2 and 3 2 = 4 since 2 4 3. The set Z 5 is an example of a finite field. CP: Make up the addition and multiplication tables for Z 6. Explain why this is NOT a field. Can you complete (and prove) the following Theorem: Z n is a field if and only if n is... 3
Equality: Two complex numbers are equal iff they have the same real and imaginary parts, i.e. if z = a + bi = w = c + di then we can conclude that a = c and b = d. Proof: Roots of Unity: A complex number α 1 is called an n-th root of unity if α n = 1. For example, if ω 3 = 1, and ω 1 then we can write, 4
Example: To show just a little of the power of complex numbers, we seek to find a simple closed formula for ( n ( n ( n ( n + + +... + 0) 3) 6) n) where n is an integer divisible by 6. We begin by noting that if ω is a complex cube root of unity then 1+ω k +ω 2k can only take the values Now expand out (1 + ω) n and (1 + ω 2 ) n. Hence if l is the largest integer such that 3l n we have ( n ( n ( n ( n + + +... + = 0) 3) 6) 3l) 1 ( 2 n + (1 + ω) n + (1 + ω 2 ) n). 3 Finally, if n is a multiple of 6, 5
CP: Suppose n > 1 is a multiple of 4. By expanding (1+i) n, (1 i) n, (1+1) n, and (1 1) n, find, in as simple a form as you can, the sum ( n ( n ( n ( n + + +... +. 0) 4) 8) n) Polynomial Equations: We can now solve ALL quadratic equations. Ex: Solve 5x 2 4x + 1 = 0 and z 2 3z + (3 + i) = 0. Note also that we can find new solutions to old equations such as x 3 1 = 0. Both of these are examples of the following remarkable theorem: Theorem. (Fundamental theorem of algebra, FTA) Suppose p(x) = a n x n + a n 1 x n 1 + + a 1 x + a 0 is a polynomial, whose co-efficients a n,, a 1, a 0 are all real (or complex) numbers, then the equation p(x) = 0 has at least one root in the complex numbers. Corollary: The equation p(x) = 0 has exactly n (complex) solutions in the complex numbers (counting multiplicity). (The last proviso counting multiplicity refers to polynomials which may, for example, have factors such as (x 2) 4 in which case the root x = 2 is counted four times.) The above result tells us (among other things) that we do not need to find any larger set of numbers if we want to solve polynomial equations. The complex numbers contain all the roots of every polynomial. 6
Solution of Cubics: The FTA tells about existence, but doesn t give us the machinery to actually find the roots of a polynomial. For quadratics we have the quadratic formula, what about cubics? The first thing to observe is that every cubic equation x 3 + px 2 + qx + r = 0 can be rewritten in the following form: x 3 + ax + b = 0. This is achieved by the change of variable, x = y p 3. Example: Remove the square term in: x 3 6x 2 + x + 3. Put x = y + 2 Cardano stole from Tartaglia, the secret of solving the cubic. He made the change of variable x = u v. It is technically easier to put x = u + v. Example: Find the real root of x 3 + 3x = 1. 7
Strange things can happen when we apply this method to cubics which have three real roots. Example: Solving x 3 6x + 4 = 0 which has x = 2 as one of it roots. Square Roots: Solve z 2 = 3 + 4i. 8
Conjugates: When solving a quadratic equation (with real roots) over the complex numbers, you will have observed that the solutions occur in pairs, in the form a + bi and a bi. There are called conjugate pairs. We say that a bi is the conjugate of a+bi (and vise-versa). To represent this, we use the notation z = a + bi, z = a bi. This conjugate operation has the following properties: i. z ± w = z ± w ii. zw = z.w iii. ( ) z w = z. w iv. z = z if and only if z is real. v. z + z = 2Re(z). You will prove these and similar results in the tutorial exercises. Also note that repeated application of (ii) gives (a + bi) n = (a + bi) n. The Argand Plane: Complex numbers can be represented using the Argand plane, which consists of Cartesian axes similar to that which you used to represent points in the plane. The horizontal axis is used to represent the real part and the vertical axis, (sometimes called the imaginary axis), is used to represent the imaginary part. For example, the following points have been plotted: 3, 2i, 3 + 2i, 4 + i. 3 + 2i 2i 4 + i 3 Complex numbers then are 2-dimensional, in that we require two axes to represent them. Observe that a complex number z and its conjugate are simply reflections of each other in the real axis. 9
We lose the notion of comparison in the complex plane. That is, we cannot say whether one complex number is greater or lesser than another. You have already seen that complex numbers can be expressed either in Cartesian Form, a + ib, a, b R. We can also specify a complex number z by specifing the distance of z from the origin and the angle it makes with the positive real axis. This distance is called the modulus and written as z while the angle is called the argument and written as Arg(z). We insist, to remove ambiguity, that π < Arg(z) π. Pythagoras theorem gives: If z = a + bi then z = a 2 + b 2. Care must be taken to find the correct argument. It is easiest to find the related angle ˆθ such that tan ˆθ = b a and then use this to find the argument in the correct quadrant recalling that we use negative angles in the third and fourth quadrant. Ex: Find the modulus and argument of z = 1 + i 3 and w = 1 2i Properties of Modulus: The modulus function has the following properties: (i) zw = z w (ii) z = z, provided w 0. w w (iii) z n = z n (iv) z = 0 z = 0. 10
Example: (Sums of squares). We can write the integer 5 as 2 2 + 1 2 = 2 + i 2. We can also write the integer 13 as 2 2 + 3 2 = 2 + 3i 2. Hence Try doing the same for 17 and 29. CP: Use the idea above (not expansion) to prove that (a 2 +b 2 )(c 2 +d 2 ) = (ac bd) 2 +(ad+bc) 2 and conclude that, in general, the product of any two numbers which are each the sum of two integer squares, is itself the sum of the two integer squares. This begs the question: What numbers can be written as the sum of two integer squares? This is a hard problem. Experiment with prime numbers and make a conjecture. CP: Suppose A and B are two points in the complex plane corresponding to the complex numbers α = a + ib and β = c + id respectively. Explain why the triangle OAB is right-angled if and only if α β 2 = α 2 + β 2. B α β β α A O Show that if triangle OAB is right-angled then ac = bd. Deduce that if triangle OAB is right-angled then Re(αβ) = 0 11
Properties of the Argument: We can distinguish between the principal argument of z, written Arg(z), which is uniquely defined and takes values between π and π (excluding π), and the more general argument, written arg(z), which is a set of values. We have Arg(z) = arg(z) mod 2π, which means that we can recover Arg(z) from arg(z) by adding or subtracting the appropriate multiple of 2π. The Argument function has the following properties: (i) Arg(zw) = Arg(z)+Arg(w) mod 2π. (ii) Arg(z/w) = Arg(z) Arg(w) mod 2π. The proofs of these will become apparent later. Ex: Suppose α < 1, use a diagram to explain why Arg ( ) 1+α 1 α < π. 2 12
Polar Form: r sin θ r z = (r(cosθ + i sin θ) θ r cos θ From the diagram, we can see that the complex number z can be written in the form z = r(cosθ + i sin θ), where r is the modulus of z and θ is the argument of z. For example, the complex number 1 i can be written as 1 i = 2(cos( π 4 ) + i sin( π 4 )) = 2(cos π 4 i sin π 4 ). This is sometimes called the polar form of z. You will need to be able to convert a complex number from cartesian form, (a + bi), into polar form and vice-versa. For example, 3(cos π + i sin π ) = 3 + 3 3 2 i3. 2 You may have seen the abbreviation cisθ to represent cosθ + i sin θ. You should not use that here, since your tutor may not know what it is. This form is NOT generally used in books beyond High School. Moreover, as we shall see, this polar form, is really a stepping stone to a much better form which involves e. One important fact about the polar form is a remarkable result called: De Moivre s Theorem: For any real number θ, and any integer n, we have (cosθ + i sin θ) n = cosnθ + i sin nθ. The proof of this, looking at the various cases of n is given in the algebra notes. The method of proof by induction is used. Note that the result also holds for n rational which we will find useful later than finding roots. Let us see how useful this result can be: Ex: Let z = 1 i 3. Find z 12. 13
Let us write de Moivre s theorem as follows: Let f(θ) = cosθ + i sin θ, then (f(θ)) n = f(nθ). Also f(0) = 1. Euler did the following: He supposed we can differentiate the function, treating i just like a real number. If we momentarily ignore the logical difficulties involved then f (θ) = i(cos θ + i sin θ). Comparing this with d dθ (eiθ ) = ie iθ, we can see then that this function seems to have properties that are very similar to the exponential function e iθ. We will therefore define: Definition: e iθ = cosθ + i sin θ (and hence e iθ = cosθ i sin θ). This formula is sometimes called Euler s formula. We shall take it to be a definition of the complex exponential. Thus, any complex number z can be expressed in the polar form z = re iθ where r is the modulus and θ the argument of z. For example, z = 1 i = and z = 1 This last formula is quite remarkable since it links together the four fundamental constants of mathematics. In a very important sense, this is the best way to write complex numbers. We have used the term polar form in two different senses. From now on, when I say polar form, I will (generally) mean this new exponential form. You ought to be able to convert a complex number from cartesian form to polar form and vice-versa. Note the following important facts: (i) The conjugate of the complex number z = e iθ is given by z = e iθ. (ii) e iθ = e i(θ+2kπ) where k is an integer. We can write cosθ and sin θ in terms of the complex exponential as follows: cosθ = eiθ + e iθ 2 and sin θ = eiθ e iθ. 2i 14
From the polar form, we can deduce the properties of modulus and argument which we listed earlier. Let z = r 1 e iθ 1 and w = r 2 e iθ 2 then zw = r 1 r 2 e i(θ 1+θ 2 ) from which it follows that zw = z w and Arg(zw) = Arg(z) + Arg(w) mod 2π. Ex: Convert z = 2e i5π 6, w = 3e i π 3 to Cartesian form: Ex: Evaluate the product (1 + i)(1 i 3) in two ways to show that cos π 12 = 1+ 3 2 2. The rules for multiplication of complex numbers in polar form tell us that when we multiply two complex numbers together, rotation and stretching are involved. In particular, 15
since i = e i π 2, multiplying a complex number by i has the effect of rotating z anti-clockwise about the origin, through an angle of 90. iz z Ex: Find the complex number obtained by rotating (4 + 2i) anti-clockwise about the origin through π 2. More generally, to rotate complex number anticlockwise around 0 through an angle θ, we multiply it by e iθ. Ex: Rotate 3 i anticlockwise about 0 through an angle of π 4. CP: Suppose w 1, w 2 are two complex numbers such that 0 < Arg(w 1 ) < Arg(w 2 ). Show that the triangle in the complex plane whose vertices are given by the origin, w 1 and w 2 is equilateral if and only if w 2 1 + w 2 2 = w 1 w 2. (Hint: Try to write w 2 in terms of w 1 using the rotation idea.) 16
The Triangle Inequality: The modulus operation has a number of other useful properties, but two very important ones are: i. zz = z 2 and ii. (The Triangle inequality), z 1 + z 2 z 1 + z 2. I will leave you to prove the first of these and look at the second: Proof of (ii): Ex: Prove that every root of the polynomial p(z) = z 4 + z + 3 lies outside the unit circle in the complex plane. 17
CP: a. Find an upper bound on the maximum of the modulus of p(z) = 4z 3 2z + 1 over all complex numbers z which lie on the unit circle. b. Prove that z 1 + z 2 z 1 z 2. (Hint: Start with z 1 = z 1 + z 2 z 2 ). c. Hence find the minimum value of the modulus in (a). (Note that there are two things to prove here.) Get MAPLE to plot the real and imaginary parts (use trigonometric polar form) of p as z moves around the unit circle. Powers and Roots of Complex Numbers: Ex: Find (1 3i) 10. To find roots of complex numbers, we will use the polar form. Note that to find the nth root of a complex number α, we are really solving z n = α and so we will convert α into polar form. Such an equation will have n solutions! (by the Fundamental Theorem of Algebra.) To get all of these solutions we express α in polar form, using the general argument, not the principal one. An example will make this clear. Ex: Find the 7th roots of 1. 18
If we plot these complex numbers we see that they lie on a circle radius 1 and are equally spaced around that circle. z 3 z 2 z 1 z 4 z 5 z 7 z 6 Ex: Find the 5th roots of 4(1 i). 19
Applications to Trigonometry: Euler s formula gives a dramatic relationship between the exponential and trigonometric functions. We can exploit this to deduce useful relationships and identities in trigonometry. Ex: Find an expression for cos 5θ in terms of sines and cosines. Observe that one can also easily obtain corresponding formula for sin 5θ by taking the imaginary parts of both sides. Web Activity: Use Google to find some information about the Chebychev Polynomials (there are various spellings of Chebychev) and see how they are related to the above example. Ex: Taking the problem the other way round, express sin 5 θ in terms of sine and cosines of multiples of θ. 20
Such a formula is extremely useful in integration, where one might, for example, wish to integrate sin 5 θ. CP: Find a similar formula for sin 4 θ cos 6 θ. Example: Suppose 0 < θ < 2π and n is a positive integer. Show that ( ) 1 e i(n+1)θ Re = 1 1 e iθ 2 + sin(n + 1)θ 2. 2 sin θ 2 Use this to find a simple closed formula for 1 + cosθ + cos 2θ +... + cos(nθ). (Try to find a similar formula for the sine sum.) 21
Regions in the Complex Plane: In this section we see how to represent regions in the argand plane algebraically. For example, the set A = {z C : z 2} represents the set of points whose distance from the origin is less or equal to 2. (N.B. z α measures the distance between z and α.) Hence this set represents a disc radius 2 centre the origin. 2 2. Similarly, the set B = {z C : z+1 < 2} represents the open disc centre ( 1, 0) radius 3 1 1 22
The set C = {x C : 0 Arg(z) π } represents a wedge vertex at the origin, and arms 3 separated by an angle of 60. o Note that the origin in NOT included since the argument of 0 is not defined. Similarly the set D = {z C : 0 Arg(z i) π } represents a wedge centre the 6 point i as shown. 1 o Here are some further examples: Ex: Sketch {z C : z i + 1 < 2} {z C : Re(z) 0} 23
Ex: Sketch {z C : z 3 < 2} {z C : Im(z 3i) > 0} More on Polynomials: The fundamental theorem of algebra, mentioned above, tells us that in the complex plane, all polynomials have all their roots. This is a very powerful theoretical tool, but it does not explicitly tell us how to find these roots for a given polynomial. Moreover, if we know the roots then we also know how to factor the polynomial. You will need to recall a number of basic facts about polynomials from High School, which are: Remainder Theorem: If p(x) is a polynomial then the remainder r when p(x) is divided by x α is given by p(α). Factor Theorem: If p(α) = 0 then (x α) is a factor of p(x). It is important to look at what the underlying set is when we are factoring, for example, x 2 2 does NOT factor over the rational numbers, but it does over the real numbers, (x + 2)(x 2). Similarly, x 2 + 1 does not factor over the real numbers but does over the complex numbers. From the fundamental theorem of algebra, it is clear that over the complex numbers, all polynomials completely factor (at least in theory) into linear factors. Theorem: Every polynomial (with real or complex co-efficients) of degree n 1 has a factorisation into linear factors of the form: p(z) = a(z α 1 )(z α 2 ) (z α n ) where α 1, α 2,, α n are the (complex) roots of p(z). This result, still does not tell us how to factor. Nor does it tell us much about factoring over the real and rational numbers. For example, does the polynomial x 4 + 4 factor over the real numbers or rational numbers? 24
The key to factoring over the real numbers, is to firstly factor over the complex numbers since in the complex plane the polynomial falls to pieces into linear factors. Ex: Factor x 4 + 1 over the complex numbers and hence over the real numbers. Ex: Factor x 6 + 8 over the complex and real numbers. Note that if the co-efficients of the polynomial are real, then the roots occur in conjugate pairs. Theorem: Suppose p(x) is a polynomial with real co-efficients, then if α is a complex (non-real) root, then so is α. 25
Proof: From this is follows that: Theorem: A polynomial with real co-efficients can be factored into a product of real linear and/or real quadratic factors. Proof: Factor p(x) over the complex numbers in the form p(x) = a(x b 1 ) (x b r )(x α 1 )(x α 1 ) (x α s )(x α s ) where the b i s are real and the α i s are complex (non-real). By the above theorem, these must occur in conjugate pairs. Now each such pair of factors containing the conjugate pairs, can be expanded, viz: (x α)(x α) = (x 2 (α + α)x + αα). Now α+α = 2Re(α) and so is REAL, and also αα = α 2 is also REAL. Hence the quadratic we obtain has REAL co-efficients. 26
Ex: Show that z = i is a root of p(z) = z 4 2z 3 + 6z 2 2z + 5 = 0 and hence factor p over R and C. CP: (Not so hard.) Factor x 9 16x 5 x 4 + 16 over the real numbers. The story over the rational numbers is much more complicated. It is possible to have polynomials of arbitrary degree which cannot be factored over the rational numbers. For example, if p is a prime, then x p 1 +x p 2 +...+x+1 cannot be factored over the rationals. (Can you prove this?) Moreover, there is no simple test to tell whether a given polynomial can be factored over the rationals. (More on this in MATH2400 and Higher Algebra in 3rd year). 27