THE CALCULUS OF POLYNOMIALS

THE CALCULUS OF POLYNOMIALS John A. Beachy Northern Illinois University 1991 Revised 1995, 2003

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Contents 1 Polynomials 9 1.1 Roots of polynomials........................ 11 1.2 Rational roots............................ 19 1.3 Approximating real roots...................... 27 1.4 Complex roots............................ 35 1.5 Interpolating polynomials...................... 43 1.6 Mathematical Induction....................... 50 3

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Preface These notes are intended to be used as a supplement to the material usually taught in the first semester of calculus. Certain techniques used to obtain numerical approximations provide the focal point of the notes. Newton s method uses tangent lines to find successive approximations to solutions of equations. The idea of using a tangent line to approximate a function (locally) can be extended to use polynomials of higher degree. Polynomial approximations are also useful in finding the area beneath a curve. This general theme of using polynomials to approximate functions presupposes some knowledge of polynomials. The first chapter of the notes is designed to review some of the necessary background as well as to provide some new information about polynomials. It also includes sections on mathematical induction and the complex numbers. The second chapter discusses derivatives of polynomial functions, without using limits. (For those already in the know, the tangent line at a point is defined as the linear part of the Taylor expansion at the point.) The definition of the average value of a function uses as motivation the statement that is the Fundamental Theorem of Calculus (in a traditional development). This notion of an average value is then used to compute areas under curves and to define the integral of a function. The third chapter introduces the concept of a limit, in connection with sequences of approximations. Then integrals are reconsidered, using limits of sequences. It ends with some results on infinite series, including a discussion of Taylor series. To help set the tone of this course, it may be useful to make some general comments about mathematics. The field of mathematics involves the construction and study of abstract models of physical situations. The construction of a model requires the selection of explicitly stated and precisely formulated premises. These assumptions are called axioms, and the the study of the model then involves drawing conclusions from these fundamental assumptions, using as high a degree of logical rigor as possible. The rigor of mathematics is not absolute, but is rather in the process of continual development. For example, Euclid s axiomatization of geometry and his 5

6 CONTENTS study of this model of our spatial surroundings was accepted as completely rigorous for over two thousand years, but a modern geometer could point out serious flaws in the logical development of the theory. The choice of the basic assumptions to be taken as axioms usually involves an oversimplification of the facts. For example, models of the U.S. economy cannot hope to take into account every variable and still have a workable model. A mathematical model should only be viewed as the best statement of the known facts. In many cases a model should be viewed as merely the most efficient, incorporating only enough assumptions to give the desired degree of accuracy in prediction. For example, in an area small in comparison to the total surface of the earth, plane geometry gives a good approximation for questions involving relationships of figures. As soon as the problems involve large distances, spherical geometry must be used as the model. As another example, Newtonian physics is good enough for many problems in mechanics, and it is necessary to introduce the additional assumptions of quantum mechanics only if answers have to be found at the atomic level. If any distinction at all is to be made between applied mathematics and theoretical mathematics, it is perhaps most evident when talking about the process of modeling. The applied mathematician is probably more involved in the construction of models, and must ask questions about the efficiency of the models and about how closely they approximate the real world. The theoretical mathematician is concerned with developing the model, by investigating the implications of the basic assumptions or axioms. This is done by proving theorems. Of course, if a theorem is proved that is obviously contrary to nature, then it is clear to everyone that the basic assumptions do not coincide with reality. The theoretical mathematician is also concerned with the internal consistency of the models. In order to make logical deductions from the basic axioms, the language used must be extremely precise. This is done by making use of careful definitions, and symbols that are lifted out of the context of ordinary language in order to strip away any possible ambiguity. Much of the precision and clarity of mathematics is made possible by its use of formulas. The modern reader is usually unaware that this is an achievement only of the past few centuries. 1 For example, the signs + and appeared in manuscripts for the first time in 1481; parentheses first appeared in 1544; brackets and braces appeared essentially for the first time in 1593 in the works of Vieta; the sign = appeared in 1557; the modern way of writing powers was first used in 1637 by Descartes, but in 1801 Gauss still wrote xx instead of x 2. The motivation of the pure mathematician certainly comes partly from the appli- 1 The notes on symbols are from Differential and Integral Calculus, by Ostrowski.

CONTENTS 7 cations of the theorems that can developed within a given model. But perhaps more than this, it comes from the joy of creating a theory of particular simplicity, elegance and broad scope. It is certainly difficult to describe the beauty of a mathematical theory, but if one understands the theory, it is not difficult to appreciate its beauty, if for no other reason than what it shows of the intellectual creativity of human kind. In defense of the theoretical mathematician, it must be said that a theory should not be judged only on its applicability to presently known problems. The history of mathematics is filled with examples of particular theories that seemed at the time to be mere intellectual exercises devoid of any relationship to physical problems, but that later were discovered to have important applications. One particularly impressive example is provided by non-euclidean geometry, which arose from the efforts (extending over two thousand years) to prove that Euclid s parallel axiom could be deduced from his other, more obvious, axioms. This seemed to be a matter of interest only to mathematicians. Even Lobacevskii, the founder of the new geometry, was careful to label it imaginary, since he could not see any meaning for it in the actual world. In spite of this, his ideas laid the foundation for a new development of geometry, namely the creation of theories of various non-euclidean spaces. These ideas later became, in the hands of Einstein, the basis of the general theory of relativity, in which the mathematical model consists of a form of non-euclidean geometry of four-dimensional space. The generalizations and abstractions of mathematics often seem at first to be strange and difficult. But with the very general expansion of knowledge and technology that we are currently experiencing, it becomes necessary to identify and elucidate general underlying principles, in order to tie this information together. The language and concepts of mathematics help to fill this need. Acknowledgments These notes were written while teaching the honors calculus sequence during the fall of 1986 and the fall of 1991. Some of the material in the first chapter had been covered in previous semesters. In particular, I am indebted to Bill Blair, Henry Leonard, Don McAlister, Linda Sons and Bob Wheeler for their notes on various topics. J.A.B.

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Chapter 1 Polynomials The branch of mathematics that provides methods for the quantitative investigation of various processes of change, motion, and dependence of one quantity on another is called mathematical analysis, or simply analysis. A first course in calculus establishes some of the basic methods of analysis, done in relatively simple cases. The development of the methods of analysis was stimulated by problems in physics. During the 16th century the central problem of physics was the investigation of motion. The expansion of trade, and the accompanying explorations, made it necessary to improve the techniques of navigation, and these in turn depended to a large extent on developments in astronomy. In 1543 Copernicus published On the revolution of the heavenly bodies, and then the New astronomy of Kepler, containing his first and second laws for the motion of planets around the sun, appeared in 1609. The third law was published by Kepler in 1618 in his book Harmony of the world. Galileo, on the basis of his study of Archimedes and his own experiments, laid the foundations for the new mechanics, an indispensable science for the newly arising technology. During the Renaissance the Europeans finally became acquainted with Greek mathematics by way of the Arabic translations, after a period of almost one thousand years of scientific stagnation. The books of Euclid, Ptolemy, and Al-Kharizmi were translated in the 12th century from Arabic into Latin, the common scientific language of Western Europe, and at the same time the earlier Greek and Roman system of calculation was gradually replaced by the vastly superior Indian method, which also reached the Europeans via the Arabs. It was not until the 16th century that European mathematicians finally surpassed the achievements of their predecessors, with the solution by the Italians Tartaglia and Ferrari of the general cubic and fourth degree equations. The concepts of variable magnitude and function arose gradually as a result 9

10 CHAPTER 1. POLYNOMIALS of the interest of physics in laws of motion, as for example in the work of Kepler and Galileo. Galileo discovered the law of falling bodies by establishing that the distance fallen increases in direct proportion to the square of the time. The appearance in 1637 of the new geometry of Descartes marked the first definite step toward a mathematics of variable magnitudes. This combined algebraic and geometric techniques, and is now known as analytic geometry. The main content of the new geometry was the theory of conic sections: the ellipse, hyperbola, and parabola. This theory had been developed extensively by the ancient Greeks in geometric form, and the combination of this knowledge with algebraic techniques and the general idea of a variable magnitude produced analytic geometry. For the Greeks the conic sections were a subject of purely mathematical interest, but by the time of Descartes they were of practical importance for astronomy, mechanics, and technology. Kepler discovered that each planet travels around the sun in an elliptical orbit, and Galileo established that an object thrown in the air travels along a parabolic path (both of these are only first approximations). These discoveries made it necessary to calculate various magnitudes associated with the conic sections and it was the method of Descartes that solved this problem. The next decisive step was take by Newton and Leibnitz during the second half of the 16th century, and resulted in the founding of differential and integral calculus. The Greeks and later mathematicians had studied the geometric problems of drawing tangents to curves and finding areas and volumes of figures. The remarkable discovery of the relation of these problems to the problems of the new mechanics and the formulation of general methods for solving them was brought to completion in the work of Newton and Leibnitz. This relation was discovered because of the possibility, through the use of analytic geometry, of making a graphical representation of how one variable depends on another. In short, what is involved is the construction of a geometric model of relationships involving variable magnitudes. The simplest relationships are those given by polynomials such as x 3 2x + 3. The most elementary ones are the linear polynomials, which have the general form mx + b, for constants m and b. Complicated expressions like e x2 sin 3 (x) are much more difficult to work with than polynomials, and so many times it is useful to approximate such complicated expressions by using polynomials. The simplest case would be to attempt to approximate a function by a linear function of the form f (x) = mx + b. At best this is only possible for a small interval of x values, and so differential calculus focuses on the construction and use of tangent lines at various values of x. By using higher derivatives, the idea of a tangent line can be extended to the idea of polynomials of higher degree which are tangent in some sense to a given curve. These ideas are introduced in Chapter 2, and provide the motivation for much of this chapter, which focuses on the algebra of polynomials.

1.1. ROOTS OF POLYNOMIALS 11 1.1 Roots of polynomials We begin with some notation. The set {0, ±1, ±2, ±3,...} is called the set of integers. We will use Z for the set of integers, and Q for the set of rational numbers. That is, { m } Q = n m, n are integers and n = 0 where we must agree that m/n and p/q represent the same ratio if mq = np. By identifying the integer m with the fraction m/1, we can think of the set Z as a subset of Q, and so we can say, very roughly, that we have enlarged the set of integers to the set Q of rational numbers in order to have a set in which division is possible. We will use the symbol R for the set of real numbers. We will simply view them as the set of all decimal numbers. They can be thought of as the coordinates of the points on a straight line. A precise development of the construction of the set of real numbers requires more sophisticated concepts and much more time than we have available to us at this point. The rational numbers can be viewed as a subset of R, since fractions correspond to decimals that are either terminating or repeating. 1.1.1 DEFINITION. An expression of the form a m x m + a m 1 x m 1 + + a 1 x + a 0 is called a polynomial in the indeterminate x. The exponents (and subscripts) m, m-1,..., 1, 0 must be non-negative integers, and we will assume that the coefficients a m, a m 1,..., a 0 are real numbers. We say that an expression of this form is a polynomial over R. If n is the largest nonnegative integer such that a n = 0, then we say that the polynomial has degree n, and a n is called the leading coefficient of the polynomial. According to this definition, the zero polynomial has no degree, and a constant polynomial a 0 has degree 0 when a 0 = 0. It is important to note that two polynomials are equal precisely when they have the same degree and all corresponding coefficients are equal. If a(x) = a m x m + a m 1 x m 1 + + a 0 and b(x) = b n x n + b n 1 x n 1 + + b 0 are polynomials, then a(x) and b(x) can be added by just adding corresponding coefficients. Their product a(x)b(x) is a(x)b(x) = a m b n x n+m + + (a 0 b 2 + a 1 b 1 + a 2 b 0 )x 2 + (a 0 b 1 + a 1 b 0 )x + a 0 b 0.

12 CHAPTER 1. POLYNOMIALS In the above formula, two polynomials are multiplied by multiplying each term of the first by each term of the second, and then collecting similar terms. Table 1.1.1 shows an efficient way to do this, by arranging the similar terms in columns. Table 1.1.1: (2x 4 5x 3 + 3x + 1)(x 2 + 2x 1) 2x 4 5x 3 +3x +1 x 2 +2x 4 8x 4 +20x 3 12x 4 +4x 5 10x 4 +6x 2 +2x 2x 6 5x 5 +3x 3 +x 2 2x 6 x 5 18x 4 +23x 3 +7x 2 10x 4 We will sometimes need to simplify expressions such as (x + c) n. For n = 2 and n = 3 we have the following identities. (x + c) 2 = (x + c)(x + c) = x(x + c) + c(x + c) = x 2 + xc + cx + c 2 = x 2 + 2cx + c 2 (x + c) 3 = (x + c)(x + c) 2 = x(x 2 + 2cx + c 2 ) + c(x 2 + 2cx + c 2 ) = x 3 + 2cx 2 + c 2 x + cx 2 + 2c 2 x + c 3 = x 3 + 3cx 2 + 3c 2 x + c 3 The coefficients in the above identities can be found from Pascal s triangle, which is given in Table 1.1.2. In this triangle each row begins and ends with 1, and the other terms are found by adding together the two numbers immediately above the term. The last row gives the coefficients for (x + c) 6. In Section 1.6 we will give a proof of the Binomial Theorem, which computes the coefficients by using a different formula, and then we will show that we get the same answers either way. Let a(x) = a m x m +a m 1 x m 1 + +a 0 and b(x) = b n x n +b n 1 x n 1 + +b 0 be polynomials. To write down the general formula for the product a(x)b(x), it is

1.1. ROOTS OF POLYNOMIALS 13 Table 1.1.2: Pascal s triangle (to n = 6 ) 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 useful to introduce a notation for sums. It is traditional to use a Greek letter, sigma, to denote a sum. To show the terms that are being added we typically use subscripts, and then part of the notation tells what the subscripts should be. For example, we can write 2 a 0 b 2 + a 1 b 1 + a 2 b 0 = a i b 2 i to describe the coefficient of x 2 in a(x)b(x). In general, the coefficient c k of a(x)b(x) is given by the formula k c k = a i b k i, which can also be written as i=0 c k = i+ j=k i=0 a i b j. 1.1.2 PROPOSITION. If a(x) and b(x) are nonzero polynomials over R, then the product a(x)b(x) is nonzero and deg(a(x)b(x)) = deg(a(x)) + deg(b(x)). Proof. Suppose that a(x) = a m x m + + a 0 and b(x) = b n x n + + b 0, with the degree of a(x) = m and the degree of b(x) = n, so that a m = 0 and b n = 0. From the general formula for multiplication of polynomials, the leading coefficient of a(x)b(x) is a m b n, which must be nonzero since the product of nonzero real numbers is nonzero. Thus the degree of a(x)b(x) is m + n since a m b n is the coefficient of x m+n.

14 CHAPTER 1. POLYNOMIALS We are interested in solving polynomial equations, or equivalently, in finding roots of polynomials. The main result that we need is that c is a root of a given polynomial p(x) if and only if x c is a factor of p(x). To check this we need to be able to divide p(x) by x c. The next example should help you recall how to divide polynomials. Example 1.1.1 ( (2x 3 3x 2 + 5x + 1) (x + 1) ). To divide the polynomial 2x 3 3x 2 + 5x + 1 by x + 1 we can use the standard algorithm for division, as illustrated in Table 1.1.3. Table 1.1.3: (2x 3 3x 2 + 5x + 1) (x + 1) 2x 2 5x +10 x + 1 2x 3 3x 2 +5x +1 2x 3 +2x 2 5x 2 +5x 5x 2 5x 10x +1 10x +10 9 Thus 2x 3 3x 2 + 5x + 1 = (x + 1)(2x 2 5x + 10) + ( 9). Theorem 1.1.4 will show that for any polynomial p(x), the remainder when p(x) is divided by x c is p(c). That is, p(x) = (x c)q(x) + p(c). The remainder p(c) and quotient q(x) are unique. The usual proof of this theorem uses the algorithm that we followed in Example 1.1.1. The next lemma leads to a simpler proof. 1.1.3 LEMMA. For any real number c and any positive integer k, the linear term x c is a factor of x k c k.

1.1. ROOTS OF POLYNOMIALS 15 Proof. The proof consists of checking that the following factorization works. x k c k = (x c)(x k 1 + cx k 2 + + c k 2 x + c k 1 ). Working with the right hand side, we have (x c)(x k 1 + cx k 2 + + c k 2 x + c k 1 ) = x(x k 1 +cx k 2 + +c k 2 x+c k 1 ) c(x k 1 +cx k 2 + +c k 2 x+c k 1 ) = x k + cx k 1 cx k 1 + + c k 1 x c k 1 x c k = x k c k, and so the factorization is correct. 1.1.4 THEOREM (The remainder theorem). Let p(x) be a nonzero polynomial over R, and let c be any real number. Then there exists a polynomial q(x) with real coefficients such that p(x) = (x c)q(x) + p(c). Moreover, if p(x) = (x c)q 1 (x) + k, where q 1 (x) is a polynomial over R and k is a real number, then q 1 (x) = q(x) and k = p(c). Proof. If p(x) = a m x m + +a 0, then p(x) p(c) = a m (x m c m )+ +a 1 (x c). By Lemma 1.1.3, x c is a factor of each term on the right hand side of the equation, and so it must be a factor of p(x) p(c). Thus p(x) p(c) = (x c)q(x) for some polynomial q(x) over R, or equivalently, p(x) = (x c)q(x) + p(c). If p(x) = (x c)q 1 (x) + k, then (x c)(q(x) q 1 (x)) = k p(c). If q(x) q 1 (x) = 0, then by Proposition 1.1.2 the left hand side of the equation has degree 1, which contradicts the fact that the right hand side of the equation is a constant. Thus q(x) q 1 (x) = 0, which also implies that k p(c) = 0, and so the quotient and remainder are unique. Given a polynomial p(x) and a real number c, to find p(c) we simply substitute c in place of x. We can also refer to this process by saying that we evaluate p(x) at c. When evaluating a polynomial, it is often best to write it with nested parentheses, as in the following example. In the polynomial 2x 3 7x 2 + 5x + 20 we first factor x out of 2x 3 7x 2 + 5x to get 2x 3 7x 2 + 5x + 20 = (2x 2 7x + 5)x + 20.

16 CHAPTER 1. POLYNOMIALS We then factor x out of 2x 2 7x, giving 2x 3 7x 2 + 5x + 20 = ((2x 7)x + 5)x + 20. When entering a polynomial into a graphing calculator, using the nested parentheses form will sometimes speed up the graphing process, since no exponentials are used, but just addition, subtraction, and multiplication. Here is another example, in which we have inserted zeros for the appropriate coefficients. x 5 + 2x 2 3 = x 5 + 0x 4 + 0x 3 + 2x 2 + 0x 3 = ((((x + 0)x + 0)x + 2)x + 0)x 3 1.1.5 DEFINITION. Let p(x) = a m x m + + a 0 be a polynomial over R. A real number c is called a root of the polynomial p(x) if p(c) = 0. 1.1.6 COROLLARY. Let p(x) be a nonzero polynomial over R, and let c be any real number. Then c if a root of p(x) if and only if x c is a factor of p(x). Proof. Using the Remainder Theorem, we can write p(x) = (x c)q(x) + p(c), and then it follows that p(c) = 0 if and only if p(x) = (x c)q(x). 1.1.7 COROLLARY. A polynomial of degree n has at most n distinct roots. Proof. Suppose that p(x) is a polynomial of degree n. If c is a root of p(x), then by Corollary 1.1.6 we can write p(x) = (x c)q(x), for some polynomial q(x). If a is any root of p(x), then substituting shows that (a c)q(a) = 0, which implies that either q(a) = 0 or a = c. This shows that we can reduce the problem to a polynomial of lower degree, since q(x) has degree n 1. If we already know that q(x) has at most n 1 distinct roots, then p(x) can have at most n 1 distinct roots which are different from c. One of the facts from geometry is that two distinct points determine a unique straight line. In the language of polynomials, this translates into the statement that if p(x) = a 1 x + a 0 is a polynomial of degree 1, then knowing p(c 1 ) and p(c 2 ) for any two points c 1 = c 2 completely determines the coefficients a 1 and a 0.

1.1. ROOTS OF POLYNOMIALS 17 The next proposition will show that a similar result holds for quadratic polynomials. If p(x) = a 2 x 2 + a 1 x + a 0 and we are given p(c 1 ), p(c 2 ), and p(c 3 ), for three distinct points c 1, c 2, and c 3, then the coefficients a 2, a 1, and a 0 are completely determined. In geometric terms, this implies that three points (not on a line) determine a unique parabola. 1.1.8 PROPOSITION. Let a(x) and b(x) be polynomials of degree n. If a(x) and b(x) agree at n + 1 distinct points, then they must be equal. Proof. Let p(x) = a(x) b(x). If a(c i ) = b(c i ) for 1 i n + 1, then p(x) has n + 1 distinct roots. If p(x) is nonzero, then deg(p(x)) n because both p(x) and b(x) have degree n, and so Corollary 1.1.7 shows that this cannot happen. The only possibility that is left is that p(x) is the zero polynomial, which shows that a(x) = b(x). Example 1.1.2 ((x 1) 3 + 3(x 1) 2 + 3(x 1) + 1 = x 3 ). Let a(x) = x 3 and b(x) = (x 1) 3 +3(x 1) 2 +3(x 1)+1. To show that a(x) and b(x) are equal we only need to show that they agree at four distinct points. We will evaluate both polynomials at x = 1, x = 0, x = 1, and x = 2. We have a( 1) = 1, a(0) = 0, a(1) = 1, and a(2) = 8. On the other hand, we have b( 1) = 8+12 6+1 = 1, b(0) = 1 + 3 3 + 1 = 0, b(1) = 1, and b(2) = 1 + 3 + 3 + 1 = 8. Therefore x 3 = (x 1) 3 + 3(x 1) 2 + 3(x 1) + 1. PROBLEMS: Section 1.1 1. Write out the expansions of (x + c) 4 and (x + c) 5. 2. Add the next row to Pascal s triangle, as given in the text, and use it to write out the expansion of (x + 2) 7. 3. Rewrite x 5 7x 3 + 5x 2 3x + 4 using nested parentheses. 4. Rewrite x 4 + 5x 3 + 11x 2 2x 8 using nested parentheses. 5. Use Proposition 1.1.8 to show that the polynomials a(x) = x 3 + x 4 and b(x) = (x + 1) 3 3(x + 1) 2 + 4(x + 1) 6 are equal.

18 CHAPTER 1. POLYNOMIALS 6. Use Proposition 1.1.8 to show that the polynomials a(x) = x 3 +1 and b(x) = 3 2 (x 1)(x)(x + 1) (x 2)(x)(x + 1) + 1 (x 2)(x 1)(x + 1) are equal. 2 7. Find q(x) with x 6 1 = (x 3 1)q(x), showing that x 3 1 is a factor of x 6 1. 8. Find q(x) with x 9 1 = (x 3 1)q(x), showing that x 3 1 is a factor of x 9 1. 9. Suppose that m and n are positive integers for which m is a factor of n. Explain why x m 1 must be a factor of x n 1. Hint: Write n = mk and apply Lemma 1.1.3. 10. Explain why rational numbers correspond to decimals that are either repeating or terminating. Hint: If q = m/n, then when dividing m by n to put q into decimal form there are at most n different remainders. Conversely, if d is a repeating decimal, then find s, t such that 10 s d 10 t d is an integer.

1.2. RATIONAL ROOTS 19 1.2 Rational roots In this section we will work with polynomials that have integer coefficients. The easiest roots to find are those which are rational numbers, and the next proposition shows that there are only a finite number of possibilities that must be checked. Using a computer program or calculator to graph the polynomial function can help a great deal in deciding which possible roots actually work. 1.2.1 PROPOSITION (Rational roots). Let p(x) = a m x m + a m 1 x m 1 + + a 1 x + a 0 be a polynomial with integer coefficients. If r/s is a rational root of p(x) such that r and s have no factors in common, then r must be a factor of a 0 and s must be a factor of a m. Proof. If p( r ) = 0, then s a m ( r s )m + a m 1 ( r s )m 1 + + a 1 ( r s ) + a 0 = 0, then multiplying by s m gives the equation This leads to the equations a m r m + a m 1 r m 1 s + + a 1 rs m 1 + a 0 s m = 0. a m r m = s( a m 1 r m 1 a 1 rs m 2 a 0 s m 1 ) and a 0 s m = r( a m r m 1 a m 1 r m 2 s a 1 s m 1 ). Thus s is a factor of a m r m and r is a factor of a 0 s m. Since r and s are assumed to have no factors in common, we see that r must be a factor of a 0 and s must be a factor of a m. Example 1.2.1. Suppose that we need to find all integer roots of the polynomial p(x) = x 3 5x 2 + 2x 10. Using Proposition 1.2.1, all rational roots of p(x) can be found by testing only a finite number of values. By looking at the signs we can see that p(x) cannot have any negative roots, so we only need to check

20 CHAPTER 1. POLYNOMIALS the positive factors of 10. Substituting these factors into p(x) we obtain p(1) = 12, p(2) = 18 and p(5) = 0. Thus 5 is a root of p(x), and so we can use polynomial division to show that x 3 5x 2 + 2x 10 = (x 5)(x 2 + 2). Because x 2 + 2 has no real roots, this completes the factorization, and it follows that 5 is the only integer root. Example 1.2.2. In this example we will find the rational roots of p(x) = 9x 4 6x 3 + 19x 2 12x + 2. The possible numerators are the factors of 2, while the possible denominators are the factors of 9. The list of possible rational roots is ±1, ±2, ± 1 3, ±2 3, ±1 9, ±2 9. Graphing the function shows that there is a root between.3 and.4, so this suggests that 1/3 might be a root. (Since the possible roots lie between 0 and 2, on your graphing calculator first use the window 0 x 2; 1 y 1. To take a closer look, you might try the window 0 x.6;.2 y.2.) If 1/3 is a root, then we know that x 1/3 must be a factor of p(x). In order to keep to integer coefficients, it is better to check whether or not 3x 1 is a factor of p(x). In fact, using polynomial division we can see that p(x) = (3x 1)(3x 3 x 2 + 6x 2). Dividing again by 3x 1 gives the complete factorization p(x) = (3x 1) 2 (x 2 + 2), and so the only root of p(x) is 1/3. We need to find some easier methods for evaluating polynomials, and so we will take a closer look at the method using nested parentheses. The algorithm for evaluating a polynomial by using nested parentheses is usually called synthetic division. Suppose that p(x) = a m x m +a m 1 x m 1 + +a 0 and q(x) = b m 1 x m 1 + b m 2 x m 2 + + b 0 with p(x) = (x c)q(x) + p(c). Multiplying this out and equating coefficients shows that the following equations hold.

1.2. RATIONAL ROOTS 21 a m = b m 1 a m 1 = b m 2 cb m 1. a 1 = b 0 cb 1 a 0 = p(c) cb 0 Solving for the coefficients of q(x) gives b m 1 = a m b m 2 = b m 1 c + a m 1 = a m c + a m 1 b m 3 = b m 2 c + a m 2 = (a m c + a m 1 )c + a m 2. This shows that the coefficients of q(x) can be found from the partial answers in the nested parentheses procedure for evaluating p(c). Example 1.2.3. We will use nested parentheses to divide p(x) = 2x 3 7x 2 +5x +20 by x 2. As before, we can write 2x 3 7x 2 + 5x + 20 = ((2x 7)x + 5)x + 20, and then we have the following steps: Thus we have b 2 = 2 b 1 = 2 2 7 = 3 b 0 = ( 3) 2 + 5 = 1 p(2) = ( 1) 2 + 20 = 18. 2x 3 7x 2 + 5x + 20 = (x 2)(2x 2 3x 1) + 18.

22 CHAPTER 1. POLYNOMIALS 1.2.2 PROPOSITION (Synthetic division). Let p(x) = a m x m +a m 1 x m 1 + + a 0 be a polynomial with real coefficients. If p(x) = (x c)q(x) + p(c), where q(x) = b m 1 x m 1 + b m 2 x m 2 + + b 0, then b m 1 = a m b m 2 = b m 1 c + a m 1. b i = b i+1 c + a i+1. b 0 = b 1 c + a 1 and p(c) = b 0 c + a 0. Example 1.2.4 (The synthetic division algorithm). In using synthetic division to divide p(x) = a m x m +a m 1 x m 1 + +a 0 by x c, the work is usually arranged as follows. c a m a m 1 a m 2... b m 1 c b m 2 c... b m 1 b m 2 b m 3... For example, dividing 2x 3 7x 2 +5x +20 by x 3 gives the following table. 3 2 7 +5 20 +6 3 6 2 1 +2 26 We have 2x 3 7x 2 2x + 20 = (x 3)(2x 2 x + 2) + 26. To construct the table, bring the 2 down to the bottom row, then multiply by 3 and insert it as the next term in the second row. As the next step, add it to the corresponding term of of the top row, and then repeat these steps, using the answer.

1.2. RATIONAL ROOTS 23 In using the synthetic division algorithm, you must insert 0 for any missing power of x. For example, to divide x 5 + 2x 2 3 by x + 1 we would have the following steps. 1 1 0 0 2 0 3 1 1 1 1 1 1 1 1 1 1 2 Thus x 5 + 2x 2 3 = (x + 1)(x 4 x 3 + x 2 + x 1) 2. The most compact notation omits the terms in the middle row, as follows. 1 1 0 0 2 0 3 1 1 1 1 1 2 This will be useful later when we need to do some repeated divisions. Using Proposition 1.2.1, all rational roots of the polynomial p(x) can be found by testing only a finite number of values. We now restrict ourselves even further, and study some ways to find integer roots of equations. The next proposition gives a method due to Newton which can speed up this process considerably. 1.2.3 PROPOSITION. Let p(x) be any polynomial with integer coefficients. If c is any integer root of p(x) and n is any integer, then c n must be a factor of p(n). Proof. If c is an integer root of p(x), then it follows from the Remainder Theorem (see Theorem 1.1.4) that p(x) = (x c)q(x), which we can rewrite as p(x) = (c x)( q(x)). The proof of the Remainder Theorem uses the factorization x k c k = (x c)(x k 1 + cx k 2 + + c k 2 x + c k 1 ). If c is an integer, then the coefficients of x k 1 + cx k 2 + + c k 2 x + c k 1 are certainly integers, and so in the general factorization p(x) = (x c)q(x), it follows that the coefficients of q(x) are integers. If we substitute n into the equation p(x) = (c x)( q(x)), we get p(n) = (c n)( q(n)), which shows that c n is a factor of the integer p(n).

24 CHAPTER 1. POLYNOMIALS Example 1.2.5. The preceding result can be combined with Proposition 1.2.1 to find the rational roots of equations such as x 3 + 15x 2 3x 6 = 0. By Proposition 1.2.1 the possible rational roots are ±1, ±2, ±3, ±6. Letting p(x) = x 3 + 15x 2 3x 6 we find that p(1) = 7. Thus by Proposition 1.2.3, for any root c we are guaranteed that c 1 must be a factor of 7. This eliminates all of the possible values except c = 2 and c = 6. We find that p(2) = 56, so 2 is not a root. This shows, in addition, that c 2 must be a factor of 56 for any root c, but 6 still passes this test. Finally, p( 6) = 24, and so this eliminates 6 and p(x) has no rational roots. If we evaluate a polynomial p(x) at x = c by using synthetic division, we cannot tell whether or not p(c) = 0 until the final step. Our final algorithm uses synthetic division, in reverse. In testing for integer roots, it may be possible to eliminate a potential root after only a few steps of the synthetic division procedure. If the potential integer root passes a divisor test at each step, then it is a root and the answers at each step provide the coefficients of the quotient. If p(x) = a m x m +a m 1 x m 1 + +a 0 is a polynomial with integer coefficients, and c is an integer root of p(x), then p(x) = (x c)q(x), for a polynomial q(x) = b m 1 x m 1 + + b 1 x + b 0 with integer coefficients. After multiplying this out, we can solve for the coefficients of q(x). Solving for b 0 gives b 0 = a 0 c, We get the following equations, which we might call backwards synthetic division. b 0 = a 0 c b 1 = (b 0 a 1 ) c b 2 = (b 1 a 2 ) c. Since b 0, b 1, b 2,..., b m 1 are integers, it follows that c must be a factor of each of the terms a 0, (b 0 a 1, etc. Furthermore, the very last term obtained must be equal to zero. This gives a series of checks to test whether or not c is a root. Just as with synthetic division, any zero coefficients must be included in the algorithm. If c passes each test, then not only do you know that c is a root, but in addition you have found the coefficients of the quotient q(x). This proves the following proposition.

1.2. RATIONAL ROOTS 25 1.2.4 PROPOSITION (Integer roots via backwards synthetic division). Let p(x) = a m x m + a m 1 x m 1 + + a 0 be a polynomial with integer coefficients. An integer c is a root of p(x) if and only if each of the following terms is an integer and b m 1 = a m. b 0 = a 0 c b 1 = (b 0 a 1 ) c b 2 = (b 1 a 2 ) c. b m 1 = (b m 2 a m 1 ) c Example 1.2.6. In finding the integer roots of p(x) = x 3 5x 2 + 2x 10, we will use backwards synthetic division to check the potential roots 2 and 5. (Compare what we did in Example 1.2.1.) We have b 0 = ( 10) 2 = 5 b 1 = (5 2) 2 = 3/2, which eliminates 2 as a root since b 1 is not an integer. In testing 5 we get b 0 = ( 10) 5 = 2 b 1 = (2 2) 5 = 0 b 2 = (0 ( 5)) 5 = 1 and b 2 = 1 = a 3. At each step we got an integer value, and so 5 is a root, giving us x 3 5x 2 + 2x 10 = (x 5)(x 2 + 2). PROBLEMS: Section 1.2 1. Use synthetic division to write p(x) = q(x)(x c) + p(c) for (a) p(x) = 2x 3 + x 2 4x + 3; c = 1; (b) p(x) = x 3 5x 2 + 6x + 5; c = 2; (c) p(x) = x 5 7x 2 + 2; c = 2.

26 CHAPTER 1. POLYNOMIALS 2. Use backwards synthetic division to show that p(c) = 0, and to write p(x) = q(x)(x c) for (a) p(x) = x 3 5x 2 + 3x + 9; c = 3; (b) p(x) = 2x 5 + 2x 4 3x 2 + 2x + 5; c = 1; (c) p(x) = x 5 + 32; c = 2. 3. Use Proposition 1.2.1 to list all possible rational roots of p(x) = 15x 4 +8x 3 + 6x 2 9x 2. Then determine which of the possibilities are in fact roots. (You may eliminate possibilities graphically, using a computer or calculator if available, but then verify by hand calculations that you have found the rational roots.) Finally, use the information about roots for factor p(x) completely as a product of polynomials with integer coefficients. 4. Use the method outlined in the previous problem to find all rational roots of p(x) = 30x 5 7x 4 + 20x 3 4x 2 10x + 3, and use this information to write out the factorization of p(x). 5. Find all integer roots of the following equations (use any method). One suggestion is to use a calculator or computer to graph the polynomial function, then check the likely roots using synthetic division. Finally, show the complete factorization of each polynomial (into a product of polynomials with integer coefficients). (a) x 3 x 2 4x 6 = 0 (b) x 3 8x 2 3x + 91 = 0 (c) x 4 + 4x 3 104x 2 105x 108 = 0 (d) x 3 6x 2 24x + 64 = 0 (e) 3x 4 5x 3 7x 2 + 9x 2 = 0 6. Let p(x) = a m x m +a m 1 x m 1 + +a 1 x +a 0 be a polynomial with rational coefficients. Show that if b is a nonzero root of p(x), then 1/b is a root of q(x) = a 0 x m + a 1 x m 1 + + a m 1 x + a m.

1.3. APPROXIMATING REAL ROOTS 27 1.3 Approximating real roots In this section we will develop some techniques for approximating solutions to polynomial equations. We will introduce the interval bisection method, the secant method, and an algorithm for approximating square roots and cube roots. In searching for all roots of a given equation, it is helpful to have a rough bound on the size of the roots. The first proposition provides such a bound. 1.3.1 PROPOSITION. Let p(x) = a m x m + + a 1 x + a 0. If c is any root of the equation p(x) = 0, then c a m + a m 1 + + a 1 + a 0. a m Proof. If c is any root of the equation p(x) = 0, then and so Dividing by a m c m 1 gives a m c m + a m 1 c m 1 + + a 1 c + a 0 = 0, a m c m = a m 1 c m 1 a 1 c a 0. c = a m 1 a m 1 c a 1 a m 1 c m 2 a 0 a m 1 c m 1. If we assume that c 1, then 1 1, and so we have c c = a m 1 1 a m c a 1 1 a m c a 0 1 m 2 a m c m 1 a m 1 1 a m c + + a 1 1 a m c m 2 + a 0 a m a m 1 + + a 1 + a 0. a m 1 c m 1 Finally, to obtain a bound that also works when c < 1, we can simply add 1 to the bound we already have, giving This completes the proof. c 1 + a m 1 + + a 1 + a 0 a m = a m + a m 1 + + a 1 + a 0. a m

28 CHAPTER 1. POLYNOMIALS 1.3.2 ALGORITHM (The interval bisection method for approximating roots). This algorithm is based on the following result. If we are solving the polynomial equation p(x) = 0 and find two numbers x 1 < x 2 for which p(x 1 ) and p(x 2 ) have opposite signs, then there must be a solution between x 1 and x 2. That is, there is a solution in the interval [x 1, x 2 ]. The algorithm then proceeds as follows. To narrow down the interval in which there must be a solution, we evaluate p(x) at the average x 3 = (x 1 + x 2 )/2, which bisects the interval determined by x 1 and x 2. If p(x 3 ) has the opposite sign from p(x 1 ), then the root is between x 1 and x 3, and so we repeat the procedure for the interval [x 1, x 3 ]. On the other hand, if p(x 3 ) has the opposite sign from p(x 2 ), then the root is between x 2 and x 3, and so we repeat the procedure using the interval [x 3, x 2 ]. It is easy to write a program to carry out this algorithm. But using the interval bisection algorithm by hand on a calculator produces numbers that are difficult to work with, and so we will use a modification of the algorithm. Suppose that the initial interval is [0, 1]. For the first step we bisect the interval, and consider either [0,.5] or [.5, 1]. Suppose that p(.5) and p(1) have opposite signs. For the next step, rather than bisecting the interval, we consider either [.5,.7] or [.7, 1]. The first of these can be bisected in the third step, while the interval [.7, 1] can be split into [.7,.8] and [.8, 1]. This gives a search procedure that increases the accuracy by one decimal point, each time that we make the necessary three or four calculations. We will illustrate this method by solving x 2 3 = 0, correct to three decimal places. We know that the solution is x = 3, which lies between 1 and 2, and so this problem provides an easy example of the technique we want to introduce. Let p(x) = x 2 3, so that we are solving the equation p(x) = 0. The first three computations in the next paragraph will find that p(1.7) =.11 and p(1.8) = +.24, so that the answer must be between 1.7 and 1.8. It takes three more computations to show that the answer lies between 1.73 and 1.74, and an additional three computations to obtain the answer 1.732, correct to three decimal places. After knowing that the answer is between 1 and 2, we do the computation for 1.5, half way between the two previous estimates. Because p(1.5) =.75 is negative, the answer must be between 1.5 and 2.0. To simplify matters and deal only with one decimal point at a time, instead of using an estimate of 1.75, which would be half way between 1.5 and 2, we use 1.7, and then when p(1.7) =.11 we go to p(1.8) = +.24.

1.3. APPROXIMATING REAL ROOTS 29 The following table shows the remaining steps in the procedure. a 1.75 1.73 1.74 1.735 1.733 1.732 a 2 +.0625.0071 +.0276 +.010225 +.003289.000176 A graphing calculator or computer can be used in a similar way. By changing the domain of the viewing window to give successively smaller and smaller intervals, it is possible to obtain more and more accurate estimates of where the graph of the polynomial crosses the x-axis. Example 1.3.1. We will solve x 4 3 = 0, correct to five decimal places. We let p(x) = x 4 3. (The values for p(x) have been rounded off to eight decimal places.) a 1.5 1.3 1.4 p(a) 2.0625.1439 +.8416 a 1.35 1.33 1.32 1.31 p(a) +.32150625 +.12900721 +.03595776.05500079 a 1.315 1.317 1.316 p(a).00978090 +.00845209.00067480 a 1.3165 1.3163 1.3162 1.3161 p(a) +.00388605 +.00206109 +.00114892 +.00023696 a 1.31605 1.31607 1.31608 p(a).00021894.00003659 +.0000546 This shows that answer we are looking for is 1.31607. Example 1.3.2. In this example we will approximate a solution to the equation 2x 3 + x 2 = 0. Let p(x) = 2x 3 + x 2. The possible rational solutions to the equation are x = ±1, ±2, ± 1, and substituting each of these values 2 for x shows that the solution(s) must be irrational. But p(0) = 2 and

30 CHAPTER 1. POLYNOMIALS p(1) = 1, so there must be a solution to the equation between x = 0 and x = 1. In the table below we have not listed all of the steps in the search procedure. We have listed only those values for which the function changes sign, to show how we can narrow down the region in which the root lies. x.8.9 f (x).176.358 x.83.84 f (x).026426.025408 x.835.836 f (x).0006342.0045541 x.8351.8352 f (x).0001159.0004026 x.83512.83513 f (x).0000121.0000397 1.3.3 ALGORITHM (The secant method for approximating roots). As in the interval bisection algorithm, to solve p(x) = 0 we begin with two numbers x 1 < x 2 for which p(x 1 ) and p(x 2 ) have opposite signs. The equation of the line joining the two points (x 1, p(x 1 )) and (x 2, p(x 2 )) is y = m(x x 1 ) + p(x 1 ), where m = p(x 2) p(x 1 ) x 2 x 1. If we set this equation equal to zero and solve, we get x = x 1 p(x 1) m and then x = x 1 p(x 1)(x 2 x 1 ) p(x 2 ) p(x 1 ). Let the solution to this equation be x 3, so that x 3 = x 1 p(x 1)(x 2 x 1 ) p(x 2 ) p(x 1 ). We still have to find p(x 3 ), in order to determine whether the root is in the interval [x 1, x 3 ] or in the interval [x 3, x 2 ]. We then repeat the procedure on the new interval.

1.3. APPROXIMATING REAL ROOTS 31 To implement this algorithm on a calculator capable of storing functions with two variables, you can store p(x), p(y), p(z), and f (x, y) = x (p(x))(y x) ((p(y)) (p(x))). Store the endpoints of the interval as x and y, and then store the new approximation as z. After evaluating p(z) to determine its sign, you must decide whether to store your new approximation as x or as y, making sure that the root is in the new interval [x, y]. Example 1.3.3. In this example we will compare the secant method with the interval bisection method that we used in Example 1.3.2 to find a root of p(x) = 2x 3 +x 2 between 0 and 1. The secant method produces the following table. x y p(x) p(y) 0 1 2 1.66666667 1.74074074 1.80851064 1.13445961 1.83120654 1.02022509 1.83455273 1.00295162 1.83503963 1.00042884 1 The table in Example 1.3.2 contains only about one third of the computations actually that are actually necessary, so when we compare it with the above table, we see that the secant method is a major improvement. Notice that the value of y never changed, since the successive approximations all remained less than the actual root. We are free to choose any value of y for which p(y) is positive, and so it helps to use trial and error in finding the second endpoint. We now repeat the computations,

32 CHAPTER 1. POLYNOMIALS making appropriate modifications in y. x y p(x) p(y) 0 1 2 1.66666667 1.74074074 1.80851064.9.13445961.358.83349060.84.00844657.025408.83511467.836.00003981.00455411.83512234.835123.00000003.00000338 The table shows that to six decimal places the root is.835122. Both the interval bisection method and the secant method are quite simple to use. However, for finding square roots there is another algorithm due to Newton that is almost as simple but much more efficient. 1.3.4 ALGORITHM (Newton s method for approximating square roots). If a is an approximate value for c, then the actual value must lie between a and c/a, since a c a = c. (If a = c/a, then a a = c and a is actually equal to c.) If a = c/a, then one of a and c/a is too large and one is too small, so the average (a + c/a)/2 of these two values will give a better approximation to c. This procedure, called Newton s method, can be continued to give more and more accurate approximations. If a n is an approximation for c, then the next approximation is given by the formula a n+1 = 1 2 (a n + can ). If you have a calculator capable of storing functions, you can store f (x) = (x + c x) 2. If a is an approximation for c, then the next approximation in Newton s method is f (a). This makes it very easy to carry out the steps in Newton s method.

1.3. APPROXIMATING REAL ROOTS 33 Example 1.3.4. We will use Newton s method to approximate 3 to five decimal places. We first use trial and error to get an answer accurate to one decimal place, giving a first approximation of a 1 = 1.7. The average of 1.7 and 3/1.7 is 1.7323529 = a 2, and then we can use this value to compute the next approximation a 3. Note that the calculations have been done on an eight digit calculator. Step 1 Step 2 Step 3 a 1.7 1.7323529 1.7320508 3/a 1.7647058 1.7317487 1.7320508 avg 1.7323529 1.7320508 1.7320508 We can end with the approximation a 3 = 1.7320508, since this carries Newton s method to the limits of the calculator. This procedure arrives at an accurate answer much faster than the modified interval bisection algorithm that we used earlier. In fact, each step in the algorithm roughly doubles the number of decimal places to which the approximation is accurate. Example 1.3.5. In Newton s method, fractions can be used instead of decimal approximations. For example, squaring 7/4 gives 49/16, which is very close to 3. Using a 1 = 7/4 as the first approximation to 3 gives the sequence of approximations a 1 = 7/4, a 2 = 97/56, and a 3 = 18817/10864. To approximate 2 we can begin with a 1 = 7/5. Newton s method then gives a 2 = 99/70 and a 3 = 19601/13860. PROBLEMS: Section 1.3 1. Use the (modified) interval bisection method to approximate 3 12 to four decimal places. (Work with the polynomial p(x) = x 3 12.) 2. Use the (modified) interval bisection method to approximate 17 to three decimal places.

34 CHAPTER 1. POLYNOMIALS 3. Use Newton s method to approximate 17 to six decimal places. 4. Check the calculations in Example 1.3.5. 5. Use Newton s method to find a fraction that approximates 17. Begin with a 1 = 4, and then compute a 2 and a 3 (as fractions). 6. Use Newton s method to find a fraction that approximates 10. Begin with a 1 = 3, and then compute a 2, a 3, and a 4. 7. Use the (modified) interval bisection method to find the root of the equation 4x 3 x 2 = 0 that lies between x = 0 and x = 1. Show that your answer is accurate to at least four decimal places. 8. Check the calculations in Example 1.3.3. 9. Use the secant method to solve the equation 4x 3 x 2 = 0. Obtain an answer accurate to six decimal places. 10. Use the secant method to solve the equation x 5 x 2 6 = 0. Obtain an answer accurate to six decimal places.

1.4. COMPLEX ROOTS 35 1.4 Complex roots If x is any real number, then x 2 + 1 1. This means that the equation x 2 + 1 = 0 has no real root. In this section we will construct a set of numbers that enlarges the set of real numbers to include a root of this equation. A set of numbers that contains a root i of the equation x 2 +1 = 0 and is closed under addition, subtraction, multiplication, and division must include all numbers of the form a + bi, where a and b are real numbers. The addition and multiplication must be given by (a + bi) + (c + di) = (a + c) + (b + d)i, and (a + bi)(c + di) = ac + (bc + ad)i + bdi 2 = (ac bd) + (ad + bc)i. Here we have used the fact that i 2 = 1 since i 2 + 1 = 0. We will simply invent a symbol i for which i 2 = 1, and then consider all pairs of real numbers a and b, in the form a + bi. This construction can be done much more formally, but at this level we hope that the reader will just accept the invention of the symbol i. 1.4.1 DEFINITION. The set C = {a + bi a, b R and i 2 = 1} is called the set of complex numbers, with addition and multiplication defined as follows: (a + bi) + (c + di) = (a + c) + (b + d)i, (a + bi)(c + di) = (ac bd) + (ad + bc)i. We say that complex numbers a+bi and c+di are equal (written a+bi = c+di) if and only if a = c and b = d. If c + di is nonzero, that is, if c = 0 or d = 0, then division by c + di is possible: a + bi c + di = (a + bi)(c di) (c + di)(c di) ac + bd bc ad = + c 2 + d2 c 2 + d i. 2 It is often helpful to use a geometric model for the set of complex numbers. The number a + bi can be viewed as the ordered pair (a, b) in the plane. (See Figure 1.4.1.) Note that i corresponds to the pair (0, 1). In polar coordinates, a + bi is represented by (r, θ), where r = a 2 + b 2 cos θ = a/r sin θ = b/r.

36 CHAPTER 1. POLYNOMIALS 1 i θ a + bi r b a 1 i Figure 1.4.1: The value r is called the absolute value of a+bi, and we write a+bi = a 2 + b 2. This definition is the same as for real numbers, since if x R, then x represents the distance from x to 0, The polar form for complex numbers is very useful. With this notation we have a + bi = r(cos θ + i sin θ). In polar form the product of two complex numbers is given by r(cos θ + i sin θ) t(cos φ + i sin φ) = rt((cos θ cos φ sin θ sin φ) + i(sin θ cos φ + cos θ sin φ)) = rt(cos(θ + φ) + i sin(θ + φ)). This simplification of the product comes from the trigonometric formulas for the cosine and sine of the sum of two angles, which we recall below. cos(θ + φ) = cos θ cos φ sin θ sin φ sin(θ + φ) = sin θ cos φ + cos θ sin φ To multiply two complex numbers represented in polar form, we multiply their absolute values and add their angles. A repeated application of this formula to cos θ + i sin θ gives the following theorem. 1.4.2 THEOREM (De Moivre). For any positive integer n, (cos θ + i sin θ) n = cos(nθ) + i sin(nθ).