# The Field of Complex Numbers

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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

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

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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