APPENDIX F Complex Numbers Operatios with Complex Numbers Complex Solutios of Quadratic Equatios Polar Form of a Complex Number Powers ad Roots of Complex Numbers Operatios with Complex Numbers Some equatios have o real solutios. For istace, the quadratic equatio x 1 0 Equatio with o real solutio has o real solutio because there is o real umber x that ca be squared to produce 1. To overcome this deficiecy, mathematicias created a expaded system of umbers usig the imagiary uit i, defied as i 1 Imagiary uit where i 1. By addig real umbers to real multiples of this imagiary uit, you obtai the set of complex umbers. Each complex umber ca be writte i the stadard form a bi. Defiitio of a Complex Number For real umbers a ad b, the umber a bi is a complex umber. If b 0, a bi is called a imagiary umber, ad bi is called a pure imagiary umber. To add (or subtract) two complex umbers, you add (or subtract) the real ad imagiary parts of the umbers separately. Additio ad Subtractio of Complex Numbers If a bi ad c di are two complex umbers writte i stadard form, their sum ad differece are defied as follows. Sum: a bi c di a c b di Differece: a bi c di a c b di F1
F APPENDIX F Complex Numbers The additive idetity i the complex umber system is zero (the same as i the real umber system). Furthermore, the additive iverse of the complex umber a bi is a bi a bi Additive iverse So, you have a bi a bi 0 0i 0. EXAMPLE 1 Addig ad Subtractig Complex Numbers a. Remove paretheses. i i i i i i Group like terms. 1 i 5 i Write i stadard form. b. i 4 i i 4 i Remove paretheses. 4 i i Group like terms. 4 Write i stadard form. c. i 5 i i 5 i 5 i i 0 i i STUDY TIP Rather tha tryig to memorize the multiplicatio rule at the right, you ca simply remember how the Distributive Property is used to multiply two complex umbers. The procedure is similar to multiplyig two polyomials ad combiig like terms. Notice i Example 1(b) that the sum of two complex umbers ca be a real umber. May of the properties of real umbers are valid for complex umbers as well. Here are some examples. Associative Properties of Additio ad Multiplicatio Commutative Properties of Additio ad Multiplicatio Distributive Property of Multiplicatio over Additio Notice below how these properties are used whe two complex umbers are multiplied. a bic di ac di bic di ac adi bci bdi ac adi bci bd1 ac bd adi bci ac bd ad bci Distributive Property Distributive Property Defiitio of i Commutative Property Associative Property
APPENDIX F Complex Numbers F EXAMPLE Multiplyig Complex Numbers a. i i 9 6i 6i 4i 9 41 Product of biomials i 1 9 4 1 Simplify. Write i stadard form. b. i 9 6i 6i 4i 9 1i 41 Product of biomials i 1 9 4 1i 5 1i Group like terms. Write i stadard form. Notice i Example (a) that the product of two complex umbers ca be a real umber. This occurs with pairs of complex umbers of the form a bi ad a bi, called complex cojugates. a bia bi a abi abi b i To write the quotiet of a bi ad c di i stadard form, where c ad d are ot both zero, multiply the umerator ad deomiator by the complex cojugate of the deomiator to obtai a bi a bi c c di c di c di a b 1 a b ac bd bc adi c d. EXAMPLE Writig a Quotiet of Complex Numbers i Stadard Form i i 4 4 i 4 i 4 i 8 4i 1i 6i 16 4i 8 6 16i 16 4 16i 0 1 10 4 5 i Multiply umerator ad deomiator by complex cojugate of deomiator. Expad. i 1 Simplify. Write i stadard form.
F4 APPENDIX F Complex Numbers Complex Solutios of Quadratic Equatios Whe usig the Quadratic Formula to solve a quadratic equatio, you ofte obtai a result such as, which you kow is ot a real umber. By factorig out i 1, you ca write this umber i stadard form. 1 1 i The umber i is called the pricipal square root of. STUDY TIP The defiitio of pricipal square root uses the rule for a > 0 ad b < 0. This rule is ot valid if both a ad b are egative. For example, whereas ab ab 55 5151 5i5i 5i 5i 5 55 5 5. To avoid problems with multiplyig square roots of egative umbers, be sure to covert to stadard form before multiplyig. EXAMPLE 4 a. b. Pricipal Square Root of a Negative Number If a is a positive umber, the pricipal square root of the egative umber a is defied as a ai. Writig Complex Numbers i Stadard Form 1 i1i 6i 61 6 48 7 48i 7i 4i i i c. 1 1 i 1 i i 1 i 1 i EXAMPLE 5 Complex Solutios of a Quadratic Equatio Solve x x 5 0. Solutio x ± 45 ± 56 6 ± 14i 6 1 ± 14 i Quadratic Formula Simplify. Write i i-form. Write i stadard form.
APPENDIX F Complex Numbers F5 ( 1, ) or 1 + i (, 1) or i Figure F.1 Imagiary 1 (, ) or + i 1 1 Real Polar Form of a Complex Number Just as real umbers ca be represeted by poits o the real umber lie, you ca represet a complex umber z a bi as the poit a, b i a coordiate plae (the complex plae). The horizotal is called the real ad the vertical is called the imagiary, as show i Figure F.1. The absolute value of a complex umber a bi is defied as the distace betwee the origi 0, 0 ad the poit a, b. The Absolute Value of a Complex Number The absolute value of the complex umber z a bi is give by a bi a b. If the complex umber a bi is a real umber that is, if b 0, the this defiitio agrees with that give for the absolute value of a real umber. ( a, b) Imagiary a 0i a 0 a. To work effectively with powers ad roots of complex umbers, it is helpful to write complex umbers i polar form. I Figure F., cosider the ozero complex umber a bi. By lettig be the agle from the positive real (measured couterclockwise) to the lie segmet coectig the origi ad the poit a, b, you ca write b a r θ Real a r cos ad b r si where r a b. Cosequetly, you have a bi r cos r si i from which you ca obtai the polar form of a complex umber. Figure F. Polar Form of a Complex Number The polar form of the complex umber z a bi is give by z rcos i si where a r cos, b r si, r a b, ad ta ba. The umber r is the modulus of z, ad is called a argumet of z. NOTE The polar form of a complex umber is also called the trigoometric form. Because there are ifiitely may choices for, the polar form of a complex umber is ot uique. Normally, is restricted to the iterval 0 although o occasio it is coveiet to use < 0. <,
F6 APPENDIX F Complex Numbers EXAMPLE 6 Writig a Complex Number i Polar Form Write the complex umber z i i polar form. Solutio The absolute value of z is r i 16 4 ad the agle is give by Figure F. Imagiary 4π 1 z = i 1 4 z = 4 Real ta b a. Because ta ad because z i lies i Quadrat III, choose to be So, the polar form is 4. z rcos i si 4 cos 4 4 i si. See Figure F.. The polar form adapts icely to multiplicatio ad divisio of complex umbers. Suppose you are give two complex umbers z ad z r cos i si 1 r 1 cos 1 i si 1. The product of z 1 ad is z z 1 z r 1 r cos 1 i si 1cos i si r 1 r cos 1 cos si 1 si isi 1 cos cos 1 si. Usig the sum ad differece formulas for cosie ad sie, you ca rewrite this equatio as z 1 z r 1 r cos1 i si1. This establishes the first part of the followig rule. Try to establish the secod part o your ow. Product ad Quotiet of Two Complex Numbers Let z ad z r cos i si 1 r 1 cos 1 i si 1 be complex umbers. z 1 z r 1 r cos1 i si1 Product z 1 r 1 cos1 z r i si1, z 0 Quotiet
APPENDIX F Complex Numbers F7 Note that this rule says that to multiply two complex umbers you multiply moduli ad add argumets, whereas to divide two complex umbers you divide moduli ad subtract argumets. EXAMPLE 7 Multiplyig Complex Numbers i Polar Form Fid the product z 1 z of the complex umbers. z 1 cos i si, z 8 cos 11 11 i si 6 6 Solutio z 1 z cos i si 16 cos 11 16 cos 5 5 i si 16 cos i si 16i 160 i1 8 cos 11 6 6 i si i si 11 6 11 6 Check this result by first covertig to the stadard forms z 4 4i ad the multiplyig algebraically. z 1 1 i ad EXAMPLE 8 Dividig Complex Numbers i Polar Form Fid the quotiet z 1 z of the complex umbers. z 1 4cos 00 i si 00, z 8cos 75 i si 75 Solutio z 1 4cos 00 i si 00 z 8cos 75 i si 75 4 8 cos 5 i si 5 i cos00 75 i si00 75 i
F8 APPENDIX F Complex Numbers Powers ad Roots of Complex Numbers To raise a complex umber to a power, cosider repeated use of the multiplicatio rule. z rcos i si z r cos i si z r cos i si This patter leads to the followig importat theorem, which is amed after the Frech mathematicia Abraham DeMoivre (1667 1754). THEOREM F.1 DeMoivre s Theorem If z rcos i si is a complex umber ad is a positive iteger, the z rcos i si r cos i si. EXAMPLE 9 Fidig Powers of a Complex Number Use DeMoivre s Theorem to fid 1 i 1. NOTE Notice i Example 9 that the aswer is a real umber. Solutio First covert to polar form. 1 i cos i si The, by DeMoivre s Theorem, you have 1 i 1 cos i si 1 1 cos 1 i si 1 4096cos 8 i si 8 4096. Recall that a cosequece of the Fudametal Theorem of Algebra is that a polyomial equatio of degree has solutios i the complex umber system. Each solutio is a th root of the equatio. The th root of a complex umber is defied as follows. Defiitio of th Root of a Complex Number The complex umber u a bi is a th root of the complex umber z if z u a bi.
APPENDIX F Complex Numbers F9 STUDY TIP The th roots of a complex umber are useful for solvig some polyomial equatios. For istace, you ca use DeMoivre s Theorem to solve the polyomial equatio x 4 16 0 by writig 16 as 16cos i si. To fid a formula for a th root of a complex umber, let u be a th root of z, where u scos i si ad z rcos i si. By DeMoivre s Theorem ad the fact that u z, you have s cos i si rcos i si. Takig the absolute value of each side of this equatio, it follows that s r. Substitutig back ito the previous equatio ad dividig by r, you get cos i si cos i si. So, it follows that cos cos ad si si. Because both sie ad cosie have a period of, these last two equatios have solutios if ad oly if the agles differ by a multiple of. Cosequetly, there must exist a iteger k such that k k By substitutig this value for. ito the polar form of u, you get the followig result. Imagiary THEOREM F. th Roots of a Complex Number For a positive iteger, the complex umber z rcos i si has exactly distict th roots give by r cos k i si where k 0, 1,,..., 1. k r π π Real Whe k exceeds 1, the roots begi to repeat. For istace, if k, the agle Figure F.4 is cotermial with, which is also obtaied whe k 0. This formula for the th roots of a complex umber z has a ice geometric iterpretatio, as show i Figure F.4. Note that because the th roots of z all have the same magitude r, they all lie o a circle of radius r with ceter at the origi. Furthermore, because successive th roots have argumets that differ by, the roots are equally spaced alog the circle.
F10 APPENDIX F Complex Numbers EXAMPLE 10 Fidig the th Roots of a Complex Number Fid the three cube roots of z i. Solutio Because z lies i Quadrat II, the polar form for z is z i 8 cos 15 i si 15. By the formula for th roots, the cube roots have the form 68 cos 15 60k i si Fially, for k 0, 1, ad, you obtai the roots cos 45 i si 45 1 i 15 60k. cos 165 i si 165 1.660 0.660i cos 85 i si 85 0.660 1.660i. EXERCISES FOR APPENDIX F I Exercises 1 4, perform the operatio ad write the result i stadard form. 1. 5 i 6 i. 1 i 5 6i. 8 i 4 i 4. 5. 6. 7. 1i 14 7i 8. 5 8i 10i 9. 5 i 5 11 i 10. 1.6.i 5.8 4.i 11. 6 1. 5 10 1. 10 14. 75 15. 1 i i 16. 6 i i 17. 6i5 i 18. 8i9 4i 19. 14 10i14 10i 0. 57 10 1. 4 5i. i. i i 4. 1 i 1 i i 6 1i 8 5 50 8 18 4 i I Exercises 5, write the complex cojugate of the complex umber. The multiply the umber by its complex cojugate. 5. 5 i 6. 9 1i 7. 5i 8. 4 i 9. 0i 0. 15 1. 8. 1 8 I Exercises 4, write the quotiet i stadard form.. 6 4. 10 i i 5. 4 6. 4 5i 1 i 7. i 8 7i 8. i 1 i 6 7i 9. 40. i 1 41. 4. 4 5i I Exercises 4 46, perform the operatio ad write the result i stadard form. 4. 44. 1 i 1 i i i 45. 46. i 8i 8 0i i i5i i i i 5 i 1 i i 4 i
APPENDIX F Complex Numbers F11 I Exercises 47 54, use the Quadratic Formula to solve the quadratic equatio. 47. x x 0 48. x 6x 10 0 49. 4x 16x 17 0 50. 9x 6x 7 0 51. 4x 16x 15 0 5. 9x 6x 5 0 5. 16t 4t 0 54. 5s 6s 0 I Exercises 55 6, simplify the complex umber ad write it i stadard form. 55. 6i i 56. 4i i 57. 5i 5 58. i 59. 75 60. 6 61. 6. 1 i 1 i I Exercises 6 68, plot the complex umber ad fid its absolute value. 6. 5i 64. 5 65. 4 4i 66. 5 1i 67. 6 7i 68. 8 i I Exercises 69 76, represet the complex umber graphically, ad fid the polar form of the umber. 69. i 70. i 71. i 7. 1 i 7. 1 i 74. 5 i 75. 6i 76. 4 I Exercises 77 8, represet the complex umber graphically, ad fid the stadard form of the umber. 77. cos 150 i si 150 78. 5cos 15 i si 15 79. cos 00 i si 00 80. 4cos 15 i si 15 81..75 cos i si 4 4 8. 8 cos i si 1 1 I Exercises 8 86, perform the operatio ad leave the result i polar form. 8. 84. i si i si 4 85. 5 cos 140 i si 140 cos 60 i si 60 cos5 i si5 86. cos i si I Exercises 87 94, use DeMoivre s Theorem to fid the idicated power of the complex umber. Write the result i stadard form. 87. 1 i 5 88. i 6 89. 1 i 10 90. 1 i 1 91. i 7 9. 41 i 9. 94. cos cos 5 5 cos i si 4 4 10 cos i si i si I Exercises 95 100, (a) use Theorem F. o page F9 to fid the idicated roots of the complex umber, (b) represet each of the roots graphically, ad (c) write each of the roots i stadard form. 95. Square roots of 5cos 10 i si 10 96. Square roots of 16cos 60 i si 60 97. Fourth roots of 98. Fifth roots of 16 cos 4 4 i si cos 5 5 i si 6 6 99. Cube roots of 15 1 i 100. Cube roots of 41 i 4 cos 6 cos 8 I Exercises 101 108, use Theorem F. o page F9 to fid all the solutios of the equatio ad represet the solutios graphically. 101. x 4 i 0 10. x 1 0 10. x 5 4 0 104. x 4 81 0 105. x 64i 0 106. x 6 64i 0 107. x 1 i 0 108. x 4 1 i 0 i si 6 6 4