of Nebraska - Lincoln

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1 Unversty of Nebraska - Lncoln DgtalCommons@Unversty of Nebraska - Lncoln MAT Exam Expostory Papers Math n the Mddle Insttute Partnershp 008 The Square Root of Tffany Lothrop Unversty of Nebraska-Lncoln Follow ths and addtonal works at: Part of the Scence and Mathematcs Educaton Commons Lothrop, Tffany, "The Square Root of " (008). MAT Exam Expostory Papers Ths Artcle s brought to you for free and open access by the Math n the Mddle Insttute Partnershp at DgtalCommons@Unversty of Nebraska - Lncoln. It has been accepted for ncluson n MAT Exam Expostory Papers by an authorzed admnstrator of DgtalCommons@Unversty of Nebraska - Lncoln.

2 The Square Root of Whle Grolamo Cardano was workng on solvng cubc and quadratc equatons n 159, he encountered some formulas that nvolved square roots of negatve numbers. In 1545 Cardano publshed Ars Magna, where he presents the frst recorded calculatons that nvolve complex numbers. Then n 157, Rafael Bombell publshed the frst three parts of hs Algebra. He s known as the nventor of complex numbers, because he dentfes some rules for workng wth them. Bombell also shows how complex numbers are very mportant and useful. From Bombell s lst of rules for addng, subtractng and multplyng the complex numbers, he was able to analyze the cubc equatons that Cardano was tryng to solve n hs paper Ars Magna. Bombell was able to use hs rules for operatons wth complex numbers to solve the cubc equatons that produced an expresson that contaned a square root of a negatve number. The next bg dscovery n complex numbers was made by Abraham de Movre. De Movre publshed papers n 1707 and n 17 where he used trgonometrc functons to represent complex numbers. He developed the representaton [r (cos x + sn x)]. Ths was a very mportant n the development and theory of complex

3 numbers. However, t was Leonhard Euler s publcatons that really brought complex numbers to the forefront. In 1748, Euler publshed Analyss of the Infnte. Mathematcal analyss s sad to have started wth Euler where he used Bernoull s deas of functons and refnes them. From ths mathematcal analyss Euler based hs study on functons and ntroduced the formulas e π = 1 and e x = cos x + sn x. Then n 1751, Euler publshed hs theory of logarthms of complex numbers and ntroduced the symbol to represent 1. Stll, most mathematcans of that day rejected the noton of complex numbers, despte Euler s publcatons. In 1799, Caspar Wessel publshed a paper gvng a geometrcal representaton of complex numbers. Ths was not a well know paper and was not even publshed n Englsh untl 1999, 00 years after ts frst publcaton. Jean Robert Argand redscovered Wessel s work n 1806 wth hs publcaton of the Argand dagram. In ths geometrcal representaton of complex numbers, Argand nterpreted as a rotaton of 90 o. Complex numbers are of the form x + y where x and y are real numbers and s the magnary part. The dagram below s the Argand dagram that shows a graphcal representaton of a complex number. The x-axs s the real number lne and an ordered par that represents a pont on ths lne has coordnates (x, 0) or the

4 complex number x + 0. The y-axs s the magnary axs and an ordered par that represents a pont on ths lne has coordnates (0, y) or the complex number 0 + y, a pure magnary number. Ths graph gves us a two-dmensonal vew of a complex number. Usng the complex plane, or Argand dagram, de Movre s formula and Euler s formula we now have z = x + y = r(cosα + snα) = re α, where.. In ths equaton, multplcaton by results n a counter-clockwse rotaton of 90 o about the orgn, or π radans. So, when we look at the geometrc representaton of = 1, t s shown as two 90 o turns (180 o ) or π radans.

5 Complex Numbers a + b where a,b R Ex. +, 7, 5 + 0,, 5, 5, π, 7, 5.67 Real Numbers a + b, b = 0 Ex. 5, 5, π, 7, 5.67, Imagnary Numbers a + b, b 0 Ex. +, 1, Pure Imagnary Numbers a + b, a = 0, b 0 Ex. 7, a b, Ratonal Numbers Ex. 5, 7, 5.67 Irratonal Numbers Infnte & non-repeatng decmals Ex. 5, π, 4

6 Now usng our knowledge of complex and magnary numbers where = 1, = 1 we can look at basc operatons wth them. Fndng square roots of some negatve numbers: a) 5 b) 5 c) 50 = 1 5 = 1 5 = 1 5 = 5 = 5 = 5 d) e) + 18 = = = 4 7 = + = - = (1 + ) = 4 f) 4 5 g) = = = ( 1) 5 = 1 6 = -10 = 6 h) = = = = ( 1) ) 6 6 = 6 = = 6 ( 1) j) x + 19 = x = - 19 x = -16 = 6 x =± 8 x =± 1 4 = x =± 5

7 Some basc operatons wth complex numbers: Sum of Complex Numbers Dfferences of Complex Numbers (a + b ) + (c + d ) (a + b) - (c + d ) = (a + c) + (b + d) = (a - c) + (b - d) Ex. ( + 6) + (4 + ) Ex. ( + 6) - (4 + ) = ( + 4) + (6 + ) = ( - 4) + (6 ) = = Ex. ( + 6) + (4 ) Ex. ( + 6) - (4 ) ( + 6) + (4 + -) ( + 6) - (4 + -) = ( + 4) + [6 + (-)] = ( - 4) + [6 (-)] = = Product of Complex Numbers (a + b)(c + d ) Quotent of Complex Numbers a + b c + d = ac + bd + ad + cb = (a + b)(c d) (c + d)(c d) = ac + (-1)bd + (ad + bc) = ac bd ad + cb c d = ac bd + (ad + bc) = = ac ( 1)bd + (cb ad) c ( 1)d ac + bd + (cb ad) c + d 6

8 Product of Complex Numbers Quotent of Complex Numbers Ex. (6 + 4)( 5) Ex. 5 (6 + 4)( + 5) =(6)() + (4)(-5) + (6)(-5) +()(4) = ( 5)( + 5) = = ( 5)5 = 1 0(-1) ( 1) = 4 + 5( 1) = = = - = The algebra of complex numbers nvolves treatng as a number and usng the basc number and operaton propertes (such as the dstrbutve, assocatve, and commutatve propertes) to rewrte the expresson n the form a + b. We can use the nformaton about complex numbers and operatons, along wth formulas such as de Movre s formula and Euler s formula, to study. Evaluatng. We begn by reteratng that any complex number, z, can be wrtten as z = a + b. We also note that the complex numbers we are lookng for wll satsfy the equaton z = 0, whch, by the Fundamental Theorem of Algebra, has four 7

9 solutons (two of whch satsfy z - = 0, two whch satsfy z + = 0). Then, applyng Euler s formula for wrtng complex numbers, we can wrte z as: z = re = ( e ) = cos( n ) + sn( n) = (cos + sn ) n ( ) n n π From the frst part of Euler s formula we wrte = e. Then, proceedng formally (and admttedly abusng some notaton) we take the square root of both sdes: = π e π = e 1 π 4 = e = cos( 1 4 π ) + sn( 1 4 π ) y The radus of the crcle s 1 unt. sn = opposte hypotenuse 1 cos sn cos = adjacent hypotenuse x 8

10 Thus we need to evaluate the sne and cosne of π/4. Snce π 4 = 45o, the reference trangle for our calculatons s an sosceles, 45 o -45 o -90 o trangle. To fnd the length of each sde of the trangle we use Pythagorean s Theorem a + b = c ; the hypotenuse s 1 unt and we let the legs each be of length x unts. Then solvng for x we have: x + x = 1 x = 1 x = 1 x = 1 x = 1 Thus, cos π 4 = 1 and sn π 4 = 1 x = 1, so we wrte = cos 1 4 π + sn 1 4 π However, we stated prevously that we were actually lookng for four complex numbers of the form z = a + b whch satsfy z = 0. Based on the calculaton above, t makes sense to consder four possbltes for z = 1 + 1, 1 + 1, To determne whch are actual answers to whch should equal 1. ; specfcally , 1., I wll square each and solve for 9

11 Does = 1 + 1? Does = 1 + 1? = = = = = = = 1 1 = 1 1 = 1 1 = 1 = 4 1 = 1 = Does = 1 + 1? Does = 1 + 1? = ( ) = ( ) = = ( 1 ) = ( 1 ) + ( 1 + ) = ( 1 + ) 1 = 1 1 = 1 ( 1 ) = (1 ) ( 1 ) = (1 ) 1 = 4 1 = 1 = 4 1 = =

12 Therefore there are two solutons to ; they are = and = The two complex numbers whch are solutons to lead us to consder the queston: Is there a pattern for fndng the nth roots of? To look for a pattern I wll return to the case where n =, the square root of : ( ) 1 = e = e I can rotate any pont n the complex plane about the orgn by 60 o or π radans and return to the same locaton on the Argand dagram. Lkewse I can rotate the pont agan by 60 o for a total of 4π radans. Each tme I perform ths rotaton I need to take π tmes k, where k s the number of tmes I have gone around the crcle. Thus, I can wrte π = + kπ. When calculatng a root, I need to check where s less than π because the sne and cosne functons are perodc wth a 11

13 perod of π. Thus, to determne all values for (snce we are consderng the square root) we consder = 1 π + kπ, where k = 0, 1,,. So, for the specfc cases where k = 0, 1, and I calculated the followng: = 1 π + kπ = 1 π + (0)π = 1 π = π 4 < π for k = 0 = 1 π + kπ = 1 π + (1)π = 1 5π = 5π 4 < π for k = 1 = 1 π + kπ = 1 π + ()π = 1 9π = 9π 4 > π for k = Thus I need only consder the cases where k = 0 and k = 1, snce for larger values of k I have > π. Now, replacng wth π 4 and 5π 4, I can return to the formula 1

14 and calculate the followng: = cos 1 4 π + sn 1 4 π = cos 45 o + sn 45 o = = cos 5 4 π + sn 5 4 π = cos15 o + sn15 o = These are the same values for that I found before. I can extend ths dea to calculate the cube root of. I begn by wrtng ( ) 1 = e and rotatng around the unt crcle. Then I determne the = e number of rotatons, k, for whch 1/ ( + kπ) < π : = 1 π + kπ = 1 π + (0)π = 1 π = π 6 < π for k = 0 = 1 π + kπ = 1 π + (1)π = 1 5π = 5π 6 < π for k = 1 1

15 = 1 π + kπ = 1 π + ()π = 1 9π = 9π 6 = π < π for k = = 1 π + kπ = 1 π + ()π = 1 1π = 1π 6 > π for k = Thus, I only need to consder the cases where = 1 π + kπ for k = 0, 1, ; specfcally when s π 6, 5π 6 and π. Then, substtutng these values nto the formula leads to the followng calculatons: = cos 1 6 π + sn 1 6 π = cos 5 4 π + sn 5 4 π = cos 0 o + sn 0 o = cos150 o + sn150 o = + 1 = + 1 = cos π + sn π = cos70 o + sn70 o = 0 + ( 1) 14

16 Therefore, =, + 1,and + 1. Now I can generalze ths dea to fnd the soluton set for any root of. From the prevous two examples, I notced that the number of rotatons, k, s always one less than the root that I am tryng to fnd. Ths wll allow me to lst a complete set of the values of : n = 1 n π + kπ for k = 0, 1,,, (n-1). Then the complex solutons to the equaton z n + 1 = 0 (.e. the n th roots of ) are gven by n = cos n + sn n where n = π n 1 + 4k ( ) for k = 0, 1,,, (n-1). 15

17 Bblography Brown, R.G., (199). Advanced mathematcs: precalculus wth dscrete mathematcs and data analyss. Geneva, IL: Houghton Mffln Company. Brown, R.G., Dolcan, M.P., Sorgenfrey R.H., & Kane, R.B., (1994). Algebra and Trgonometry: Structure and Method. Geneva, IL: Houghton Mffln Company. Answers Corporaton. (006). Complex Number. Retreved on July 17, 006 from O Connor, J., & Robertson, E. ( 006). Abraham de Movre. Retreved on July 17, 006 from O Connor, J., & Robertson, E. ( 006). Caspar Wessel. Retreved on July 17, 006 from O Connor, J., & Robertson, E. ( 006). Grolamo Cardano. Retreved on July\ 17, 006 from O Connor, J., & Robertson, E. ( 006). Index for the Chronology. Retreved on July 17, 006 from O Connor, J., & Robertson, E. ( 006). Jean Robert Argand. Retreved on July 17, 006 from O Connor, J., & Robertson, E. ( 006). Johann Carl Fredrch Gauss. Retreved on July 17, 006 from 16

18 O Connor, J., & Robertson, E. ( 006). Leonhard Euler. Retreved on July 17, 006 from O Connor, J., & Robertson, E. ( 006). Rafael Bombell. Retreved on July 17, 006 from Wessten, Erc W. (00). Argand Dagram. Retreved on July 17,006 from MathWorld A Wolfram Web Resource. Wessten, Erc W. (004). Complex Number. Retreved on July 17,006 from MathWorld A Wolfram Web Resource. Wessten, Erc W. (000). de Movre s Identty. Retreved on July 17,006 from MathWorld A Wolfram Web Resource. Wessten, Erc W. (004). Euler Formula. Retreved on July 17,006 from MathWorld A Wolfram Web Resource. Wessten, Erc W. (00). Trgonometry Angles P/4. Retreved on July 17, 006 from MathWorld A Wolfram Web Resource. 17