Appendix F: Complex Numbers

Transcription

1 Appedix F Complex Numbers F1 Appedix F: Complex Numbers Use the imagiary uit i to write complex umbers, ad to add, subtract, ad multiply complex umbers. Fid complex solutios of quadratic equatios. Write the trigoometric forms of complex umbers. Fid powers ad th roots of complex umbers. 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. The real umber a is called the real part of the complex umber a + bi, ad the umber bi (where b is a real umber) is called the imagiary part of the complex umber. Defiitio of a Complex Number For real umbers a ad b, the umber a + bi is a complex umber. If b 0, the a + bi is called a imagiary umber. A umber of the form bi, where b 0, 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, the their sum ad differece are defied as follows. Sum: (a + bi) + (c + di) = (a + c) + (b + d)i Differece: (a + bi) (c + di) = (a c) + (b d)i

2 F Appedix 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. ( i) + ( + i) = i + + i Remove paretheses. = + 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 I Example 1(b), otice 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 how these properties are used whe two complex umbers are multiplied. (a + bi)(c + di) = a(c + di) + bi(c + di) Distributive Property = ac + (ad)i + (bc)i + (bd)i Distributive Property = ac + (ad)i + (bc)i + (bd)( 1) i = 1 = ac bd + (ad)i + (bc)i Commutative Property = (ac bd) + (ad + bc)i Associative Property The procedure above is similar to multiplyig two polyomials ad combiig like terms, as i the FOIL method. Remark Rather tha tryig to memorize the multiplicatio rule at the left, 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.

3 Appedix F Complex Numbers F EXAMPLE Multiplyig Complex Numbers a. ( + i)( i) = ( i) + i( i) Distributive Property = 9 6i + 6i 4i Distributive Property = 9 6i + 6i 4( 1) i = 1 = Simplify. = 1 Write i stadard form. b. ( + i) = ( + i)( + i) Square of a biomial = ( + i) + i( + i) Distributive Property = 9 + 6i + 6i + 4i Distributive Property = 9 + 6i + 6i + 4( 1) i = 1 = 9 + 1i 4 Simplify. = 5 + 1i Write i stadard form. I Example (a), otice 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 + bi)(a bi) = a abi + abi b i = a b ( 1) = a + b 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 c + di = a + bi c + di( c di c di) = Multiply umerator ad deomiator by complex cojugate of deomiator. (ac + bd) + (bc ad)i c + d. Write i stadard form. EXAMPLE Writig Complex Numbers i Stadard Form + i 4 i = + i 4 i( 4 + Multiply umerator ad deomiator by complex cojugate of deomiator. 4 + i) 8 + 4i + 1i + 6i = 16 4i Expad. = i = + 16i 0 i = 1 Simplify. = i Write i stadard form. 5

4 F4 Appedix 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. Pricipal Square Root of a Negative Number If a is a positive umber, the the pricipal square root of the egative umber a is defied as a = ai. EXAMPLE 4 Writig Complex Numbers i Stadard Form a. 1 = i 1i = 6i = 6( 1) = 6 b = 48i 7i = 4 i i = i c. ( 1 + ) = ( 1 + i) = ( 1) i + ( ) (i ) = 1 i + ( 1) = i EXAMPLE 5 Complex Solutios of a Quadratic Equatio Remark The defiitio of pricipal square root uses the rule ab = a b for a > 0 ad b < 0. This rule is ot valid whe both a ad b are egative. For example, 5 5 = 5( 1) 5( 1) = 5i 5i = 5i = 5i = 5 whereas ( 5)( 5) = 5 = 5. To avoid problems with multiplyig square roots of egative umbers, be sure to covert to stadard form before multiplyig. Solve x x + 5 = 0. Solutio x = ( ) ± ( ) 4()(5) () = ± 56 6 = ± 14i 6 Quadratic Formula Simplify. Write 56 i stadard form. = 1 ± 14 i Write i stadard form.

5 Appedix F Complex Numbers F5 Trigoometric 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. (, 1) or i ( 1, ) or 1 + i Imagiary (, ) or + i 1 1 Figure F.1 1 Real Whe the complex umber a + bi is a real umber (that is, b = 0), this defiitio agrees with that give for the absolute value of a real umber. a + 0i = a + 0 = a To work effectively with powers ad roots of complex umbers, it is helpful to write complex umbers i trigoometric form. I Figure F., cosider the ozero complex umber a + bi. By lettig θ be the agle from the positive real (measured couterclockwise) to the lie segmet coectig the origi ad the poit (a, b), you ca write ( a, b) Imagiary a = r cos θ ad b = r si θ where r = a + b. Cosequetly, you have a + bi = r cos θ + (r si θ)i from which you ca obtai the trigoometric form of a complex umber. b α r θ Real Trigoometric Form of a Complex Number The trigoometric form of the complex umber z = a + bi is give by z = r(cos θ + i si θ) Figure F. where a = r cos θ, b = r si θ, r = a + b, ad ta θ = b a. The umber r is the modulus of z, ad θ is called a argumet of z. The trigoometric form of a complex umber is also called the polar form. Because there are ifiitely may choices for θ, the trigoometric form of a complex umber is ot uique. Normally, θ is restricted to the iterval 0 θ < π, although o occasio it is coveiet to use θ < 0.

6 F6 Appedix F Complex Numbers EXAMPLE 6 Trigoometric Form of a Complex Number Write the complex umber z = i i trigoometric form. Solutio The absolute value of z is r = i = ( ) + ( ) = 16 = 4 ad the agle θ is give by ta θ = b a = =. Because ta(π ) = ad because z = i lies i Quadrat III, choose θ to be θ = π + π = 4π. So, the trigoometric form is z = r(cos θ + i si θ) = 4 ( cos 4π + i si 4π ). See Figure F.. The trigoometric form adapts icely to multiplicatio ad divisio of complex umbers. Cosider the two complex umbers z 1 = r 1 (cos θ 1 + i si θ 1 ) ad z = r (cos θ + i si θ ). The product of z 1 ad z is z 1 z = r 1 r (cos θ 1 + i si θ 1 )(cos θ + i si θ ) = r 1 r [(cos θ 1 cos θ si θ 1 si θ ) + i(si θ 1 cos θ + cos θ 1 si θ )]. Usig the sum ad differece formulas for cosie ad sie, you ca rewrite this equatio as z 1 z = r 1 r [cos(θ 1 + θ ) + i si(θ 1 + θ )]. This establishes the first part of the rule show below. The secod part is left for you to verify (see Exercise 109). 4π 1 1 z = i Figure F. Imagiary 4 z = 4 Real Product ad Quotiet of Two Complex Numbers Let z 1 = r 1 (cos θ 1 + i si θ 1 ) ad z = r (cos θ + i si θ ) be complex umbers. z 1 z = r 1 r [cos(θ 1 + θ ) + i si(θ 1 + θ )] Product z 1 = r 1 [cos(θ z r 1 θ ) + i si(θ 1 θ )], z 0 Quotiet

7 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 Fid the product z 1 z of the complex umbers. z 1 = ( cos π + i si π ), Solutio z 1 z = ( cos π + i si π ) 8 ( cos 11π 6 z = 8 11π 11π ( cos + i si 6 6 ) + i si 11π 6 ) = 16 [ cos ( π + 11π 6 ) + i si ( π + 11π Multiply moduli ad 6 )] add argumets. = 16 [ cos 5π + i si 5π ] = 16 [ cos π + i si π 5π ] ad π are cotermial. = 16[0 + i(1)] = 16i Check this result by first covertig to the stadard forms z 1 = 1 + i ad z = 4 4i ad the multiplyig algebraically. EXAMPLE 8 Dividig Complex Numbers Fid the quotiet z 1 z of the complex umbers. z 1 = 4(cos 00 + i si 00 ), z = 8(cos 75 + i si 75 ) Solutio z 1 4(cos 00 + i si 00 ) = z 8(cos 75 + i si 75 ) = 4 8 [cos(00 75 ) + i si(00 75 )] Divide moduli ad subtract argumets. = [cos 5 + i si 5 ] = [ + i ( )] = i Appedix F Complex Numbers F7

8 F8 Appedix 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 = r(cos θ + i si θ) z = r (cos θ + i si θ) z = r (cos θ + i si θ) This patter leads to the ext theorem, which is amed after the Frech mathematicia Abraham DeMoivre ( ). THEOREM F.1 DeMoivre s Theorem If z = r(cos θ + i si θ) is a complex umber ad is a positive iteger, the z = [r(cos θ + i si θ)] = r (cos θ + i si θ). EXAMPLE 9 Fidig Powers of a Complex Number Use DeMoivre s Theorem to fid ( 1 + i) 1. Solutio First covert the complex umber to trigoometric form. 1 + i = ( cos π + i si π ) The, by DeMoivre s Theorem, you have ( 1 + i) 1 = [ ( cos π + i si π )] 1 = 1 [ cos ( 1 π ) + i si ( 1 π )] = 4096(cos 8π + i si 8π) = 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 below. Defiitio of th Root of a Complex Number The complex umber u = a + bi is a th root of the complex umber z whe z = u = (a + bi).

9 To fid a formula for a th root of a complex umber, let u be a th root of z, where u = s(cos β + i si β) ad z = r(cos θ + i si θ). By DeMoivre s Theorem ad the fact that u = z, you have s (cos β + i si β) = r(cos θ + i si θ). Takig the absolute value of each side of this equatio, it follows that s = r. Substitutig r for s i the previous equatio ad dividig by r, you get cos β + i si β = cos θ + i si θ. So, it follows that cos β = cos θ ad si β = si θ. Because both sie ad cosie have a period of π, these last two equatios have solutios if ad oly if the agles differ by a multiple of π. Cosequetly, there must exist a iteger k such that β = θ + πk β = θ + πk. By substitutig this value for β ito the trigoometric form of u, you get the result stated i the ext theorem. Appedix F Complex Numbers F9 THEOREM F. th Roots of a Complex Number For a positive iteger, the complex umber z = r(cos θ + i si θ) has exactly distict th roots give by r ( cos θ + πk + i si θ + πk ) where k = 0, 1,,..., 1. For k > 1, the roots begi to repeat. For istace, whe k =, the agle θ + π = θ + π is cotermial with θ, which is also obtaied whe k = 0. The 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 π, the roots are equally spaced alog the circle. r Imagiary π π Real Figure F.4

10 F10 Appedix 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 trigoometric form for z is z = + i = 8(cos 15 + i si 15 ). By the formula for th roots, the cube roots have the form 8 ( k k cos + i si ). Fially, for k = 0, 1, ad, you obtai the roots (cos 45 + i si 45 ) = 1 + i (cos i si 165 ) i (cos 85 + i si 85 ) i. F Exercises Performig Operatios 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. ( + i) (6 + 1i) 5. ( + 8) + (5 50) 6. (8 + 18) (4 + i) 7. 1i (14 7i) 8. + ( 5 + 8i) + 10i 9. ( + 5 i) + ( i) 10. (1.6 +.i) + ( i) ( 10) 14. ( 75) 15. (1 + i)( i) 16. (6 i)( i) 17. 6i(5 i) 18. 8i(9 + 4i) 19. ( i)( 14 10i) 0. ( + 5)(7 10) 1. (4 + 5i). ( i). ( + i) + ( i) 4. (1 i) (1 + i) Writig a Complex Cojugate I Exercises 5, write the complex cojugate of the complex umber. The multiply the umber by its complex cojugate i i 7. 5i i 9. 0i Writig i Stadard Form I Exercises 4, write the quotiet i stadard form i i i 7. + i i i i i i 1 i i i 1 ( i)(5i) (4 5i) 4. + i Performig Operatios I Exercises 4 46, perform the operatio ad write the result i stadard form i 1 i i i + i + 8i 44. i + i + 5 i i i 4 i

11 Appedix F Complex Numbers F11 Usig the Quadratic Formula I Exercises 47 54, use the Quadratic Formula to solve the quadratic equatio. 47. x x + = x + 6x + 10 = x + 16x + 17 = x 6x + 7 = x + 16x + 15 = x 6x 5 = t 4t + = s + 6s + = 0 Writig i Stadard Form I Exercises 55 6, simplify the complex umber ad write it i stadard form i + i 56. 4i i 57. 5i ( i) 59. ( 75) 60. ( ) i 6. 1 (i) Absolute Value of a Complex Number I Exercises 6 68, plot the complex umber ad fid its absolute value. 6. 5i i i i i Writig i Trigoometric Form I Exercises 69 76, represet the complex umber graphically, ad fid the trigoometric form of the umber. 69. i i i i 7. (1 + i) ( i) 75. 6i Writig i Stadard Form I Exercises 77 8, represet the complex umber graphically, ad fid the stadard form of the umber. 77. (cos i si 150 ) 78. 5(cos 15 + i si 15 ) 79. (cos 00 + i si 00 ) 80. 4(cos 15 + i si 15 ) ( cos π 4 + i si π 4 ) 8. 8 ( cos π 1 + i si π 1) Performig Operatios I Exercises 8 86, perform the operatio ad leave the result i trigoometric form. 8. [ ( cos π + i si π )][ 4 ( cos π 6 + i si π 6)] 84. [ ( cos π + i si π )][ 6 ( cos π 4 + i si π 4)] 85. [ 5 (cos i si 140 )][ (cos 60 + i si 60 )] 86. cos(5π ) + i si(5π ) cos π + i si π Usig DeMoivre s Theorem 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) ( + i) ( 1 + i) (1 i) ( + i) (1 i) 9. ( cos 5π 4 + i si 5π 4 ) [ ( cos π + i si π )] 8 Fidig th Roots I Exercises , (a) use Theorem F. 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 5(cos 10 + i si 10 ) 96. Square roots of 16(cos 60 + i si 60 ) 97. Fourth roots of 16 ( cos 4π + i si 4π ) 98. Fifth roots of ( cos 5π 6 + i si 5π 6 ) 99. Cube roots of 15 (1 + i) 100. Cube roots of 4 (1 i) Solvig a Equatio I Exercises , use Theorem F. to fid all the solutios of the equatio ad represet the solutios graphically x 4 i = x + 1 = x = x 4 81 = x + 64i = x 6 64i = x (1 i) = x 4 + (1 + i) = Proof Give two complex umbers z 1 = r 1 (cos θ 1 + i si θ 1 ) ad z = r (cos θ + i si θ ) show that z 1 z = r 1 r [cos(θ 1 θ ) + i si(θ 1 θ )], z 0.