A.5. Complex Numbers AP-12. The Development of the Real Numbers

Transcription

1 AP- A.5 Comple Numbers Comple numbers are epressions of the form a + ib, where a and b are real numbers and i is a smbol for -. Unfortunatel, the words real and imaginar have connotations that somehow place - in a less favorable position in our minds than. As a matter of fact, a good deal of imagination, in the sense of inventiveness, has been required to construct the real number sstem, which forms the basis of the calculus (see Appendi A.4). In this appendi we review the various stages of this invention. The further invention of a comple number sstem is then presented. The Development of the Real Numbers The earliest stage of number development was the recognition of the counting numbers,, 3, Á, which we now call the natural numbers or the positive integers. Certain simple arithmetical operations can be performed with these numbers without getting outside the sstem. That is, the sstem of positive integers is closed under the operations of addition and multiplication. B this we mean that if m and n are an positive integers, then m + n = p and mn = q are also positive integers. Given the two positive integers on the left side of either equation in (), we can find the corresponding positive integer on the right side. More than this, we can sometimes specif the positive integers m and p and find a positive integer n such that m + n = p. For instance, 3 + n = 7 can be solved when the onl numbers we know are the positive integers. But the equation 7 + n = 3 cannot be solved unless the number sstem is enlarged. The number zero and the negative integers were invented to solve equations like 7 + n = 3. In a civilization that recognizes all the integers Á, -3, -, -, 0,,, 3, Á, an educated person can alwas find the missing integer that solves the equation m + n = p when given the other two integers in the equation. Suppose our educated people also know how to multipl an two of the integers in the list (). If, in Equations (), the are given m and q, the discover that sometimes the can find n and sometimes the cannot. Using their imagination, the ma be () ()

2 A.5 Comple Numbers AP-3 FIGURE A.3 With a straightedge and compass, it is possible to construct a segment of irrational length. inspired to invent still more numbers and introduce fractions, which are just ordered pairs m> nof integers m and n. The number zero has special properties that ma bother them for a while, but the ultimatel discover that it is hand to have all ratios of integers m> n, ecluding onl those having zero in the denominator. This sstem, called the set of rational numbers, is now rich enough for them to perform the rational operations of arithmetic:. (a) addition. (a) multiplication (b) subtraction (b) division on an two numbers in the sstem, ecept that the cannot divide b zero because it is meaningless. The geometr of the unit square (Figure A.3) and the Pthagorean theorem showed that the could construct a geometric line segment that, in terms of some basic unit of length, has length equal to. Thus the could solve the equation = b a geometric construction. But then the discovered that the line segment representing is an incommensurable quantit. This means that cannot be epressed as the ratio of two integer multiples of some unit of length. That is, our educated people could not find a rational number solution of the equation =. There is no rational number whose square is. To see wh, suppose that there were such a rational number. Then we could find integers p and q with no common factor other than, and such that p = q. Since p and q are integers, p must be even; otherwise its product with itself would be odd. In smbols, p = p where is an integer. This leads to p = q, p which sas q must be even, sa q = q, where q is an integer. This makes a factor of both p and q, contrar to our choice of p and q as integers with no common factor other than. Hence there is no rational number whose square is. Although our educated people could not find a rational solution of the equation =, the could get a sequence of rational numbers (3), 7 5, 4 9, 39 69, Á, (4) whose squares form a sequence, 49 5, 68 84, 57, 8,56, Á, (5) that converges to as its limit. This time their imagination suggested that the needed the concept of a limit of a sequence of rational numbers. If we accept the fact that an increasing sequence that is bounded from above alwas approaches a limit (Theorem 6, Section.) and observe that the sequence in (4) has these properties, then we want it to have a limit L. This would also mean, from (5), that L =, and hence L is not one of our rational numbers. If to the rational numbers we further add the limits of all bounded increasing sequences of rational numbers, we arrive at the sstem of all real numbers. The word real is placed in quotes because there is nothing that is either more real or less real about this sstem than there is about an other mathematical sstem.

3 AP-4 The Comple Numbers Imagination was called upon at man stages during the development of the real number sstem. In fact, the art of invention was needed at least three times in constructing the sstems we have discussed so far:. The first invented sstem: the set of all integers as constructed from the counting numbers.. The second invented sstem: the set of rational numbers m> n as constructed from the integers. 3. The third invented sstem: the set of all real numbers as constructed from the rational numbers. These invented sstems form a hierarch in which each sstem contains the previous sstem. Each sstem is also richer than its predecessor in that it permits additional operations to be performed without going outside the sstem:. In the sstem of all integers, we can solve all equations of the form + a = 0, where a can be an integer.. In the sstem of all rational numbers, we can solve all equations of the form a + b = 0, provided a and b are rational numbers and a Z In the sstem of all real numbers, we can solve all of Equations (6) and (7) and, in addition, all quadratic equations a + b + c = 0 having a Z 0 and b - 4ac Ú 0. You are probabl familiar with the formula that gives the solutions of Equation (8), namel, (6) (7) (8) = -b ; b - 4ac, a (9) and are familiar with the further fact that when the discriminant, b - 4ac, is negative, the solutions in Equation (9) do not belong to an of the sstems discussed above. In fact, the ver simple quadratic equation + = 0 is impossible to solve if the onl number sstems that can be used are the three invented sstems mentioned so far. Thus we come to the fourth invented sstem, the set of all comple numbers a + ib. We could dispense entirel with the smbol i and use the ordered pair notation (a, b). Since, under algebraic operations, the numbers a and b are treated somewhat differentl, it is essential to keep the order straight. We therefore might sa that the comple number sstem consists of the set of all ordered pairs of real numbers (a, b), together with the rules b which the are to be equated, added, multiplied, and so on, listed below. We will use both the (a, b) notation and the notation a + ib in the discussion that follows. We call a the real part and b the imaginar part of the comple number (a, b).

4 A.5 Comple Numbers AP-5 We make the following definitions. Equalit a + ib = c + id Two comple numbers (a, b) if and onl if and (c, d) are equal if and onl a = c and b = d. if a = c and b = d. Addition sa + ibd + sc + idd The sum of the two comple = sa + cd + isb + dd numbers (a, b) and (c, d) is the comple number sa + c, b + dd. Multiplication sa + ibdsc + idd The product of two comple = sac - bdd + isad + bcd numbers (a, b) and (c, d) is the comple number sac - bd, ad + bcd. csa + ibd = ac + isbcd The product of a real number c and the comple number (a, b) is the comple number (ac, bc). The set of all comple numbers (a, b) in which the second number b is zero has all the properties of the set of real numbers a. For eample, addition and multiplication of (a, 0) and (c, 0) give which are numbers of the same tpe with imaginar part equal to zero. Also, if we multipl a real number (a, 0) and the comple number (c, d), we get In particular, the comple number (0, 0) plas the role of zero in the comple number sstem, and the comple number (, 0) plas the role of unit or one. The number pair (0, ), which has real part equal to zero and imaginar part equal to one, has the propert that its square, has real part equal to minus one and imaginar part equal to zero. Therefore, in the sstem of comple numbers (a, b) there is a number = s0, d whose square can be added to unit = s, 0d to produce zero = s0, 0d, that is, The equation sa, 0d + sc, 0d = sa + c, 0d, sa, 0d # sc, 0d = sac, 0d, sa, 0d # sc, dd = sac, add = asc, dd. s0, ds0, d = s -, 0d, s0, d + s, 0d = s0, 0d. + = 0 therefore has a solution = s0, d in this new number sstem. You are probabl more familiar with the a + ib notation than ou are with the notation (a, b). And since the laws of algebra for the ordered pairs enable us to write sa, bd = sa, 0d + s0, bd = as, 0d + bs0, d, while (, 0) behaves like unit and (0, ) behaves like a square root of minus one, we need not hesitate to write a + ib in place of (a, b). The i associated with b is like a tracer element

5 AP-6 that tags the imaginar part of a + ib. We can pass at will from the realm of ordered pairs (a, b) to the realm of epressions a + ib, and conversel. But there is nothing less real about the smbol s0, d = i than there is about the smbol s, 0d =, once we have learned the laws of algebra in the comple number sstem of ordered pairs (a, b). To reduce an rational combination of comple numbers to a single comple number, we appl the laws of elementar algebra, replacing i wherever it appears b -. Of course, we cannot divide b the comple number s0, 0d = 0 + i0. But if a + ib Z 0, then we ma carr out a division as follows: c + id a + ib = sc + iddsa - ibd sa + ibdsa - ibd The result is a comple number + i with = ac + bd ad - bc a, = + b a + b, and a + b Z 0, since a + ib = sa, bd Z s0, 0d. The number a - ib that is used as multiplier to clear the i from the denominator is called the comple conjugate of a + ib. It is customar to use z (read z bar ) to denote the comple conjugate of z; thus z = a + ib, z = a - ib. = sac + bdd + isad - bcd a + b. Multipling the numerator and denominator of the fraction sc + idd>sa + ibd b the comple conjugate of the denominator will alwas replace the denominator b a real number. P(, ) r O FIGURE A.4 This Argand diagram represents z = + i both as a point P(, ) and as a vector OP. EXAMPLE (d) = Argand Diagrams Arithmetic Operations with Comple Numbers (a) s + 3id + s6 - id = s + 6d + s3 - di = 8 + i (b) s + 3id - s6 - id = s - 6d + s3 - s -ddi = i (c) s + 3ids6 - id = sds6d + sds -id + s3ids6d + s3ids -id + 3i 6 - i = + 3i 6 + i 6 - i 6 + i + 4i + 8i + 6i 36 + i - i - 4i = 6 + i 40 = - 4i + 8i - 6i = + 4i + 6 = 8 + 4i = i There are two geometric representations of the comple number z = + i:. as the point P(, ) in the -plane. as the vector OP from the origin to P. In each representation, the -ais is called the real ais and the -ais is the imaginar ais. Both representations are Argand diagrams for + i (Figure A.4). In terms of the polar coordinates of and, we have = r cos u, = r sin u,

6 A.5 Comple Numbers AP-7 and z = + i = rscos u + i sin ud. (0) We define the absolute value of a comple number to be the length r of a vector OP + i from the origin to P(, ). We denote the absolute value b vertical bars; thus, ƒ + i ƒ = +. If we alwas choose the polar coordinates r and u so that r is nonnegative, then r = ƒ + i ƒ. The polar angle u is called the argument of z and is written u = arg z. Of course, an integer multiple of p ma be added to u to produce another appropriate angle. The following equation gives a useful formula connecting a comple number z, its conjugate z, and its absolute value ƒ z ƒ, namel, z # z = ƒ z ƒ. Euler s Formula The identit e iu = cos u + i sin u, called Euler s formula, enables us to rewrite Equation (0) as z = re iu. This formula, in turn, leads to the following rules for calculating products, quotients, powers, and roots of comple numbers. It also leads to Argand diagrams for e iu. Since cos u + i sin u is what we get from Equation (0) b taking r =, we can sa that e iu is represented b a unit vector that makes an angle u with the positive -ais, as shown in Figure A.5. e i cos i sin e i cos i sin r arg z O O (cos, sin ) (a) (b) FIGURE A.5 Argand diagrams for e iu = cos u + i sin u (a) as a vector and (b) as a point. Products To multipl two comple numbers, we multipl their absolute values and add their angles. Let z = r e iu, z = r e iu, ()

7 AP-8 z z z r r z r r O FIGURE A.6 When z and z are multiplied, ƒ z z ƒ = r # r and arg sz z d = u + u. so that Then and hence ƒ z ƒ = r, arg z = u ; ƒ z ƒ = r, arg z = u. z z = r e iu # r e iu = r r e isu + u d ƒ z z ƒ = r r = ƒ z ƒ # ƒ z ƒ arg sz z d = u + u = arg z + arg z. Thus, the product of two comple numbers is represented b a vector whose length is the product of the lengths of the two factors and whose argument is the sum of their arguments (Figure A.6). In particular, from Equation () a vector ma be rotated counterclockwise through an angle u b multipling it b e iu. Multiplication b i rotates 90, b - rotates 80, b -i rotates 70, and so on. () 0 z i 4 6 z z 3 Then 3 z 3 i EXAMPLE Finding a Product of Comple Numbers Let z = + i, z = 3 - i. We plot these comple numbers in an Argand diagram (Figure A.7) from which we read off the polar representations z = e ip>4, z = e -ip>6. z z = ep a ip 4 - ip 6 b = ep aip b = acos p + i sin p b L i. FIGURE A.7 To multipl two comple numbers, multipl their absolute values and add their arguments. The notation ep (A) stands for e A. Quotients Suppose r Z 0 in Equation (). Then Hence z ` z ` = r r = ƒ z ƒ ƒ z ƒ z z = r eiu r e iu That is, we divide lengths and subtract angles for the quotient of comple numbers. EXAMPLE 3 Let z = + i and z = 3 - i, as in Eample. Then + i 3 - i = eip>4 = e -ip>6 e5pi> L acos 5p L i. = r r e isu - u d. and arg a z z b = u - u = arg z - arg z. + i sin 5p b

8 A.5 Comple Numbers AP-9 Powers If n is a positive integer, we ma appl the product formulas in Equation () to find z n = z # z # Á # z. n factors With z = re iu, we obtain z n = sre iu d n = r n e isu + u + Á + ud n summands = r n e inu. (3) The length r = ƒ z ƒ is raised to the nth power and the angle u = arg z is multiplied b n. If we take r = in Equation (3), we obtain De Moivre s Theorem. De Moivre s Theorem scos u + i sin ud n = cos nu + i sin nu. (4) If we epand the left side of De Moivre s equation above b the Binomial Theorem and reduce it to the form a + ib, we obtain formulas for cos nu and sin nu as polnomials of degree n in cos u and sin u. EXAMPLE 4 If n = 3 in Equation (4), we have scos u + i sin ud 3 = cos 3u + i sin 3u. The left side of this equation epands to cos 3 u + 3i cos u sin u - 3 cos u sin u - i sin 3 u. The real part of this must equal cos 3u and the imaginar part must equal sin 3u. Therefore, cos 3u = cos 3 u - 3 cos u sin u, sin 3u = 3 cos u sin u - sin 3 u. Roots If z = re iu is a comple number different from zero and n is a positive integer, then there are precisel n different comple numbers w 0, w, Á, w n -, that are nth roots of z. To see wh, let w = re ia be an nth root of z = re iu, so that w n = z or r n e ina = re iu. Then r = n r is the real, positive nth root of r. For the argument, although we cannot sa that na and u must be equal, we can sa that the ma differ onl b an integer multiple of p. That is, na = u + kp, k = 0, ;, ;, Á.

9 AP-0 Therefore, w 3 r /3 O 3 r 3 z re i 3 w 0 a = n u + k p n. Hence, all the nth roots of z = re iu are given b n re iu = n r ep i an u + k p n b, k = 0, ;, ;, Á. (5) FIGURE A.8 z = re iu. 6 FIGURE A.9-6. w w w The three cube roots of 4 w 0 w 3 The four fourth roots of There might appear to be infinitel man different answers corresponding to the infinitel man possible values of k, but k = n + m gives the same answer as k = m in Equation (5). Thus, we need onl take n consecutive values for k to obtain all the different nth roots of z. For convenience, we take k = 0,,, Á, n -. All the nth roots of re iu lie on a circle centered at the origin and having radius equal to the real, positive nth root of r. One of them has argument a = u>n. The others are uniforml spaced around the circle, each being separated from its neighbors b an angle equal to p>n. Figure A.8 illustrates the placement of the three cube roots, w 0, w, w, of the comple number z = re iu. EXAMPLE 5 Finding Fourth Roots Find the four fourth roots of -6. Solution As our first step, we plot the number -6 in an Argand diagram (Figure A.9) and determine its polar representation re iu. Here, z = -6, r = +6, and u = p. One of the fourth roots of 6e ip is e ip>4. We obtain others b successive additions of p>4 = p> to the argument of this first one. Hence, and the four roots are 4 6 ep ip = ep i a p 4, 3p 4, 5p 4, 7p 4 b, w 0 = ccos p 4 + i sin p d = s + id 4 w = ccos 3p 4 + i sin 3p 4 d = s - + id w = ccos 5p 4 + i sin 5p 4 d = s - - id w 3 = ccos 7p 4 + i sin 7p 4 d = s - id. The Fundamental Theorem of Algebra One might sa that the invention of - is all well and good and leads to a number sstem that is richer than the real number sstem alone; but where will this process end? Are

10 A.5 Comple Numbers AP- we also going to invent still more sstems so as to obtain 4-, -, 6 and so on? But it turns out this is not necessar. These numbers are alread epressible in terms of the comple number sstem a + ib. In fact, the Fundamental Theorem of Algebra sas that with the introduction of the comple numbers we now have enough numbers to factor ever polnomial into a product of linear factors and so enough numbers to solve ever possible polnomial equation. The Fundamental Theorem of Algebra Ever polnomial equation of the form a n z n + a n - z n - + Á + a z + a 0 = 0, in which the coefficients a 0, a, Á, a n are an comple numbers, whose degree n is greater than or equal to one, and whose leading coefficient a n is not zero, has eactl n roots in the comple number sstem, provided each multiple root of multiplicit m is counted as m roots. A proof of this theorem can be found in almost an tet on the theor of functions of a comple variable. EXERCISES A.5 Operations with Comple Numbers In Eercises 5 0, graph the points z = + i that satisf the given. How computers multipl comple numbers Find sa, bd # conditions. sc, dd = sac - bd, ad + bcd. 5. a. ƒ z ƒ = b. ƒ z ƒ 6 c. ƒ z ƒ 7 a. s, 3d # s4, -d b. s, -d # s -, 3d 6. ƒ z - ƒ = 7. ƒ z + ƒ = c. s -, -d # s, d 8. ƒ z + ƒ = ƒ z - ƒ 9. ƒ z + i ƒ = ƒ z - ƒ (This is how comple numbers are multiplied b computers.) 0. ƒ z + ƒ Ú ƒ z ƒ. Solve the following equations for the real numbers, and. Epress the comple numbers in Eercises 4 in the form re iu, a. s3 + 4id - s - id = + i with r Ú 0 and -p 6 u p. Draw an Argand diagram for each b. a + i - i b + + i = + i calculation. + i. A + -3B. - i c. s3 - ids + id = s - id + i - + i s + 3ids - id - i3 3. How ma the following comple numbers be obtained from z = + i geometricall? Sketch. Powers and Roots a. z b. s -zd Use De Moivre s Theorem to epress the trigonometric functions in Eercises 5 and 6 in terms of cos u and sin u. c. -z d. > z 5. cos 4u 6. sin 4u 4. Show that the distance between the two points z and z in an Argand diagram is ƒ z - z ƒ. 7. Find the three cube roots of. Graphing and Geometr

11 AP- 8. Find the two square roots of i. 9. Find the three cube roots of -8i. 0. Find the si sith roots of 64.. Find the four solutions of the equation z 4 - z + 4 = 0.. Find the si solutions of the equation z 6 + z 3 + = Find all solutions of the equation = Solve the equation 4 + = 0. Theor and Eamples 5. Comple numbers and vectors in the plane Show with an Argand diagram that the law for adding comple numbers is the same as the parallelogram law for adding vectors. 6. Comple arithmetic with conjugates Show that the conjugate of the sum (product, or quotient) of two comple numbers, z and z, is the same as the sum (product, or quotient) of their conjugates. 7. Comple roots of polnomials with real coefficients come in comple-conjugate pairs ƒ ƒ ƒ ƒ ƒ ƒ ƒ ƒ ƒ ƒ ƒ ƒ ƒ ƒ ƒ ƒ ƒ ƒ ƒ a. Etend the results of Eercise 6 to show that ƒszd = ƒszd if ƒszd = a n z n + a n - z n - + Á + a z + a 0 is a polnomial with real coefficients a 0, Á, a n. b. If z is a root of the equation ƒszd = 0, where ƒ(z) is a polnomial with real coefficients as in part (a), show that the conjugate z is also a root of the equation. (Hint: Let ƒszd = u + i = 0; then both u and are zero. Use the fact that ƒszd = ƒszd = u - i. ) 8. Absolute value of a conjugate Show that z = z ƒ. 9. When z = z If z and z are equal, what can ou sa about the location of the point z in the comple plane? 30. Real and imaginar parts Let Re(z) denote the real part of z and Im(z) the imaginar part. Show that the following relations hold for an comple numbers z, z, and z. a. z + z = Reszd b. z - z = iimszd c. Reszd z d. z + z = z + z + Resz z d e. z + z z + z