{VERSION 5 0 "IBM INTEL NT" "5.0" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 1 }{CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 1 }{CSTYLE "2D Comment" 2 18 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 256 "" 0 1 0 128 128 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 257 "" 0 1 128 128 128 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 258 "" 0 1 0 0 128 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 259 "" 0 1 0 0 128 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 260 "" 0 1 0 0 128 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 261 "" 0 1 0 0 128 1 0 0 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 262 "" 0 1 0 0 128 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 263 "" 0 1 0 128 0 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 264 "" 0 1 0 128 0 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 265 "" 0 1 0 0 128 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 266 "" 0 1 0 0 128 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 267 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 268 "" 0 1 0 128 128 1 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 269 "" 0 1 0 128 128 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 270 "" 0 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 271 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 272 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 273 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 274 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 275 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 276 "" 0 1 0 0 128 1 0 0 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 277 "" 0 1 0 0 128 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 278 "" 0 1 0 128 128 1 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 279 "" 0 1 0 128 128 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 280 "" 0 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 281 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 282 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 283 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 284 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 285 "" 0 1 0 128 128 1 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 286 "" 0 1 0 128 128 1 0 1 0 0 0 0 0 0 0 0 }{PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Heading 1" -1 3 1 {CSTYLE "" -1 -1 "Times" 1 18 0 0 0 1 2 1 2 2 2 2 1 1 1 1 }1 1 0 0 8 4 1 0 1 0 2 2 0 1 }{PSTYLE "Heading 2" -1 4 1 {CSTYLE "" -1 -1 "Time s" 1 14 0 0 0 1 2 1 2 2 2 2 1 1 1 1 }1 1 0 0 8 2 1 0 1 0 2 2 0 1 } {PSTYLE "Normal" -1 256 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 1 2 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }} {SECT 0 {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 264 53 "High School Modul es > Precalculus by Gregory A. Moore" }}{PARA 3 "" 0 "" {TEXT -1 4 " \+ " }{TEXT 263 24 "Roots of Complex Numbers" }}{PARA 0 "" 0 "" {TEXT -1 121 "\nThis worksheet goes through the development of the concept o f roots of complex numbers geometrically and algebraically.\n" }} {PARA 0 "" 0 "" {TEXT 256 153 "[Directions : Execute the Code Resource section first. Although there will be no output immediately, these de finitions are used later in this worksheet.]" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 1 " " }{TEXT 258 9 " 0. Co de" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "restart; with(plots): \+ " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 955 "ComplexPlotY := proc ()\n local k,Radial,Pt,c,c1,c2,shade,r,u,Lines; \n u:= 0;\n c1 : = COLOR(RGB, .67, .54, .27 );\n c2 := COLOR(RGB, .72,.6,.3 );\n f or k from 1 to nargs do\n shade := evalf(rand()/10^12,2)/4 + .3; \n c := COLOR(RGB, shade+.3,shade+.2, shade); \n Radial|| k := complexplot( [0,args[k]], \n scaling = constrained, \+ color = c1, thickness = 1);\n Lines||k := plot( [[ 0, Im(args[k] )], [Re(args[k]), Im(args[k])],\n [Re(args[k] ), 0]], \n color = c2, linestyle = 2); \n\n \+ u := max( u, abs(Re(args[k])),abs(Im(args[k])) ); \n od;\n\n \+ r := evalf(u/12,2);\n c := COLOR(RGB, .8, .66, .25 ); \n for k fro m 1 to nargs do \n Pt||k := plottools[disk]( [Re(args[k]), Im(ar gs[k])],r, \n color = c ); od;\n display( [ seq( Radial||k, k = 1..nargs), seq( Pt||k, k = 1..nargs),\n \+ seq( Lines||k, k = 1..nargs)] ); \nend proc:\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 955 "ComplexPlotB := proc()\n local k,A,Pt,c, cg,shade,r,u,Lines; \n u:= 0;\n cg := COLOR(RGB, .7,.7,.8);\n fo r k from 1 to nargs do\n shade := evalf(rand()/10^12,2)/4 + .3; \n c := COLOR(RGB, shade,shade, shade+2); \n A||k := comp lexplot( [0,args[k]], linestyle = 2, \n scaling = constr ained, color = c);\n Lines||k := plot( [[ 0, Im(args[k])], [Re(a rgs[k]), Im(args[k])],\n [Re(args[k]), 0]], c olor = cg); \n\n u := max( u, abs(Re(args[k])),abs(Im(args[k])) \+ ); \n od;\n\n r := evalf(u/45,2);\n for k from 1 to nargs do \n \+ shade := evalf(rand()/10^12,2)/5 + .1;\n c := COLOR(RGB, s hade, shade, .7); \n Pt||k := plottools[rectangle]( \n [ Re(args[k])-r, Im(args[k])+r],[Re(args[k])+r, Im(args[k])-r], \n \+ color = c ); od;\n\n\n display( [seq( A||k, k = \+ 1..nargs), seq( Pt||k, k = 1..nargs),\n seq( Lines||k, k = 1..nargs)] ); \nend proc:\n\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 377 "UnityRootPlot := proc(n)\n local c;\n c := COLOR(RGB , .85, .65, .3 );\n display( [ComplexPlotB( 1),\n Comple xPlotY( seq( cos(2*k*Pi/n) + I*sin(2*k*Pi/n), k = 1..n) ),\n \+ seq( plottools[arc]( [0,0], .95-k*.045, 0..(2*k*Pi/n), \n \+ color = c,thickness = 3 ), k = 1..(n-1)), \n plottools[cir cle]([0,0],1, color = gold)]);\nend proc:\n\n\n" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 122 "UnityRoots := proc(n)\n local k,S;\n \+ S := solve(z^n - 1= 0):\n for k from 1 to n do print(S[k]); od;\n end proc:\n\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 204 "reset := \+ proc()\nglobal a,b,w,z,R,theta, phi,n,k;\na := 'a': b := 'b': z := 'z ': theta := 'theta': R := 'R':\nw:='w': z:='z': phi := 'phi': n:= 'n': k:= 'k': \nreturn (a,b,z,w,R,n,k,theta, phi):\nend proc:\n" }}}} {SECT 0 {PARA 4 "" 0 "" {TEXT -1 2 " " }{TEXT 260 29 "1. The Idea of Complex Roots" }}{PARA 0 "" 0 "" {TEXT -1 64 "\nAs we know there are \+ two square roots of every real number ...\n" }}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 19 "solve( x^2 = 9, x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "3^2; (-3)^2;" }}}{PARA 0 "" 0 "" {TEXT -1 25 "\nOr co mplex numbers ...\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "solv e( x^2 = -9 );" }}}{PARA 0 "" 0 "" {TEXT -1 123 "\nSo it should not be surprising that there are three numbers which have same cube. In othe r words, 8 has three cube roots.\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 58 "z1 := 2; \nz2 := -1 + sqrt(3)*I; \nz3 := -1 - sqr t(3)*I;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 60 "z1^3: % = evalc( %); \nz2^3: % = evalc(%);\nz3^3: % = evalc(%);" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 111 "display( ComplexPlotB(z1^3),\n Comp lexPlotY( z1,z2,z3), \n polarplot( abs(z1), color = tan ));" }}}{PARA 0 "" 0 "" {TEXT -1 65 "Nor should it be surprising that a num ber has four fourth roots. " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 115 "z1 := sqrt(2) + I*sqrt(2); \nz2 := sqrt(2) - I*sqrt(2);\nz3 \+ := -sqrt(2) + I*sqrt(2);\nz4 := -sqrt(2) - I*sqrt(2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 83 "z1^4: % = evalc(%); \nz2^4: % = eva lc(%); \nz3^4: % = evalc(%); \nz4^4: % = evalc(%); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 114 "display( ComplexPlotB(z1^3),\n \+ ComplexPlotY( z1,z2,z3,z4), \n polarplot( abs(z1), color = t an ));" }}}{PARA 0 "" 0 "" {TEXT -1 42 "\nEven complex numbers have fo ur 4th roots." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 92 "z1 := .8 \+ + 1.1*I; \nz2 := .8 - 1.1*I; \nz3 := - .8 + 1.1*I; \nz4 := - \+ .8 - 1.1*I; " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 82 "z1^4: % = evalc(%); \nz2^4: % = evalc(%); \nz3^4: % = evalc(%); \nz4^4: % = eva lc(%);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 114 "display( Comple xPlotB(z1^3),\n ComplexPlotY( z1,z2,z3,z4), \n polar plot( abs(z1), color = tan ));" }}}{PARA 0 "" 0 "" {TEXT -1 83 "\nTher e is clearly some pattern to these roots, which we will explore in som e depth." }}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 2 " " }{TEXT 265 41 "2. Expanding One Root of Unity into Many" }}{PARA 0 "" 0 "" {TEXT -1 180 "\nNotice that the number 1 raised to the 7th power is 1. Thus, it is a 7th root of 1. We would like to find more numbers which are also 7th roots of 1. In fact, there are six more.\n" }}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 14 "phi := 2*Pi/7;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 53 "u := cos(phi) + I*sin(phi);\nu^7:\n% = round(evalf(%) );" }}}{PARA 0 "" 0 "" {TEXT -1 21 "\nGiven an angle like " }{XPPEDIT 18 0 "2*Pi/7" "6#*(\"\"#\"\"\"%#PiGF%\"\"(!\"\"" }{TEXT -1 60 ", there is only one integer which can be multiplied to give " }{XPPEDIT 18 0 "2*Pi" "6#*&\"\"#\"\"\"%#PiGF%" }{TEXT -1 1 "." }}{EXCHG }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "phi*x = 2*Pi;\nsolve(%,x);" }}} {PARA 0 "" 0 "" {TEXT -1 40 "\nHowever, there are other angles, like \+ " }{XPPEDIT 18 0 "2*Pi/7" "6#*(\"\"#\"\"\"%#PiGF%\"\"(!\"\"" }{TEXT -1 62 ", that can be multiplied by 7 to give other even multiples of \+ " }{XPPEDIT 18 0 "2*Pi" "6#*&\"\"#\"\"\"%#PiGF%" }{TEXT -1 37 " - whic h are angularly equivalent to " }{XPPEDIT 18 0 "2*Pi" "6#*&\"\"#\"\"\" %#PiGF%" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "for k from 1 to 7 do `(7)`*k*'phi' \+ = 7*k*phi; od;" }}}{PARA 0 "" 0 "" {TEXT -1 95 "\nSo there are 7 diffe rent angles that can be multiplied by 7 to give an angle equivalent to 0. " }}{PARA 0 "" 0 "" {TEXT -1 136 "\nNow, if we recall one of the b asic properties of DeMoivre's Theorem and apply it to a number u, whic h is on the unit circle, we have :\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 85 "u:='u':\nu = ( cos(theta) + I*sin(theta) );\nu^n = \+ ( cos(n*theta) + I*sin(n*theta) );" }}}{PARA 0 "" 0 "" {TEXT -1 155 " \nNow if we put these ideas together, we find 7 different complex numb ers which give the same result when raised to the 7th power. Here are \+ seven numbers. \n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 61 "phi := \+ 2*Pi/7:\nfor k from 1 to 7 do\n phi||k := phi*k; od;" }}}{PARA 0 " " 0 "" {TEXT -1 62 "\nBut this list of numbers gives the same results, since 0 and " }{XPPEDIT 18 0 "2*Pi" "6#*&\"\"#\"\"\"%#PiGF%" }{TEXT -1 23 " are equivalent angles." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "for k from 0 to 6 do\n phi||k := phi*k; od;" }}}{PARA 0 "" 0 "" {TEXT -1 96 "\nIf we convert these numbers to complex numbers hav ing these angles as their arguments, we get :" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 58 "for k from 0 to 6 do\n cos(phi||k) + I*sin(p hi||k); od;" }}}{PARA 0 "" 0 "" {TEXT -1 44 "\nHere is a diagram showi ng these 7 numbers.\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "Uni tyRootPlot(7);" }}}{PARA 0 "" 0 "" {TEXT -1 2 "\n " }}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 2 " " }{TEXT 266 27 "3. Complex Roots of Unity" }}{PARA 0 "" 0 "" {TEXT -1 51 "\nRoots of unity are solutions to the e quation(s) :\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "z := 'z': \nz^n - 1 = 0;\nz^n = 0;" }}}{PARA 0 "" 0 "" {TEXT -1 130 " \n\nThese \+ have a nice geometric interpretation - equally spaced spokes of a whee l - with the first spoke along the positive x axis." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "UnityRoots(3);\nUnityRootPlot(3);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "UnityRoots(4);\nUnityRootPlo t(4);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "UnityRoots(5);\nUn ityRootPlot(5);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "UnityRoo ts(6);\nUnityRootPlot(6);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "UnityRoots(13);\nUnityRootPlot(13);" }}}{PARA 0 "" 0 "" {TEXT -1 110 "\nThere is an algebraic interpretation to these complex roots. No tice that the equation z^n -1 can be factored." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "z^7 - 1: \n% = (z-1)*simplify( %/(z-1) );" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "z^13 - 1: \n% = (z-1)*simpli fy( %/(z-1) );" }}}{PARA 0 "" 0 "" {TEXT -1 224 "\nSo an equation with this kind of polynomial can be factored - showing that there is alway s a root of 1. However, there are six more roots of the resuling sixth degree polynomial - these are the six other roots we saw above!" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "z^7 - 1 = 0;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "(z-1)*simplify( lhs(%)/(z-1) ) = 0; " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "UnityRoots(7);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "UnityRootPlot(7);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "r6 := cos(2*Pi/7) - I*sin(2* Pi/7);\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "z^6+z^5+z^4+z^3 +z^2+z+1;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "subs( z= r6, % );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "evalc(%): round(%);" }}}{PARA 0 "" 0 "" {TEXT -1 43 "\nSo the 6 roots (excluding 1) are roo ts of " }{TEXT 267 4 "both" }{TEXT -1 21 " of these equations. " }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "z^7 - 1 = 0;\nz^6+z^5+z^4+z^ 3+z^2+z+1 = 0;" }}}{PARA 0 "" 0 "" {TEXT -1 3 "\n\n " }}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 2 " " }{TEXT 277 48 "4. One Root Using DeMo ivre's Theorem in Reverse" }}{PARA 0 "" 0 "" {TEXT -1 39 "\nYou might \+ recall De Moivre's formula.\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "demoivre := \nz^n = (r^n)*( cos(n*theta) + I*sin(n*theta) );" }} }{PARA 0 "" 0 "" {TEXT -1 159 "\nNow we want to use it in reverse - wi th some care. The opposite of an nth power is a nth root, and the oppo site of multiplying by n is dividing by n.\n \n " }{TEXT 278 7 "E xample" }{TEXT 279 3 " : " }{TEXT 280 32 "Find a 5th root of w = 24 + \+ 10i." }}{PARA 0 "" 0 "" {TEXT -1 6 "\n " }{TEXT 281 6 "Step 1" } {TEXT -1 25 ". Express w in polar form" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "w := 24 + 10*I;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "R := abs(w);\ntheta := argument(w);\nevalf(theta);" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "w = R*'(cos(theta) + I*sin (theta))';" }}}{PARA 0 "" 0 "" {TEXT -1 5 "\n " }{TEXT 282 6 "Step \+ 2" }{TEXT -1 120 ". Find the 5th root of R. (The opposite of raising \+ \n the absolute value to the 5th power is to take the 5th root). " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "r := R^(1/5);" }}}{PARA 0 "" 0 "" {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT 283 7 " 3. " } {TEXT -1 25 " Divide the argument by 5" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "phi := theta/5;" }}}{PARA 0 "" 0 "" {TEXT 284 9 " \n \+ 4. " }{TEXT -1 43 "Construct the root using these new values. " }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "z := r*('cos'(phi) + I*'sin' (phi));" }}}{PARA 0 "" 0 "" {TEXT -1 69 "\n\nIndeed, if we raise this \+ number to the 5th power, we get back to w." }}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 91 "'z'^5 = z^5; \n phi := evalf(theta/5):\n z := r*(' cos'(phi) + I*'sin'(phi)):\n simplify(z^5);" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 2 " " }{TEXT 276 42 "5. \+ Expanding One Arbitrary Root into Many" }}{PARA 0 "" 0 "" {TEXT -1 79 "\nWe found one root, but we know there are others. Lets see how to fi nd them.\n\n " }{TEXT 286 22 " Example 5.1 : " }{TEXT -1 113 " \+ Find the 5th roots of -1\n\nLets start with a \"complex\" number, w = \+ -1. We can express this number in polar form.\n" }}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 8 "w := -1;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "R := abs(w);\ntheta := argument(w);" }}}{PARA 0 "" 0 "" {TEXT -1 73 " \nWe can find a single 5th root by dividing the angle by 5. (s ince R = 1)" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "z1 := cos(the ta/5) + I*sin(theta/5);" }}}{PARA 0 "" 0 "" {TEXT -1 69 "\nIt's also a pparent from the diagram that 5 times the angle of z1 is " }{XPPEDIT 18 0 "Pi;" "6#%#PiG" }{TEXT -1 1 "." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 213 "display(ComplexPlotY(z1), ComplexPlotB(w), \nplottoo ls[circle]([0,0],1, color = gold),\nplottools[arc]([0,0],.8, 0..Pi/5, \+ color = gold, thickness = 3),\nplottools[arc]([0,0],.9, 0..Pi, color = gold, thickness = 3) );" }}}{PARA 0 "" 0 "" {TEXT -1 123 "\nNow here \+ is the key new idea. We need to find other angles, which when multipli ed by 5, give theta plus even multiples of " }{XPPEDIT 18 0 "Pi;" "6#% #PiG" }{TEXT -1 102 ". We can simply solve an equation like this. We s olve all of these kinds of equations in the same way." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "theta + 3*2*Pi = 5*x; \nphi||3 := solve ( %, x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 171 "for k from 1 t o 5 do\n theta + (k-1)*2*Pi = 5*x; \n phi||k := solve( %, x);\n z ||k := cos(phi||k) + I*sin(phi||k);\n print(`--------------------- --------------`);\nod;" }}}{PARA 0 "" 0 "" {TEXT -1 30 "\nLet's see wh at it looks like." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "Complex PlotY(z1,z2,z3,z4,z5);" }}}{PARA 0 "" 0 "" {TEXT -1 81 "\n\n\nIndeed, \+ if we raise any of these numbers to the 5th power, we get back w = -1. " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "for k from 1 to 5 do\n \+ z||k^5 = round(evalf(z||k^5)) ;\nod;" }}}{PARA 0 "" 0 "" {TEXT -1 6 " \n\n " }{TEXT 285 14 "Example 5.2 : " }{TEXT -1 102 " Find the 4th \+ roots of i\n\nLets start with a \"complex\" number, w = i. And go thro ugh these same steps.\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "w \+ := I;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "R := abs(w);\nthet a := argument(w);" }}}{PARA 0 "" 0 "" {TEXT -1 73 " \nWe can find a si ngle 4th root by dividing the angle by 5. (since R = 1)" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "z1 := cos(theta/4) + I*sin(theta/4) ;" }}}{PARA 0 "" 0 "" {TEXT -1 69 "\nIt's also apparent from the diagr am that 4 times the angle of z1 is " }{XPPEDIT 18 0 "Pi/2" "6#*&%#PiG \"\"\"\"\"#!\"\"" }{TEXT -1 1 "," }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 215 "display(ComplexPlotY(z1), ComplexPlotB(w), \nplottools[circle ]([0,0],1, color = gold),\nplottools[arc]([0,0],.8, 0..Pi/8, color = g old, thickness = 3),\nplottools[arc]([0,0],.9, 0..Pi/2, color = gold, \+ thickness = 3) );" }}}{PARA 0 "" 0 "" {TEXT -1 58 "\nSolve the equati ons which give the same resulting angle." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 171 "for k from 1 to 4 do\n theta + (k-1)*2*Pi = 4*x; \+ \n phi||k := solve( %, x);\n z||k := cos(phi||k) + I*sin(phi||k) ;\n print(`-----------------------------------`);\nod;" }}}{PARA 0 " " 0 "" {TEXT -1 30 "\nLet's see what it looks like." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 96 "display( ComplexPlotY(z1,z2,z3,z4), Complex PlotB(w), \nplottools[circle]([0,0],1, color = gold));" }}}{PARA 0 "" 0 "" {TEXT -1 186 "\n\nPlease take a moment to study this diagram beca use it illustrates the principle of complex roots. The blue square at \+ i (0,1), is the original number. You can see the first 4th root at " } {XPPEDIT 18 0 "Pi;" "6#%#PiG" }{TEXT -1 326 "/8 in the first quadrant \+ - and we know that angle is exactly 1/4 of the angle of i. Now other t hree roots are spread around the circle - cutting the circle into four equal pieces - but starting at the first root - not at an angle of ze ro.\n\n\n\nAnd as before, if we raise any of these numbers to the 4th \+ power, we get back w = i." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "for k from 1 to 4 do\n z||k^5 = round(evalf(z||k^4)) ; od;" }}} {PARA 0 "" 0 "" {TEXT -1 2 "\n " }}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 2 " " }{TEXT 261 34 "6. Complete Set of Complex Roots " }}{PARA 0 " " 0 "" {TEXT -1 283 "\nWe have seen how we can use DeMoivre's formula \+ in reverse to find a single root of a complex number. We have also see n how a single complex root can be expanded to n distinct complex root s. Now we sew this all up, and use these ideas to find the nth roots f or any complex number. \n\n" }}{PARA 0 "" 0 "" {TEXT -1 4 " " } {TEXT 268 7 "Example" }{TEXT 269 3 " : " }{TEXT 270 42 "Find ALL of th e 5th roots of w = 24 + 10i." }}{PARA 0 "" 0 "" {TEXT -1 6 "\n " } {TEXT 271 1 "1" }{TEXT -1 25 ". Express w in polar form" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "w := 24 + 10*I;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "R := abs(w);\ntheta := argument(w);\nevalf( theta);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "w = R*'(cos(thet a) + I*sin(theta))';" }}}{PARA 0 "" 0 "" {TEXT -1 5 "\n " }{TEXT 272 1 "2" }{TEXT -1 179 ". Find the 5th root of R - which is never a p roblem since R is always non-negative. (The opposite of raising \n \+ the absolute value to the 5th power, is to take the 5th root)." }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "r := R^(1/5);" }}}{PARA 0 " " 0 "" {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT 273 7 " 3. " }{TEXT -1 77 " Divide the argument by 5 (the opposite of multiplying by 5 is \+ dividing by 5)" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "phi := the ta/5;" }}}{PARA 0 "" 0 "" {TEXT 274 9 " \n 4. " }{TEXT -1 45 "Const ruct the first root using these values. " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "z||1 := r*('cos'(phi) + I*'sin'(phi));" }}}{PARA 0 " " 0 "" {TEXT -1 1 "\n" }{TEXT 275 7 " 5. " }{TEXT -1 59 "Construct \+ the complete set of roots by adding multiples of " }{XPPEDIT 18 0 "2*P i" "6#*&\"\"#\"\"\"%#PiGF%" }{TEXT -1 28 " to theta, then divide by n. " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 97 "for k from 0 to 4 do\n \+ z||k :=r*('cos'((theta + 2*Pi*k)/n) + I*'sin'((theta + 2*Pi*k)/n)); \nod;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 139 "for k from 0 to 4 do\n z||k :=r*( 'cos'(evalf( (theta + 2*Pi*k)/5 ) )\n \+ + I*'sin'(evalf( (theta + 2*Pi*k)/5 ) ) );\nod;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 138 "for k from 0 to 4 do\n z||k := evalf( r*( 'cos'( (theta + 2*Pi*k)/5 )\n + I *'sin'( (theta + 2*Pi*k)/5 ) ));\nod;" }}}{PARA 0 "" 0 "" {TEXT -1 71 "\n\n\nIf we raise any of these numbers to the 5th power, we get back \+ to w!" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 64 "z0 := 1.912667497+. 1513354546*I;\n'z0^5' = round(simplify(z2^5));" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 65 "z1 := .4471181897+1.865820114*I;\n'z1^5' = rou nd(simplify(z2^5));\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 65 "z2 := -1.636333258+1.001804792*I;\n'z2^5' = round(simplify(z2^5));" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 65 "z3 := -1.458427760-1.2466707 03*I;\n'z3^5' = round(simplify(z2^5));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 64 "z4 := .7349753320-1.772289659*I;\n'z4^5' = round(simp lify(z2^5));" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 2 "\n " }}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 2 " " }{TEXT 262 47 "7. A Geometric Representation of Complex Roots" }}{PARA 0 "" 0 "" {TEXT -1 59 "\nHere are some geometric representations of complex roots.\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "z := 'z':" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 62 "w := 3 + 3*I; n := 3;\nro ots_of_unity := solve( z^3 = w, z);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 152 "display( ComplexPlotB( w),\n ComplexPlotY( seq( roots_of_unity[k], k = 1..n) ), \n polarplot( abs(root s_of_unity[1]), color = gray ));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 62 "w := 1 + 3*I; n := 3;\nroots_of_unity := solve( z^3 = w, z);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 152 "display( ComplexPlotB( w),\n \+ ComplexPlotY( seq( roots_of_unity[k], k = 1..n) ), \n p olarplot( abs(roots_of_unity[1]), color = gray ));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 160 "display( ComplexPlotB( w),\n Com plexPlotY( seq( roots_of_unity[k], k = 1..n) ), \n polarplot( evalf( abs( roots_of_unity[1])), color = gray ));" }}}{PARA 0 "" 0 " " {TEXT -1 2 "\n " }}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 1 " " }{TEXT 259 19 " 8. General Formula" }}{PARA 0 "" 0 "" {TEXT -1 69 "\nHere is \+ the formula for the powers of complex number in polar form.\n" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "reset();\nz^n = (r^n)*( cos( n*theta) + I*sin(n*theta) );" }}}{PARA 0 "" 0 "" {TEXT -1 70 "\n\nHere is the formula for the powers of complex number in polar form.\n" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "w = R*('cos'(theta) + I*'sin '(theta));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 70 "z = (R^(1/n)) *( cos((theta + 2*Pi*k)/n) + I*sin((theta + 2*Pi*k)/n)); " }}}{PARA 0 "" 0 "" {TEXT -1 37 "\nFor example, for n = 7, we have ...." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 99 "for k from 0 to 6 do \nz||k = (R^( 1/n))*( cos((theta + 2*Pi*k)/n) + I*sin((theta + 2*Pi*k)/n)); od;" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{PARA 0 "" 0 "" {TEXT 257 35 "\n \251 2002 Waterloo Maple Inc" }}}{MARK "0 1" 46 } {VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }