Complex Numbers OBJECTIVES

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(1)calter_c21_001-027v2. 28-01-2008. 18:12. Page 1. 21 Complex Numbers. ◆◆◆. OBJECTIVES. ◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆. When you have completed this chapter, you should be able to: • Simplify radicals having negative radicands. • Write complex numbers in rectangular, polar, trigonometric, and exponential forms. • Evaluate powers of j. • Find the sums, differences, products, quotients, powers, and roots of complex numbers. • Solve quadratic equations having complex roots. • Factor polynomials that have complex factors. • Add, subtract, multiply, and divide vectors using complex numbers. • Solve alternating current problems using complex numbers. ◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆. So far we have dealt entirely with real numbers, which are all of the numbers that correspond to points on the number line; that is, the rational and irrational numbers. The real numbers include integers, fractions, roots, and the negatives of these numbers. In this chapter we consider imaginary and complex numbers. We learn what they are, how to express them in several different forms, and how to perform the basic operations with them. Why bother with a new type of number when, as we have seen, the real numbers can do so much? The complex number system is a natural extension of the real number system. We use complex numbers mainly to manipulate vectors, which are so important in many branches of technology. Complex numbers (despite their unfortunate name) actually simplify computations with vectors, as we see especially with the alternating current calculations at the end of this chapter.. 1.

(2) calter_c21_001-027v2. 28-01-2008. 2. 18:12. Page 2. Chapter 21. 21–1. ◆. Complex Numbers. Complex Numbers in Rectangular Form. The Imaginary Unit Recall that in the real number system, the equation x2 1 had no solution because there was no real number such that its square was 1. Now we extend our number system to allow the quantity 11 to have a meaning. We define the imaginary unit as the square root of 1 and represent it by the symbol j. Imaginary unit: i ⬅ 21 Complex Numbers A complex number is any number, real or imaginary, that can be written in the form The letter i is often used for the imaginary unit. In technical work, however, we save i for electric current. Further, j (or i) is sometimes written before the b, and sometimes after. So the number j5 may also be written 5i, 5j, or i5.. a jb where a and b are real numbers, and j 11 is the imaginary unit. ◆◆◆. Example 1: The following numbers are complex numbers: 4 j2. 7 j8. 5.92 j2.93. 83. j 27. ◆◆◆. Real and Imaginary Numbers When b 0 in a complex number a jb, we have a real number. When a 0, the number is called a pure imaginary number. ◆◆◆ Example 2: (a) The complex number 48 is also a real number. (b) The complex number j9 is a pure imaginary number.. ◆◆◆. The two parts of a complex number are called the real part and the imaginary part. This is called the rectangular form of a complex number.. Complex Number. a jb real part. 218 imaginary part. Addition and Subtraction of Complex Numbers To combine complex numbers, separately combine the real parts, then combine the imaginary parts, and express the result in the form a jb.. Addition of Complex Numbers. 1a jb2 1c jd2 1a c2 j 1b d2. 219. Subtraction of Complex Numbers. 1a jb2 1c jd2 1a c2 j 1b d2. 220. ◆◆◆. Example 3: Add or subtract, as indicated.. (a) (b) (c) (d). j3 j5 j8 j2 (6 j5) 6 j3 (2 j5) (4 j3) (2 4) j(5 3) 2 j2 (6 j2) (4 j) (6 4) j[2 (1)] 10 j3. ◆◆◆.

(3) calter_c21_001-027v2. 28-01-2008. Section 21–1. ◆. 18:12. Page 3. 3. Complex Numbers in Rectangular Form. Powers of j We often have to evaluate powers of the imaginary unit, especially the square of j. Since j 21 then j 2 21 # 21 1 Higher powers are easily found. j3 j4 j5 j6. . j2j (j2)2 j4j j4j2. . (1)j (1)2 (1)j (1)j2. j 1 j 1. We see that the values are starting to repeat: j 5 j, j 6 j 2, and so on. The first four values keep repeating.. Powers of j. ◆◆◆. j 21 j4 1. j 2 1. j3 i. j5 j. .... 217. Example 4: Evaluate j17.. Solution: Using the laws of exponents, we express j17 in terms of one of the first four powers of j. j17 j16j ( j 4)4j (1)4j j. ◆◆◆. Multiplication of Imaginary Numbers Multiply as with ordinary numbers, but use Eq. 217 to simplify any powers of j. ◆◆◆. Example 5:. (a) (b) (c) (d). 5 j3 j15 j2 j4 j 2 8 (1)8 8 3 j4 j5 j j360 (j )60 j60 ( j 3)2 j 232 (1)9 9. ◆◆◆. Multiplication of Complex Numbers Multiply complex numbers as you would any algebraic expressions: Replace j 2 by 1, and put the expression into the form a jb.. Multiplication of Complex Numbers ◆◆◆. 1a jb21c jd2 1ac bd2 j1ad bc2. Example 6:. (a) 3(5 j2) 15 j6 (b) ( j3)(2 j4) j6 j 212 j6 (1)12 12 j6 (c) (3 j2)(4 j5) 3(4) 3( j5) (j2)(4) (j2)( j5) 12 j15 j8 j 210 12 j15 j8 (1)10 2 j23. 221. Note that when the exponent n is a multiple of 4, then j n 1..

(4) calter_c21_001-027v2. 28-01-2008. 4. 18:12. Page 4. Chapter 21. ◆. Complex Numbers. (d) (3 j5)2 (3 j5)(3 j5) 9 j15 j15 j 225 9 j 30 25 16 j30. ◆◆◆. To multiply radicals that contain negative quantities in the radicand, first express all quantities in terms of j, then proceed to multiply. Always be sure to convert radicals to imaginary numbers before performing other operations, or contradictions can result. ◆◆◆. We will see in the next section that multiplication and division are easier in polar form.. Example 7: Multiply 14 by 14.. Solution: Converting to imaginary numbers, we obtain 24 # 24 (j2)(j2) j24 Since j 2 1, j 24 4. ◆◆◆. It is incorrect to write. Common Error. 24 24 2(4) (4) 216 4 Our previous rule of 1a # 1b 1ab applied only to positive a and b.. The Conjugate of a Complex Number The conjugate of a complex number is obtained by changing the sign of its imaginary part. ◆◆◆. A computer algebra system can find the conjugate of a complex number. See problem 68 in Exercise 1.. Example 8:. (a) The conjugate of 2 j 3 is 2 j 3. (b) The conjugate of 5 j 4 is 5 j 4. (c) The conjugate of a jb is a jb.. ◆◆◆. Multiplying any complex number by its conjugate will eliminate the j term. ◆◆◆. This equation has the same form as the difference of two squares.. Example 9: (2 j3)(2 j 3) 4 j6 j6 j 29 4 (1)(9) 13. ◆◆◆. Division of Complex Numbers Division involving single terms, real or imaginary, is shown by examples. ◆◆◆. Example 10:. (a) j8 2 j4 (b) j6 j3 2 (c) (4 j6) 2 2 j 3 ◆◆◆. ◆◆◆. Example 11: Divide 6 by j3.. Solution: j18 j18 6 6 # j3 2 j2 j3 j3 j3 9 j9. ◆◆◆. To divide by a complex number, multiply dividend and divisor by the conjugate of the divisor. This will make the divisor a real number, as in the following example..

(5) calter_c21_001-027v2. 28-01-2008. Section 21–1. ◆◆◆. ◆. 18:12. Page 5. 5. Complex Numbers in Rectangular Form. Example 12: Divide 3 j4 by 2 j.. Solution: We multiply numerator and denominator by the conjugate (2 j) of the denominator. 3 j4 3 j4 # 2 j 2j 2j 2j 6 j3 j8 j24 4 j2 j2 j2 2 j11 5 11 2 j 5 5. This is very similar to rationalizing the denominator of a fraction containing radicals (Sec. 13–2).. ◆◆◆. In general, the following equation applies:. Division of Complex Numbers. a jb ac bd bc ad 2 j 2 c jd c d2 c d2. 222. A CAS will give complex roots of a quadratic equation. See problem 69 in Exercise 1.. Quadratics with Complex Roots When we solved quadratic equations in Chapter 14, the roots were real numbers. Some quadratics, however, will yield complex roots, as in the following example.. y 30. ◆◆◆. Example 13: Solve for x to three significant digits:. y = 2x2 − 5x + 9. 2x 5x 9 0 2. 20. Solution: By the quadratic formula (Eq. 100), x . 5 225 4(2)(9) 4. . 5 247 4. 10. 5 j6.86 1.25 j1.72 4. A graph of the function y 2x2 5x 9 (Fig. 21–1) shows that the curve does not cross the x axis and hence has no real zeros.. −2 ◆◆◆. FIGURE 21–1. Complex Factors It is sometimes necessary to factor an expression into complex factors, as shown in the following example. ◆◆◆. Example 14: Factor the expression a2 4 into complex factors.. Solution: a2 4 a2 j 24 a2 ( j2)2 (a j2)(a j2). Exercise 1. ◆. ◆◆◆. Complex Numbers in Rectangular Form. Write as imaginary numbers. 1. 29. 2.. 281. 3.. 4 A 16 . 4.. 1 A 5 . 0. 2. 4. x.

(6) calter_c21_001-027v2. 6. 28-01-2008. 18:12. Page 6. Chapter 21. ◆. Complex Numbers. Write as a complex number in rectangular form. 5. 4 24 7.. 9 5 A 4. 6. 225 3 8.. 1 2 A 3. Combine and simplify. 9. 29 24 11. (3 j2) (4 j3) 13. (a j3) (a j5). 10. 24a2 a225 12. (1 j2) ( j 6) 14. ( p jq) (q jp). 15. a. 16. (84 j 91) (28 j 72). j j 1 1 b a b 2 3 4 6. 17. (2.28 j1.46) (1.75 j2.66) Evaluate each power of j. 18. j 11 20. j 10 22. j 14. 19. j 5 21. j 21. Multiply and simplify. 23. 25. 27. 29. 31. 33. 35. 37. 39.. 7 j2 j3 j5 4 j2 j 3 j 4 ( j 5)2 2(3 j4) j4(5 j 2) (3 j5)(2 j6) (6 j3)(3 j8) (5 j2)2. 24. 26. 28. 30. 32. 34. 36. 38. 40.. 9 j3 j j4 j 5 j4 j3 5 ( j3)2 3(7 j5) j5(2 j 3) (5 j4)(4 j 2) (5 j3)(8 j 2) (3 j6)2. Write the conjugate of each complex number. 41. 2 j3 43. p jq 45. jm n. 42. 5 j 7 44. j5 6 46. 5 j8. Divide and simplify. 47. 49. 51. 53. 55.. j8 4 j12 j6 (4 j2) 2 (2 j3) (1 j) ( j7 2) ( j3 5). 48. 50. 52. 54. 56.. 9 j3 j44 j2 8 (4 j) (5 j6) (3 j 2) (9 j3) (2 j4). Quadratics with Complex Roots Solve for x to three significant digits. 57. 3x2 5x 7 0 59. x2 2x 6 0. 58. 2x2 3x 5 0 60. 4x2 x 8 0.

(7) calter_c21_001-027v2. 28-01-2008. Section 21–2. ◆. 18:12. Page 7. 7. Graphing Complex Numbers. Complex Factors Factor each expression into complex factors. 61. x2 9 63. 4y2 z2. 21–2. 62. b2 25 64. 25a2 9b2. Graphing Complex Numbers. The Complex Plane A complex number in rectangular form is made up of two parts: a real part a and an imaginary part jb. To graph a complex number, we can use a rectangular coordinate system (Fig. 21–2) in which the horizontal axis is the real axis and the vertical axis is the imaginary axis. Such a coordinate system defines what is called the complex plane. Real numbers are graphed as points on the horizontal axis; pure imaginary numbers are graphed as points on the vertical axis. Complex numbers, such as 3 j5 (Fig. 21–2), are graphed elsewhere within the plane. Imaginary 6 axis 5. 3 + j5. 4 3 2 1 0 −3 −2 −1−1. 1 2 3 4 5 Real axis. −2 −3. FIGURE 21–2 plane.. The complex. Graphing a Complex Number To plot a complex number a jb in the complex plane, simply locate a point with a horizontal coordinate of a and a vertical coordinate of b. ◆◆◆. Example 15: Plot the complex numbers (2 j3), (1 j2), (3 j2), and (1 j3).. Solution: The points are plotted in Fig. 21–3. Imaginary axis. 2 + j3. 3 −1 + j2. Such a plot is called an Argand diagram, named for Jean Robert Argand (1768–1822).. 2 1. −3 −2 −1 0 −1 −3 − j2. 1. 2. 3 Real axis. −2 −3. FIGURE 21–3. 1 − j3 ◆◆◆.

(8) calter_c21_001-027v2. 28-01-2008. 8. 18:12. Page 8. Chapter 21. ◆. Exercise 2. Complex Numbers. ◆. Graphing Complex Numbers. Graph each complex number. 1. 3. 5. 7. 9.. 2 j5 3 j2 j5 2.7 j3.4 1.46 j2.45. 21–3 In some books, the terms polar form and trigonometric form are used interchangeably. Here, we distinguish between them. Imaginary a + bi = r ␪. b r. 2. 4. 6. 8.. 1 j3 2j 2.25 j3.62 5.02 j. Complex Numbers in Trigonometric and Polar Forms. Polar Form In Chapter 17 we saw that a point in a plane could be located by polar coordinates, as well as by rectangular coordinates, and we learned how to convert from one set of coordinates to the other. We will do the same thing now with complex numbers, converting between rectangular and polar form. In Fig. 21–4, we plot a complex number a jb. We connect that point to the origin with a line of length r at an angle of ␪ with the horizontal axis. Now, in addition to expressing the complex number in terms of a and b, we can express it as follows in terms of r and ␪:. ␪ 0. a. Real. FIGURE 21–4 Polar form of a complex number shown on a complex plane.. Polar Form. a jb rl u. 228. Since cos ␪ a/r and sin ␪ b/r, we have the following equations:. a r cos ␪. 224. b r sin ␪. 225. The radius r is called the absolute value. It can be found from the Pythagorean theorem.. r 2a2 b2. 226. The angle ␪ is called the argument of the complex number. Since tan ␪ b/a:. u arctan. ◆◆◆. b a. Example 16: Write the complex number 2 j3 in polar form.. Solution: The absolute value is r 222 32 ⬵ 3.61. 227.

(9) calter_c21_001-027v2. 28-01-2008. Section 21–3. ◆. 18:12. Page 9. 9. Complex Numbers in Trigonometric and Polar Forms. The argument is found from 3 2 ␪ ⯝ 56.3°. tan ␪ . So 2 j 3 3.61l 56.3°. ◆◆◆. Trigonometric Form If we substitute the values of a and b from Eqs. 224 and 225 into a jb, we get a jb (r cos ␪) j (r sin ␪) Factoring gives us the following equation:. Trigonometric Form. ◆◆◆. a jb r(cos ␪ j sin ␪) rectangular form. trigonometric form. The angle ␪ is sometimes written ␪ arg z, which means “␪ is the argument of the complex number z.” Also, the expression in parentheses in Eq. 223 is sometimes abbreviated as cis ␪, so cis ␪ cos ␪ j sin ␪.. 223. Example 17: Write the complex number 2 j3 in polar and trigonometric forms.. Solution: We already have r and ␪ from Example 16. r 3.61. and. ␪ 56.3°. That is, 2 j3 3.61l 56.3° So, by Eq. 224, 2 j3 3.61(cos 56.3° j sin 56.3°). ◆◆◆. Example 18: Write the complex number 6(cos 30° j sin 30°) in polar and rectangular forms.. ◆◆◆. Solution: By inspection, r 6 and. ␪ 30°. So a r cos ␪ 6 cos 30° 5.20 and b r sin ␪ 6 sin 30° 3.00 Thus our complex number in rectangular form is 5.20 j3.00, and in polar form, 6l 30° .. ◆◆◆. Arithmetic Operations in Polar Form It is common practice to switch back and forth between rectangular and polar forms during a computation, using that form in which a particular operation is easier. We saw that addition and subtraction are fast and easy in rectangular form, and we now show that multiplication, division, and raising to a power are best done in polar form..

(10) calter_c21_001-027v2. 28-01-2008. 10. 18:12. Page 10. Chapter 21. ◆. Complex Numbers. Products We will use trigonometric form to work out the formula for multiplication and then express the result in the simpler polar form. Let us multiply r(cos ␪ j sin ␪) by r⬘(cos ␪⬘ j sin ␪⬘). [r(cos ␪ j sin ␪)][r⬘(cos ␪⬘ j sin ␪⬘)] rr⬘(cos ␪ cos ␪⬘ j cos ␪ sin ␪⬘ j sin ␪ cos ␪⬘ j 2 sin ␪ sin ␪⬘) rr⬘[(cos ␪ cos ␪⬘ sin ␪ sin ␪⬘) j(sin ␪ cos ␪⬘ cos ␪ sin ␪⬘)] But, by Eq. 168, cos ␪ cos ␪⬘ sin ␪ sin ␪⬘ cos(␪ ␪⬘) and by Eq. 167, sin ␪ cos ␪⬘ cos ␪ sin ␪⬘ sin(␪ ␪⬘) So r(cos ␪ j sin ␪) r⬘(cos ␪⬘ j sin ␪⬘) rr⬘[cos(␪ ␪⬘) j sin (␪ ␪⬘)] Switching now to polar form, we get the following equation:. Products. rl u # rl u rrl u u. 229. The absolute value of the product of two complex numbers is the product of their absolute values, and the argument is the sum of the individual arguments. ◆◆◆. Example 19: Multiply 5l 30° by 3l 20°.. Solution: The absolute value of the product will be 5(3) 15, and the argument of the product will be 30° 20° 50°. So 5l 30° # 3l 20° 15l 50°. ◆◆◆. The angle ␪ is not usually written greater than 360°. Subtract multiples of 360° if necessary. ◆◆◆. Example 20: Multiply 6.27l 300° by 2.75l 125°.. Solution: 6.27l 300° 2.75l 125° (6.27) (2.75)l 300° 125° 17.2l 425° 17.2l 65°. ◆◆◆. Quotients The rule for division of complex numbers in trigonometric form is similar to that for multiplication.. You might try deriving this yourself.. Quotient. rl u rl u. . r l u u r. 230. The absolute value of the quotient of two complex numbers is the quotient of their absolute values, and the argument is the difference (numerator minus denominator) of their arguments..

(11) calter_c21_001-027v2. 28-01-2008. Section 21–3. ◆◆◆. ◆. 18:12. Page 11. 11. Complex Numbers in Trigonometric and Polar Forms. Example 21: 6l 70° 2l 50°. 3l 20°. ◆◆◆. Powers To raise a complex number to a power, we merely have to multiply it by itself the proper number of times, using Eq. 229. (rl u)2 rl u # rl u r # rl u u r2l 2u (rl u)3 rl u # r2l 2u r # r2l u 2u r3l 3u Do you see a pattern developing? In general, we get the following equation:. DeMoivre’s Theorem. (rl u)n rnl nu. DeMoivre’s theorem is named after Abraham DeMoivre (1667–1754).. 231. When a complex number is raised to the nth power, the new absolute value is equal to the original absolute value raised to the nth power, and the new argument is n times the original argument. ◆◆◆. Example 22: (2l 10°)5 25l 5 (10°) 32l 50°. ◆◆◆. Roots We know that 14 1, so, conversely, the fourth root of 1 is 1. 2 4 11 But we have seen that j 4 1, so shouldn’t it also be true that 2 4 1j. ?. In fact, both 1 and j are fourth roots of 1, and there are two other roots as well. In this section we learn how to use DeMoivre’s theorem to find all roots of a number. First we note that a complex number r兾␪ in polar form is unchanged if we add multiples of 360° to the angle ␪ (in degrees). Thus rl u rl u 360° rl u 720° rl u 360k where k is an integer. Rewriting DeMoivre’s theorem with an exponent 1/p (where p is an integer), we get (rl u )1/p (rl u 360k)1/p r1/pl (u 360k)/p so (rl u )1/p r1/pl (u 360k)/p. (1). We use this equation to find all of the roots of a complex number by letting k take on the values k 0, 1, 2, . . . , (p 1). We’ll see in the following example that values of k greater than ( p 1) will give duplicate roots. ◆◆◆. Example 23: Find the fourth roots of 1.. Solution: We write the given number in polar form; thus 1 1l 0° . Then, using Eq. (1) with p 4, 11/4 (1l 0 )1/4 11/4l (0 360k)/4 1l 90k.

(12) calter_c21_001-027v2. 28-01-2008. 12. 18:12. Page 12. Chapter 21. Complex Numbers. ◆. We now let k 0, 1, 2, . . . . Root k 0 1 2 3 4 5. Polar Form 1l 0° 1l 90° 1l 180° 1l 270° 1l 360° 1l 450°. Rectangular Form 1 j 1 j 1l 0° 1 1l 90° j. Notice that the roots repeat for k 4 and higher, so we look no further. Thus the fourth roots of 1 are 1, j, 1, and j. We check them by noting that each raised to the fourth power equals 1. ◆◆◆ Thus the number 1 has four fourth roots. In general, any number has two square roots, three cube roots, and so on. There are p pth roots of a complex number. If a complex number is in rectangular form, we convert to polar form before taking the root. ◆◆◆. Example 24: Find 2256 3 j192.. Solution: We convert the given complex number to polar form. r 2(256)2 (192)2 320. ␪ arctan(192/256) 36.9°. Then, using Eq. (1), we have 2256 3 j192 (320l 36.9° )1/3 (320)1/3l (36.9° 360k)/3 6.84l 12.3° 120k We let k 0, 1, 2, to obtain the three roots. Root k. Polar Form. Rectangular Form. 0 1 2. 6.84l 12.3° 0° 6.84l 12.3° 6.84l 12.3° 120° 6.84l 132.3° 6.84l 12.3° 240° 6.84l 252.3°. 6.68 j1.46 4.60 j5.06 2.08 j6.52. Exercise 3. ◆. Complex Numbers in Trigonometric and Polar Forms. Write each complex number in polar and trigonometric forms. Round to three significant digits, where necessary, in this exercise.. 1. 2. 3. 4. 5. 6.. 5 j4 3 j7 4 j3 8 j4 5 j2 4 j 7. ◆◆◆.

(13) calter_c21_001-027v2. 28-01-2008. Section 21–3. ◆. 18:12. Page 13. Complex Numbers in Trigonometric and Polar Forms. 7. 9 j5 8. 7 j3 9. 4 j 7 Write in rectangular and in polar forms. 10. 11. 12. 13. 14. 15.. 4(cos 25° j sin 25°) 3(cos 57° j sin 57°) 2(cos 110° j sin 110°) 9(cos 150° j sin 150°) 7(cos 12° j sin 12°) 5.46(cos 47.3° j sin 47.3°). Write in rectangular and trigonometric forms. 16. 5l 28° 17. 9l 59° 18. 4l 63° 19. 7l 53° 20. 6l 125°. Multiplication and Division Multiply. 21. 22. 23. 24. 25. 26.. 3(cos 12° j sin 12°) by 5(cos 28° j sin 28°) 7(cos 48° j sin 48°) by 4(cos 72° j sin 72°) 5.82(cos 44.8° j sin 44.8°) by 2.77(cos 10.1° j sin 10.1°) 5l 30° by 2l 10° 8l 45° by 7l 15° 2.86l 38.2° by 1.55l 21.1°. Divide. 27. 28. 29. 30. 31. 32.. 8(cos 46° j sin 46°) by 4(cos 21° j sin 21°) 49(cos 27° j sin 27°) by 7(cos 15° j sin 15°) 58.3(cos 77.4° j sin 77.4°) by 12.4(cos 27.2° j sin 27.2°) 24l 50° by 12l 30° 50l 72° by 5l 12° 71.4l 56.4° by 27.7l 15.2°. Powers and Roots Evaluate. 33. 34. 35. 36. 37. 38. 39. 40. 41. 42.. [2(cos 15° j sin 15°)]3 [9(cos 10° j sin 10°)]2 (7l 20° )2 (1.55l 15° )3 257l 52° 222l 12° 238 3 l 73° 215 3 l 89° 2135 j204 256.3 3 j28.5. 13.

(14) calter_c21_001-027v2. 28-01-2008. 14. 18:12. Page 14. Chapter 21. 21–4. ◆. Complex Numbers. Complex Numbers in Exponential Form. Euler’s Formula We have already expressed a complex number in rectangular form a jb in polar form rl u and in trigonometric form r(cos ␪ j sin ␪) Our next (and final) form for a complex number is exponential form, given by Euler’s formula.. Euler’s Formula. This formula is named after Leonard Euler (1707–83). We give it without proof here.. re j␪ r(cos ␪ j sin ␪). 232. Here e is the base of natural logarithms (approximately 2.718), discussed in Chapter 20, and ␪ is the argument expressed in radians. ◆◆◆. Example 25: Write the complex number 5(cos 180° j sin 180°) in exponential form.. Solution: We first convert the angle to radians: 180° ␲ rad. So 5(cos ␲ j sin ␲) 5e j␲ Setting ␪ ␲ and r 1 in Euler’s formula gives. Common Error. e cos ␲ j sin ␲ j␲. ◆◆◆. Make sure that ␪ is in radians when using Eq. 232.. 1 j0 1 The constant e arises out of natural growth, j is the square root of 1, and ␲ is the ratio of the circumference of a circle to its diameter. Thus we get the astounding result that the irrational number e raised to the product of the imaginary unit j and the irrational number ␲ give the integer 1.. Example 26: Write the complex number 3e j 2 in rectangular, trigonometric, and polar forms.. ◆◆◆. Solution: Changing ␪ to degrees gives us ␪ 2 rad 115° and from Eq. 232, r3 From Eq. 224, a r cos ␪ 3 cos 115° 1.27 and from Eq. 225, b r sin ␪ 3 sin 115° 2.72 So, in rectangular form, 3e j2 1.27 j2.72 In polar form, 3e j2 3l 115° In trigonometric form, 3e j2 3(cos 115° j sin 115°). ◆◆◆.

(15) calter_c21_001-027v2. 28-01-2008. Section 21–4. ◆. 18:12. Page 15. 15. Complex Numbers in Exponential Form. Products We now find products, quotients, powers, and roots of complex numbers in exponential form. These operations are quite simple in this form, as we merely have to use the laws of exponents. Thus by Eq. 29: r1e j␪1 r2 e j␪2 r1r2 e j(␪1 ␪2). Products ◆◆◆. 233. Example 27:. (a) 2e j3 5e j4 10e j7 (c) 13.5ej3 2.75e j5 37.1e j2. (b) 2.83e j2 3.15e j4 8.91e j6 ◆◆◆. Quotients By Eq. 30:. Quotients. ◆◆◆. Example 28:. (a). 8ej5 2ej3 4ej2. (c). 5.82ej4 0.592ej3 9.83ej7. r1eju1 r2e. ju2. r1 j(u u ) e 1 2 r2. . (b). 234. 63.8ej2 4.66ej3 13.7ej5 ◆◆◆. Powers and Roots By Eq. 31:. Powers and Roots. ◆◆◆. (re j␪)n r ne jn␪. 235. Example 29: (b) (3.85ej2)3 57.1ej6. (a) (2e j3)4 16e j12 (c) (0.233ej3)2 . Exercise 4. ◆. ej6 ej6 20.1e j6 2 0.0497 (0.223). ◆◆◆. Complex Numbers in Exponential Form. Express each complex number in exponential form. 1. 2 j3. 2. 1 j 2. 3. 3(cos 50° j sin 50°). 4. 12l 14. 5. 2.5l p/6. 6. 7 cos. 7. 5.4l p/12. 8. 5 j 4. 冢. p p j sin 3 3. 冣.

(16) calter_c21_001-027v2. 28-01-2008. 16. 18:12. Page 16. Chapter 21. ◆. Complex Numbers. Express in rectangular, polar, and trigonometric forms. 9. 5e j3 10. 7e j5 11. 2.2e j1.5 12. 4e j2. Operations in Exponential Form Multiply. 13. 9e j2 2e j4 15. 7e j 3e j3 17. 1.7e j5 2.1e j2. 14. 8e j 6e j3 16. 6.2e j1.1 5.8e j 2.7 18. 4e j7 3e j5. Divide. 19. 18e j6 by 6e j3 21. 55e j9 by 5e j6 23. 21e j2 by 7e j. 20. 45e j4 by 9e j2 22. 123e j6 by 105e j2 24. 7.7e j4 by 2.3e j2. Evaluate. 25. (3e j5)2 27. (2e j )3. 21–5 Imaginary 2 + j3. 3 R. 0. 2 Real. FIGURE 21–5 A vector represented by a complex number. Imaginary 2 + j3 3 2 1 0 −1. Vector Operations Using Complex Numbers. Vectors Represented by Complex Numbers One of the major uses of complex numbers is that they can represent vectors and, as we will soon see, can enable us to manipulate vectors in ways that are easier than we learned when studying oblique triangles. Take the complex number 2 j3, for example, which is plotted in Fig. 21–5. If we connect that point with a line to the origin, we can think of the complex number 2 j3 as representing a vector R having a horizontal component of 2 units and a vertical component of 3 units. The complex number used to represent a vector can, of course, be expressed in any of the forms of a complex number. Vector Addition and Subtraction Let us place a second vector on our diagram, represented by the complex number 3 j (Fig. 21–6). We can add them graphically by the parallelogram method, and we get a resultant 5 j2. But let us add the two original complex numbers, 2 j3 and 3 j. (2 j3) (3 j) 5 j2. 5 + j2 Real. 3 4 5 3−j. 26. (4e j2)3. This result is the same as what we obtained graphically. In other words, the resultant of two vectors is equal to the sum of the complex numbers representing those vectors. ◆◆◆. Example 30: Subtract the vectors 25l 48° 18l 175° .. FIGURE 21–6 Vector addition Solution: Vector addition and subtraction are best done in rectangular form. Converting the with complex numbers.. first vector, we have a1 25 cos 48° 16.7 b1 25 sin 48° 18.6.

(17) calter_c21_001-027v2. 28-01-2008. Section 21–5. ◆. 18:12. Page 17. 17. Vector Operations Using Complex Numbers. so 25l 48° 16.7 j18.6 Similarly, for the second vector, a2 18 cos 175° 17.9 b2 18 sin 175° 1.57 so 18l 175° 17.9 j1.57 Combining, we have (16.7 j18.6) (17.9 j1.57) (16.7 17.9) j(18.6 1.57) 34.6 j17.0 These vectors are shown in Fig. 21–7. y. 25. 10. 48. ˚. 20. 25. 48˚. 8 −1. 175. ˚. 18 175˚. −20. −10. 0. 10. 20. 30. x. FIGURE 21–7. ◆◆◆. Multiplication and Division of Vectors Multiplication or division is best done if the vector is expressed as a complex number in polar form. ◆◆◆. Example 31: Multiply the vectors 2l 25° and 3l 15° .. Solution: By Eq. 229, the product vector will have a magnitude of 3(2) 6 and an angle of 25° 15° 40°. The product is thus 6l 40° ◆◆◆. ◆◆◆. Example 32: Divide the vector 8l 44° by 4l 12° .. Solution: By Eq. 230, 8 (8l 44° ) (4l 12° ) l 44° 12° 2l 32° 4. ◆◆◆. Example 33: A certain current I is represented by the complex number 1.15l 23.5° amperes, and a complex impedance Z is represented by 24.6l 14.8° ohms. Multiply I by Z to obtain the voltage V. ◆◆◆. Solution: V IZ (1.15l 23.5° )(24.6l 14.8° ) 28.3l 38.3°. (V). ◆◆◆.

(18) calter_c21_001-027v2. 28-01-2008. 18. 18:12. Page 18. Chapter 21. ◆. Complex Numbers. The j Operator Let us multiply a complex number rl u by j. First we must express j in polar form. By Eq. 224, j 0 j1 1(cos 90° j sin 90°) 1l 90° After we multiply by j, the new absolute value will be r1r and the argument will be ␪ 90° So rl u j rl u 90° Thus the only effect of multiplying our original complex number by j was to increase the angle by 90°. If our complex number represents a vector, we may think of j as an operator that causes the vector to rotate through one-quarter revolution.. Exercise 5. ◆. Vector Operations Using Complex Numbers. Express each vector in rectangular and polar forms.. 1. 2. 3. 4. 5. 6.. Magnitude. Angle. 7.00 28.0 193 34.2 39.0 59.4. 49.0° 136° 73.5° 1.10 rad 2.50 rad 58.0°. Combine the vectors. 7. 8. 9. 10. 11. 12. Round to three significant digits, where needed, in this exercise set.. (3 j2) (5 j4) (7 j3) (4 j) 58l 72° 21l 14° 8.60l 58° 4.20l 160° 9(cos 42° j sin 42°) 2(cos 8° j sin 8°) 8(cos 15° j sin 15°) 5(cos 9° j sin 9°). Multiply the vectors. 13. (7 j3)(2 j5) 14. (2.50l 18° )(3.70l 48° ) 15. 2(cos 25° j sin 25°) 6(cos 7° j sin 7°) Divide the vectors. 16. (25 j2) (3 j4) 17. (7.70l 47° ) (2.50l 15° ) 18. 5(cos 72° j sin 72°) 3(cos 31° j sin 31°).

(19) calter_c21_001-027v2. 28-01-2008. Section 21–6. 21–6. ◆. 18:12. Page 19. 19. Alternating Current Applications. Alternating Current Applications. Rotating Vectors in the Complex Plane We have already shown that a vector can be represented by a complex number, rl u . For example, the complex number 5.00l 28° represents a vector of magnitude 5.00 at an angle of 28° with the real axis. A phasor (a rotating vector) may also be represented by a complex number Rl vt by replacing the angle ␪ by ␻t, where ␻ is the angular velocity and t is time. ◆◆◆. Glance back at Sec. 17–5 where we first introduced alternating current.. Example 34: The complex number 11l 5t. represents a phasor of magnitude 11 rotating with an angular velocity of 5 rad/s.. ◆◆◆. In Sec. 17–5 we showed that a phasor of magnitude R has a projection on the y axis of R sin ␻t. Similarly, a phasor Rl vt in the complex plane (Fig. 21–8) will have a projection on the imaginary axis of R sin ␻t. Thus R sin ␻t imaginary part of R␻t Similarly, R cos ␻t real part of Rl vt. Imaginary. ␻ R ␪. R sin ␻t 0. Real R cos ␻t. Thus, either a sine or a cosine wave can be represented by a complex number Rl vt , depending upon whether we project onto the imaginary or the real axis. It usually does not matter which we choose, because the sine and cosine functions are identical except for a phase difference of 90°. What does matter is that we are consistent. Here we will follow the convention of FIGURE 21–8 projecting the phasor onto the imaginary axis, and we will drop the phrase “imaginary part of.” Thus we say that R sin ␻t Rl vt Similarly, if there is a phase angle ␾, R sin(␻t ␾) Rl vt ␾ One final simplification: It is customary to draw phasors at t 0, so that ␻t vanishes from the expression. Thus we write R sin(␻t ␾) Rl f Effective or Root Mean Square (rms) Values Before we write currents and voltages in complex form, we must define the effective value of current and voltage. The power P delivered to a resistor by a direct current I is P I 2R. We now define an effective current Ieff that delivers the same power from an alternating current as that delivered by a direct current of the same number of amperes. P Ieff2R Figure 21–9 shows an alternating current i Im sin ␻t and i2, the square of that current. The mean value of this alternating quantity is found to deliver the same power in a resistor as a steady quantity of the same value. P Ieff 2R (mean i2)R But the mean value of i2 is equal to Im2/2, half the peak value, so Ieff 2Im 2>2 Im> 22. Thus the effective current Ieff is equal to the peak current Im divided by 12. Similarly, the effective voltage Veff is equal to the peak voltage Vm divided by 12.. We get the effective value by taking the square root of the mean value squared. Hence it is also called a root mean squared value. Thus Ieff is often written Irms instead..

(20) calter_c21_001-027v2. 28-01-2008. 20. 18:12. Page 20. Chapter 21. ◆. Complex Numbers. i2. i. Im mean i 2 0. Im2. t Im2 2. t. 0 (b) i2 = Im2sin2 ωt. (a) i = Im sin ωt. FIGURE 21–9. Example 35: An alternating current has a peak value of 2.84 A. What is the effective current?. ◆◆◆. Solution: Ieff . Im 22. . 2.84 22. 2.01 A. ◆◆◆. Alternating Current and Voltage in Complex Form Next we will write a sinusoidal current or voltage expression in the form of a complex number because computations are easier in that form. At the same time, we will express the magnitude of the current or voltage in terms of effective value. The effective current Ieff is the quantity that is always implied when a current is given, rather than the peak current Im. It is the quantity that is read when using a meter. The same is true of effective voltage. To write an alternating current in complex form, we give the effective value of the current and the phase angle. The current is written in boldface type: I. Thus we have the following equations:. The current is represented by i Im sin (vt f2). Complex Voltage and Current. I Ieff l f2 . Im 2 2. l f2. Similarly, the voltage is represented by v Vm sin (vt fi) V Veff l f1 . Vm 2 2. A76. A75. l f1. Note that the complex expressions for voltage and current do not contain t. We say that the voltage and current have been converted from the time domain to the phasor domain. ◆◆◆. Example 36: The complex expression for the alternating current i 2.84 sin(␻t 33°) A. is.

(21) calter_c21_001-027v2. 28-01-2008. Section 21–6. ◆. 18:12. Page 21. 21. Alternating Current Applications. I Ieff l 33° . 2.84 22. l 33° A. 2.01l 33° A ◆◆◆. ◆◆◆. Example 37: The sinusoidal expression for the voltage V 84.2l 49° V. is 84.2 22 sin(␻t 49°) V 119 sin(␻t 49°) V. ◆◆◆. Complex Impedance In Sec. 7–7 we drew a vector impedance diagram in which the impedance was the resultant of two perpendicular vectors: the resistance R along the horizontal axis and the reactance X along the vertical axis. If we now use the complex plane and draw the resistance along the real axis and the reactance along the imaginary axis, we can represent impedance by a complex number. where R circuit resistance X circuit reactance XL XC (A98) (A99) ƒ Z ƒ magnitude of impedance 2R2 X2 X ␾ phase angle arctan (A100) R Example 38: A circuit has a resistance of 5 in series with a reactance of 7 , as shown in Fig. 21–10(a). Represent the impedance by a complex number.. ◆◆◆. 8 6. X. 4 R=5Ω. X=7Ω. Z. 2 ␾. Z (a). 0. R 2. 4. 6. (b). FIGURE 21–10. Solution: We draw the vector impedance diagram as shown in Fig. 21–10(b). The impedance in rectangular form is Z 5 j7 The magnitude of the impedance is 252 72 8.60 The phase angle is arctan. 7 54.5° 5. So we can write the impedance as a complex number in polar form as follows: Z 8.60l 54.5°. ◆◆◆. Note that the “complex” expressions are the simpler. This happy situation is used in the ac computations to follow, which are much more cumbersome without complex numbers..

(22) calter_c21_001-027v2. 28-01-2008. 22. 18:12. Page 22. Chapter 21. ◆. Complex Numbers. Ohm’s Law for Alternating Current (ac) We stated at the beginning of this section that the use of complex numbers would make calculations with alternating current almost as easy as for direct current. We do this by means of the following relationship: V ⴝ ZI. Ohm’s Law for ac. A103. Note the similarity between this equation and Ohm’s law, Eq. A62. It is used in the same way. Example 39: A voltage of 142 sin 200t is applied to the circuit of Fig. 21–10(a). Write a sinusoidal expression for the current i. ◆◆◆. Solution: We first write the voltage in complex form. By Eq. A75, V. 142 22. l 0° 100l 0°. Next we find the complex impedance Z. From Example 38, Z 8.60l 54.5° The complex current I, by Ohm’s law for ac, is I. 100l 0° V 11.6l 54.5° Z 8.60l 54.5°. Then the current in sinusoidal form is (Eq. A76) i 11.6 22 sin(200t 54.5°) 16.4 sin(200t 54.5°) The current and voltage phasors are plotted in Fig. 21–11, and the instantaneous current and voltage are plotted in Fig. 21–12. Note the phase difference between the voltage and current waves. This phase difference, 54.5°, converted to time, is 54.5° 0.951 rad 200t t 0.951/200 0.00475 s 4.75 ms We say that the current lags the voltage by 54.5° or 4.75 ms. V = 100 0° 0. 54.5°. v(V) i(A) 150. 15 i = 16.4 sin (200t − 54.5°). I = 11.6 −54.5°. FIGURE 21–11. 100. 10. 50. 5. 0. 4.75 ms 10. 20. 30. t(ms). −50 v = 142 sin 200t. −100 −15. FIGURE 21–12. ◆◆◆.

(23) calter_c21_001-027v2. 28-01-2008. Section 21–6. ◆. Exercise 6. 18:12. Page 23. 23. Alternating Current Applications. ◆. Alternating Current Applications. Express each current or voltage in complex form. 1. i 250 sin(␻t 25°) 3. 57 sin(␻t 90°) 5. 144 sin ␻t. 2. 1.5 sin(␻t 30°) 4. i 2.7 sin ␻t 6. i 2.7 sin(␻t 15°). Express each current or voltage in sinusoidal form. 7. V 150l 0° 9. V 300l 90° 11. I 7.5l 0°. 8. V 1.75l 70° 10. I 25l 30° 12. I 15l 130°. Express the impedance of each circuit as a complex number in rectangular form and in polar form. 13. 15. 17. 19.. Fig. 21–13(a) Fig. 21–13(c) Fig. 21–13(e) Fig. 21–13(g). 14. Fig. 21–13(b) 16. Fig. 21–13(d) 18. Fig. 21–13(f). R = 1550 Ω R = 155 Ω. XL = 72 Ω. (a). XC = 18 Ω. (b). XL = 1240 Ω. (c). (d) R = 552 Ω. R = 72 Ω. XL = 4300 Ω. XC = 42 Ω. XL = 148 Ω. XC = 3400 Ω XC = 720 Ω. (e). (f). (g). FIGURE 21–13. 20. Write a sinusoidal expression for the curr§ent i in each part of Fig. 21–14. R = 125 Ω. v = 75 sin ␻t. i. Xc = 217 Ω (a). FIGURE 21–14. R = 3550 Ω. XL = 415 Ω. v = 177 sin ␻t. i. Xc = 4780 Ω (b). XL = 1100 Ω.

(24) calter_c21_001-027v2. 24. 28-01-2008. 18:12. Page 24. Chapter 21. ◆. Complex Numbers. 21. Write a sinusoidal expression for the voltage ␷ in each part of Fig. 21–15. i = 1.7 sin (␻t + 25°). R = 4.78 Ω. R = 176 Ω. i = 43 sin (␻t − 30°). v. XL = 2.35 Ω. v. Xc = 7.21 Ω (b). XL = 308 Ω (a). FIGURE 21–15. 22. Find the complex impedance Z in each part of Fig. 21–16. i = 4.0 sin (␻t + 45°). v = 10 sin ␻t. i = 95.0 sin (␻t − 28°). Z. v = 270 sin ␻t. (a). Z. (b). FIGURE 21–16. ◆◆◆. CHAPTER 21 REVIEW PROBLEMS. ◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆. Express the following as complex numbers in rectangular, polar, trigonometric, and exponential forms. 1. 3 24 2. 29 5 3. 2 249 4. 3.25 211.6 Evaluate. 5. j17 6. j25 Combine the complex numbers. Leave your answers in rectangular form. 7. (7 j3) (2 j5) 8. 4.8l 28° 2.4l 72° 9. 52(cos 50° j sin 50°) 28(cos 12° j sin 12°) 10. 2.7e j7 4.3e j5 Multiply. Leave your answer in the same form as the complex numbers. 11. (2 j)(3 j5) 12. (7.3l 21° )(2.1l 156° ) 13. 2(cos 20° j sin 20°) ⴢ 6(cos 18° j sin 18°) 14. (93e j2)(5e j7).

(25) calter_c21_001-027v2. 28-01-2008. 18:12. Page 25. 25. Review Problems. Divide and leave your answer in the same form as the complex numbers. 15. (9 j3) (4 j) 16. (18l 72° ) (6l 22° ) 17. 16(cos 85° j sin 85°) 8(cos 40° j sin 40°) 18. 127e j8 4.75e j5 Graph each complex number. 19. 7 j4 20. 2.75l 44° 21. 6(cos 135° j sin 135°) 22. 4.75e j2.2 Evaluate each power. 23. (4 j3)2 24. (5l 12° )3 25. [5(cos 10° j sin 10°)]3 26. [2e j3]5 27. Express in complex form: i 45 sin(␻t 32°) 28. Express in sinusoidal form: V 283l 22° 29. Write a complex expression for the current i in Fig. 21–17. i(A) 1.25. 0.05 0. 0.1. 0.2. 0.3. 0.4. 0.5. 0.6. 0.7 t (s). −1.25. FIGURE 21–17. Writing 30. Suppose that you are completing a job application to an electronics company which asks you: “Explain, in writing, how you would use complex numbers in an electrical calculation, and illustrate your explanation with an example.” How would you respond?.

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