The Field of Complex Numbers


 Doris Fleming
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1 The Field of Complex Numbers S. F. Ellermeyer The costructio of the system of complex umbers begis by appedig to the system of real umbers a umber which we call i with the property that i = 1. (Note that there is o real umber whose square is 1.) The system of complex umbers cosists of all umbers of the form a + bi where a ad b are real umbers. We de e additio ad multiplicatio for complex umbers i such a way that the rules of additio ad multiplicatio are cosistet with the rules for real umbers. Thus, if z 1 = a 1 + b 1 i ad z = a + b i, we de e ad z 1 + z = (a 1 + a ) + (b 1 + b ) i z 1 z = (a 1 + b 1 i) (a + b i) = a 1 a + a 1 b i + a b 1 i + b 1 b i = (a 1 a b 1 b ) + (a 1 b + a b 1 ) i. To be formal, we de e the system of complex umbers to cosist of the set C = f(a; b) j a ad b are real umbersg with additio ad multiplicatio de ed for z 1 = (a 1 ; b 1 ) ad z = (a ; b ) i C by z + z = (a 1 + a ; b 1 + b ) ad z 1 z = (a 1 a b 1 b ; a 1 b + a b 1 ). Note that this de itio of the complex umbers (i terms of ordered pairs of real umbers) is similar to the formal de itio that we gave for rigs of polyomials the di ereces beig that elemets of rigs of polyomials 1
2 cosist of i ite sequeces of elemets (oly a ite umber of which are o zero) ad that the rule of multiplicatio is de ed di eretly. The umber i C is formally de ed as i = (0; 1). I order to show that C (with the operatios of additio ad multiplicatio de ed above) is a eld, me must show that C is a commutative rig with uity ad that all o zero members of C are uits. 0.1 C is closed uder additio ad multiplicatio Let z 1 ad z C. The z 1 = (a 1 ; b 1 ) ad z = (a ; b ) where a 1 ; a ; b 1 ; ad b are real umbers. Also (by the de itios of additio ad multiplicatio), we have z + z = (a 1 + a ; b 1 + b ) ad z 1 z = (a 1 a b 1 b ; a 1 b + a b 1 ). Sice a 1 + a, b 1 + b, a 1 a b 1 b, ad a 1 b + a b 1 are all real umbers, we see that z 1 + z ad z 1 z are both members of C. Thus C is closed uder additio ad multiplicatio. 0. Associative Laws of Additio ad Multiplicatio Let z 1 = (a 1 ; b 1 ), z = (a ; b ), ad z = (a ; b ) be members of C. The z 1 + (z + z ) = (a 1 ; b 1 ) + (a + a ; b + b ) = (a 1 + (a + a ) ; b 1 + (b + b )) = ((a 1 + a ) + a ; (b 1 + b ) + b ) = (z 1 + z ) + z which shows that the associative property of additio is satis ed. (Note that the associative property of additio for real umbers is used i showig this.) Also z 1 (z z ) = (a 1 ; b 1 ) (a a b b ; a b + a b ) = (a 1 (a a b b ) b 1 (a b + a b ) ; a 1 (a b + a b ) + b 1 (a a b b )) = (a 1 a a a 1 b b a b 1 b a b 1 b ; a 1 a b + a 1 a b + a a b 1 b 1 b b )
3 ad (z 1 z ) z = (a 1 a b 1 b ; a 1 b + a b 1 ) (a ; b ) = ((a 1 a b 1 b ) a (a 1 b + a b 1 ) b ; (a 1 a b 1 b ) b + (a 1 b + a b 1 ) a ) = (a 1 a a a 1 b b a b 1 b a b 1 b ; a 1 a b + a 1 a b + a a b 1 b 1 b b ) which shows that the associative property of multiplicatio is satis ed. 0. Commutative Properties of Additio ad Multiplicatio Let z 1 = (a 1 ; b 1 ) ad z = (a ; b ) be members of C. The z 1 + z = (a 1 + a ; b 1 + b ) = (a + a 1 ; b + b 1 ) = z + z 1 which shows that the commutative property of additio is satis ed, ad z 1 z = (a 1 ; b 1 ) (a ; b ) = (a 1 a b 1 b ; a 1 b + a b 1 ) = (a a 1 b b 1 ; b a 1 + b 1 a ) = (a ; b ) (a 1 ; b 1 ) = z z 1 which shows that the commutative property of multiplicatio is satis ed. 0. Existece of Additive ad Multiplicative Idetity Elemets The complex umber 0 = (0; 0) serves as a additive idetity elemet for C because if z = (a; b) is ay complex umber the z + 0 = (a; b) + (0; 0) = (a + 0; b + 0) = (a; b) = z ad it is also obvious that 0 + z = z.
4 The complex umber 1 = (1; 0) serves as a multiplicative idetity elemet for C because if z = (a; b) is ay complex umber the 1z = (1; 0) (a; b) = (1a 0b; 1b + 0a) = (a; b) = z (ad, sice the commutative property of multiplicatio is satis ed, it is also true that z1 = z). 0.5 Existece of Additive Iverses Let z = (a; b) be a complex umber. The z = ( a; b) is also a complex umber ad z + ( z) = (a; b) + ( a; b) = (a + ( a) ; b + ( b)) = (0; 0) = 0 which shows that z (as de ed) is the additive iverse of z. 0.6 Distributive Property Let z 1 = (a 1 ; b 1 ), z = (a ; b ), ad z = (a ; b ) be members of C. The z 1 (z + z ) = (a 1 ; b 1 ) (a + a ; b + b ) = (a 1 (a + a ) b 1 (b + b ) ; a 1 (b + b ) + b 1 (a + a )) = (a 1 a + a 1 a b 1 b b 1 b ; a 1 b + a 1 b + a b 1 + a b 1 ) = ((a 1 a b 1 b ) + (a 1 a b 1 b ) ; (a 1 b + a b 1 ) + (a 1 b + a b 1 )) = z 1 z + z 1 z which shows that the distributive property is satis ed. 0.7 Existece of Multiplicative Iverses Let z = (a; b) be a complex umber with z 6= 0. (This meas that ot both a ad b are zero ad thus that a + b 6= 0.) We claim that the complex umber a z 1 = a + b ; b a + b
5 is the multiplicative iverse of z. To check that this correct, observe that a zz 1 = (a; b) a + b ; b a + b a = a + b + b a + b ; ab a + b + ab a + b = (1; 0) = 1. 1 Some Termiology, De itios, Etc. A complex umber of the form (a; 0) ca be ideti ed with the real umber a ad we make the covetio to write (a; 0) = a. (For example, (0; 0) = 0 ad (1; 0) = 1.) A complex umber of the form (0; b) is called a purely imagiary umber. Thus, for example, the umber (0; 1) = i is a purely imagiary umber. A complex umber, (a; b), for which b 6= 0 is called a imagiary umber. Thus the system of complex umbers is made up of the real umbers ad the imagiary umbers (some of which are purely imagiary). If z = (a; b) is ay complex umber, the (a; b) = (a; 0) + (0; b) = (a; 0) + (b; 0) (0; 1) = a + bi ad we usually d it more coveiet to write z = a + bi (istead of z = (a; b)) i doig calculatios. The usual rules of algebra of the real umbers carry over exactly to the complex umbers. We just eed to remember that i = (0; 1) (0; 1) = ( 1; 0) = 1. However, the view of a complex umber as a ordered pair of real umbers is useful for gaiig a visual picture of the complex umbers. Each complex umber, (a; b), ca be ideti ed with the poit (a; b) i the Cartesia Plae. Whe dealig with complex umbers, we call this the complex plae. Just as we de e the absolute value of a real umber to be the distace from that umber to the origi o the umber lie, we de e the modulus (or absolute value) of a complex umber, z = (a; b), to be the distace from z to 0 = (0; 0) i the complex plae. The otatio for the modulus of z is the same otatio used for the absolute value of a real umber. Thus if z = (a; b), the jzj = p a + b. 5
6 The locatio of a complex umber i the complex plae is completely determied by the modulus of the umber ad by a argumet of the umber. Assumig that z = a + bi 6= 0, a argumet of z is ay umber (agle) such that cos () = a jzj ad si () = b jzj. The basic geometric picture relatig z, jzj, ad is give below. b z=a+bi z θ a Ay complex umber, z = a + bi 6= 0, has i itely may argumets but they all di er by multiples of. Thus, for example, = is a argumet of the umber z = 1 + i, but so is = + = 9= ad so is = 8 = 1=. The pricipal argumet of a complex umber z 6= 0 is the uique argumet,, such that <. The pricipal argumet of z is deoted by Arg (z) ad the set of all argumets of z is deoted by arg (z). Thus arg (z) = farg (z) + k j k Zg. Usig the otio of cogruece modulo, we ca write this as arg (z) = [Arg (z)]. To illustrate these ideas, let us cosider the complex umber z = 1 + i. 6
7 For this umber, we have jzj = p = p Arg (z) = arg (z) = ; 9 ; 17 ; : : : [ This example is illustrated i the picture below. 7 ; 15 ; ; : : :. z=1+i π/ We remark that the complex umber 0 is the oly complex umber for which the argumet is ot de ed. For ay o zero complex umber, z = a + bi, with argumet (where ca be ay member of arg (z)), we observe that z = a + bi = jzj cos () + jzj si () i = jzj (cos () + si () i). I order to express this more brie y, we de e This allows us to write cis () = cos () + si () i. z = a + bi = jzj cis (). For example, we ca write 1 + i = p cis. 7
8 Both ways of expressig a complex umber (a + bi ad jzj cis ()) are useful. I particular, we will see that the expressio a + bi is useful i helpig us to gai a geometric uderstadig of additio of complex umbers; whereas the expressio jzj cis () is useful i helpig us to gai a geometric uderstadig of multiplicatio of complex umbers. Exercise 1 Let z = + i ad w = i. 1. Compute z + w.. Compute zw.. Fid w 1.. Compute z=w (which meas zw 1 ). 5. Fid jzj, Arg (z), ad arg (z) ad write z i the form z = jzj cis () (where is ay member of arg (z)). Exercise Suppose that jzj = ad Arg (z) = z = a + bi. =6. Write z i the form Exercise If A is ay set of real umbers ad b is ay real umber, the we de e the set A + b as A + b = fa + b j a Ag. Thus, for example, if A = f ; 1; ; 5g, the A + 5 = f1; 6; 8; 0g. Suppose that z is a o zero complex umber ad that k is a o zero real umber. Explai why arg (kz) = arg (z) if k > 0 ad arg (kz) = arg (z) + if k < 0. Geometric Iterpretatios of Additio ad Multiplicatio of Complex Numbers.1 Additio If z = a + bi ad w = c + di are complex umbers, the z + w = (a + c) + (b + d) i. 8
9 If we picture z ad w as vectors i the complex plae, the z +w is the vector sum of z ad w. This is illustrated i the gure below usig the example z = w = + i i z + w = 1 + i Multiplicatio If z = jzj cis () ad w = jwj cis () are two o zero complex umbers, the zw = jzj jwj (cos () + si () i) (cos () + si () i) = jzj jwj (cos () cos () si () si () + (cos () si () + si () cos ()) i) = jzj jwj (cos ( + ) + si ( + ) i) = jzj jwj cis ( + ). This shows that the modulus of zw is the product of the moduli of z ad w ad that a argumet of zw is the sum of a argumet of z ad a argumet of w. To illustrate this idea, let us take z = 1+i ad w = i. Note that jzj = p, Arg (z) = =, jwj = 1, ad Arg (w) = =. Also zw = (1 + i) i = 1 + i 9
10 ad we observe that ad that Arg (zw) = jzwj = p = jzj jwj = Arg (z) + Arg (w). Exercise Suppose that z is a o zero complex umber ad suppose that is a argumet of z. Explai why must be a argumet of z 1. (Hit: What is a argumet of the umber 1?) Exercise 5 Prove that if z = jzj cis () ad w = jwj cis () are o zero complex umbers, the z w = jzj cis ( ). jwj Exercise 6 Explai why it is ot always true (for ay two o zero complex umbers, z ad w) that Arg (zw) = Arg (z) + Arg (w) : Give a example of two complex umbers, z ad w, for which the above equatio is true ad the give a example where the above equatio is ot true. Exercise 7 If A ad B are two sets, the we de e the set A + B as A + B = fa + b j a A ad b Bg. For example, if A = f ; 5; 6g ad B = f1; ; 5; 7g, the A + B = f ; ; 1; ; 6; 7; 10; 1; 8; 11; 1g. Explai why it is true for ay two o zero complex umbers, z ad w, that arg (zw) = arg (z) + arg (w). 10
11 De Moivre s Formula ad Roots of Complex Numbers We will ow prove (by iductio) that if z = jzj cis () is a o zero complex umber ad is ay positive iteger, the z = jzj cis (). This formula is clearly true whe = 1 ad if we assume that it s truth has bee proved for some positive iteger, the we obtai z +1 = zz = jzj cis () jzj cis () = jzj +1 (cos () + si () i) (cos () + si () i) = jzj +1 ((cos () cos () si () si ()) + (cos () si () + si () cos ()) i) = jzj +1 (cos (( + 1) ) + si (( + 1) ) i) = jzj +1 cis (( + 1) ) thus showig that the truth of the formula for implies the truth of the formula for + 1. This formula is called De Moivre s formula. To illustrate DeMoivre s Formula, let us use it to compute (1 + i) 8. Sice j1 + ij = p ad Arg (1 + i) = =, we ca write 1 + i = p cis. The by De Moivre s formula, p 8 (1 + i) 8 = cis 8 = 16 (cos () + si () i) = 16 (1 + 0i) = 16. We will ow address the questio of dig the th roots of a give o zero complex umber, w = jwj cis (), where is ay give positive iteger. Our rst observatio is that oe such root is the umber z 0 = p jwjcis 11
12 because (by De Moivre s formula) p z0 p = jwjcis = jwj cis = jwj cis () = w. We will show that w has exactly distict th roots. To see this, let z be ay th root of w (z = w) ad ote that it must be true that Also ote that sice jzj = p jwj. arg (w) = f + k j k Zg, the it follows from De Moivre s formula that + k arg (z) = j k Z. For k = 0; 1; ; : : : 1, the umbers + k cis are all distict from each other but + j + k cis = cis wheever j k mod because if j k mod, the j = k + t for some iteger t ad we have + j + (k + t) cis = cis + k = cis + t + k = cis. We coclude that if w = jwj cis (), the the distict th roots of w are z = p + k jwjcis, k = 0; 1; ; : : : 1. 1
13 Note that, i the above discussio, ca be ay member of arg (w). I particular, we ca certaily take = Arg (w). The pricipal th root of w, which we deote by p w, is de ed to be the root p p w = Arg (w) jwjcis. Example 8 Let us d the three cube roots of 1. Sice j1j = 1 ad Arg (1) = 0, the pricipal cube root of 1 is p p 0 1 = 1cis = cos (0) + si (0) i = 1 ad the other two cube roots of 1 are p 0 + (1) 1cis = cos + si i = 1 p + i ad p 0 + () 1cis = cos + si i = 1 p i. 1 Let us check by computatio that + p i is a cube root of 1. We will use the formula (x + y) = x + x y + xy + y. = p! 1 + i 1 + = p 8 i = 1. p! 1 i + p 8 i p! 1 i + p! i Example 9 Let us d the two square roots of i. Sice jij = 1 ad Arg (i) = =, the pricipal square root of i is p p = p p i = 1cis = cos + si i = + i 1
14 ad the other square root of i is p = + (1) 1cis = cos 5 + si 5 i = p p i. Remark 10 For ay o zero complex umber, w, the th roots of w are spaced evely aroud the circle of radius p jwj cetered at the origi of the complex plae. Exercise 11 Fid the four fourth roots of 1 ad the four fourth roots of 16. Exercise 1 Fid the three cube roots of 1. Exercise 1 Fid the three cube roots of 8i. Reducibility of Polyomials i C [x] We have bee studyig the problem of reducibility of polyomials f F [x] where F is a give eld (or perhaps eve just a commutative rig with uity). We have see examples of polyomials that are ot reducible i Z [x] ; Q [x], ad R [x] (where R is the eld of real umbers). Oe of the very importat properties of the eld of complex umbers, C, is that all polyomials are reducible i C [x]! Furthermore (ad this is actually a equivalet statemet), all polyomials ca be factored ito liear factors i C [x]. This fact, which was discovered through the combied e orts of may famous mathematicias (Gauss, LaGrage, Weierstrass, to ame a few) over a more tha 100 year period is what is referred to as the Fudametal Theorem of Algebra. Theorem 1 (The Fudametal Theorem of Algebra) Let f (x) = a x + a 1 x 1 + : : : + a 1 x + a 0 be a polyomial i C [x]. The f ca be factored ito liear factors: f (x) = a (x r 1 ) (x r ) (x r ) where r 1 ; r : : : ; r are the (ot ecessarily distict) roots of f. 1
15 Although we will ot attempt to prove this theorem, we will illustrate it with a simple example: If we cosider the polyomial f (x) = x 5 x + x 6x 8x + = x + x (x ), we see that this polyomial is reducible i Z [x] (which is show by the factorizatio give above). However, sice both x + ad x are irreducible i Z [x], we see that f caot be factored ito liear factors i Z [x]. If we cosider f as a polyomial i R [x] (where R is the eld of real umbers), the f (x) = x + p x x + p (x ) which shows that f ca be further factored i R [x] (i compariso with Z [x]). Fially, if we cosider f as a polyomial i C [x], the p f (x) = (x i) (x + i) x x + p (x ) which is a complete factorizatio of f ito liear factors. Exercise 15 Factor the followig polyomials as completely as possible i Z [x], R [x], ad C [x]. 1. f (x) = x x + x. f (x) = x 7x 18. f (x) = x + 5x + 1 Exercise 16 If z = a + bi is a complex umber, the the complex cojugate of z, deoted by z, is de ed as z = a bi. For example, if z = 9 + 8i, the z = 9 8i. Prove that the followig facts are true for ay complex umbers z ad w: 1. z + w = z + w. zw = z w.. If z is a real umber the z = z.. If f is ay polyomial with real coe ciets ad z is a complex umber that is a root of f, the z is also a root of f. 15
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