2 Geometric interpretation of complex numbers

Transcription

1 2 Geometric iterpretatio of complex umbers 2.1 Defiitio I will start fially with a precise defiitio, assumig that such mathematical object as vector space R 2 is well familiar to the studets. Recall that ay elemet of R 2 is a (colum) vector [ ] x, y where x, y are just real umbers. Defiitio 2.1. The set of complex umbers, usually deoted as C (aother stadard otatio is C, but I will stick to the former), is, by defiitio, the vector space R 2, i.e., the set of pairs of real umbers, with operatios of additio ad multiplicatio defied as follows for ay z 1, z 2 C: z 1 + z 2 = z 1 z 2 = z 1 z 2 = [ x1 y 1 [ x1 y 1 ] + ] + [ x2 y 2 [ x2 ] [ ] x1 + x = 2, y 1 + y 2 y 2 ] [ ] x1 x = 2 y 1 y 2. x 1 y 2 + y 1 x 2 The motivatio to defie the multiplicatio exactly as it is writte above comes from aive maipulatios with umbers cotaiig square root of 1 to fid the roots of a cubic polyomial, recall the previous lecture. Recall that the stadard basis of R 2 is e 1 = [1 0], e 2 = [0 1], which literally meas that ay vector from R 2 (ad hece from C) ca be represeted i a uique way as a liear combiatio [ ] x z = = xe y 1 + ye 2. Now ote that usig the multiplicatio defied above for ay z I have e 1 z = ze 1 = z, so with a slight abuse of otatio I will deote e 1 as 1 emphasizig that it is my uit. Moreover, [ ] [ ] [ ] [ ] e 2 e 2 = e = = = = e = 1, therefore, if I itroduce the otatio i = e 2, the the last lie reads (familiar, is t it?) ad hece the usual expressio i 2 = 1, z = x + iy ca (ad should) be uderstood as a represetatio of z C with respect to the stadard basis {1, i}. No imagiary quatities aymore! Ayway, usig symbol i saves so much space ad time (istead of writig vectors) ad makes the computatios so much easier that we will always use this otatio, Math 52: Complex Aalysis by Artem Novozhilov c artem.ovozhilov@dsu.edu. Sprig

2 but a studet should always remember that behid it there are always a couple of our very familiar vectors that form the stadard basis of R 2. The importat thig is that itroduced i the way how it is writte above operatios make the set C ito a field (see the textbook or ay other source if this term is ufamiliar). I will leave all the (rigorous ad borig) details to the textbook, see Sectio 1.1, ad oly will show how to divide complex umbers (subtractio should be absolutely obvious). For this for a complex z = x + iy I itroduce its cojugate z = x iy ad ote that z z = (x + iy)(x iy) = x 2 + y 2 is a real umber (which rigorously, agai, should be uderstood as a vector i R 2 with the secod coordiate zero, or eve better as (x 2 + y 2 )e 1 ). Now if I wat to divide two complex umbers z 1 = x 1 + iy 1, z 2 = x 2 + iy 2 I do the followig: z 1 z 2 = z 1 z 2 z 2 z 2 = 1 x y2 2 z 1 z 2 = x 1x 2 + y 1 y 2 x y2 2 + i x 2y 1 x 1 y 2 x y2 2 = a + ib, which is defied if ad oly if x y2 2 0, i.e., if ad oly if z 2 0 (i.e., it is ot a zero vector). Sice I have a divisio it meas that for ay ozero z 0 I ca fid its iverse z 1, which is defied as such a complex umber as zz 1 = z 1 z = 1. Ideed, to fid a iverse meas divide 1 by z, which leads to z 1 = 1 z = z z z = x x 2 + y 2 + i y x 2 + y 2, which also prove the uiqueess of the iverse. Havig defied the iverse, I ca easily check ow all the axioms of a field, i particular the associativity of additio ad multiplicatio, the commutativity of additio ad multiplicatio, distributivity, etc, please check the textbook for exact details. What is importat here is that sice R is a field ad C is a field, all the algebraic formulas we got used to while maipulatig expressios with real umbers stays the same if real umbers replaced with complex oes. I.e., (z 1 + z 2 ) 3 = z z 2 1z 2 + 3z 1 z z 3 2 is true for ay complex z 1, z 2. Before fially turig to the geometric iterpretatio of complex umbers I would like to state as a exercise the properties of cojugate umbers: Problem 2.1. Show that for ay z, w C z ± w = z ± w, zw = z w, ( z w ) = z w, z = z, Re z = z + z, 2 Im z = z z, 2 z R z = z. 2

3 Here I itroduced ew otatio Re z ad Im z for the real ad imagiary parts of the complex umber z = x + iy, which are defied as Re z = x, Im z = y respectively. Also, a careful studet should uderstad ow that whe I write somethig like z R for a complex z, I agai abuse the otatio ad mea that z = ae 1, for a real costat a. But this abuse does ot lead to ay mistakes, everyoe does it, ad so I will be doig the same thig (to be a little more formal: Strictly speakig the set of real umbers is ot a subset of complex umbers C, which are pairs of real umbers, but the field R is isomorphic to a subfield of C, which is the set of all vectors with the secod coordiate zero; if the last setece does ot make much sese to you, do ot worry, we wo t eed it really). Problem 2.2. Let p(z) = a z + a 1 z a 1 z + a 0, be a polyomial of complex variable z. w C is by defiitio its root if p(w) = 0. Show that if all the coefficiets a, a 1,..., a 1, a 0 are real ad if w is a root the p has aother root w. Problem 2.3. Show that for ay real matrix A complex eigevalues occur i complex cojugate pairs. 2.2 Basic geometry of complex umbers Havig at our disposal the fact that complex umbers are literally the vectors from R 2 (or sometimes it is more coveiet to look at them as the poits i the plae), we start with very basic termiology, illustratig it with geometric pictures. Figure 1: Complex plae. For each complex umber z = x + iy we have a vector [x y] ad/or a poit with (rectagular or Cartesia) coordiates (x, y). The coordiate x is called the real part of z, ad y is called the imagiary part of y, the correspodig otatios were itroduced above. Note that geometrically 3

4 cojugatio is simply a reflectio with respect to the x-axis, which is, for obvious reasos, is called the real axis (the y-axis is called the imagiary axis for the same reasos). The whole plae R 2 is called the complex plae C. The complex umbers o the real axis are idetified with usual real umbers, ad the umbers o the imagiary axis are called purely imagiary. Sice z C is geometrically a vector (see the figure), we ca always calculate its legth z, which is give, by Pythagoras theorem, z = x + iy = x 2 + y 2, ad all the square roots of real umbers i these otes should be uderstood as positive square roots if ot states otherwise. The legth of the vector z is aturally called its modulus, or absolute value. I hope that all the studets remember that for our usual absolute value x of a real umber x it is true that x + y x + y (ca you prove this iequality?). It turs out the same is true for ay complex z, w C: z + w z + w. (2.1) It is ot difficult to prove it algebraically, but much icer just to see it geometrically. For this ote that additio of complex umbers is actually our usual additio of vectors (by a triagle or parallelogram law, see Fig. 2), ad hece iequality (2.1) compares the legth of oe side of a triagle with vertices z, w, z + w with the sum of the legthes of two other sides, ad therefore becomes obvious. Figure 2: Geometric iterpretatio of complex additio ad subtractio. Problem 2.. Prove (2.1) algebraically. Sice z w = z + ( w), ad w geometrically is the vector that is obtaied reflectig w with respect to the origi, the z w is exactly the distace betwee the poits z ad w (see Fig. 2, right ad ote that from ow o I will use the oe geometric iterpretatio of complex umbers, either vectors or poits, which is more coveiet for a give situatio, without explicitly metioig it). It is actually our usual Euclidia distace betwee two poits ad of course o-egative, symmetric ad satisfies the triagle iequality z w z v + w v for ay complex z, v, w C (why the last

5 iequality is true?). I more formal words complex plae C is a metric space with distace fuctio d(z, w) = z w (eve more importatly, it is a complete metric space, which should be proved by those who discussed completeess i other classes). Oe more importat remark is to ote that z 2 = x 2 + y 2 = z z, which is ofte very helpful. There is o straightforward geometric multiplicatio of complex umbers i (rectagular) coordiates (x, y), but remember that we always ca pass to polar coordiates (r, θ) (textbook uses letter ϕ istead of θ, but there is small caveat here: Mathematicias prefer to use θ, physicists ϕ, ad it is ot a big deal while we stay o the plae, whe we move, however, to the three dimesioal world ad spherical coordiates, mathematicias call the third coordiate ϕ, physicists θ ad sice they are ot symmetric, it leads to differet formulas, so be careful). Recall that the polar coordiates of a poit A are give by the distace r of this poit from the origi ad by the agle θ of the vector OA with the polar axis (if A coicides with the origi the agle is udetermied). Here is oe subtle thig: polar coordiates are ot defied uiquely sice, e.g., agles θ ad θ + 2πk, k Z are idistiguishable. I have, usig the basic trigoometry, that for the complex umber z = x + iy, its polar coordiates are r = z = x 2 + y 2, ad the correspodig agle, which is called the argumet of the complex umber z, satisfies ta θ = y x. The argumet is usually deoted Arg z. I other words z ad Arg z are the polar coordiates r, θ of the poit with rectagular coordiates (x, y). There is oly oe argumet that satisfies the coditio π < θ π, ad I will call this (uique) argumet of z as pricipal argumet, θ = arg z (I ote that i some textbooks the otatio Arg ad arg is used i the opposite way, i my otes the first capital letter will mea that the result is multivalued). Now usig ta θ = y/x I would like to determie arg z. Sice arcta maps ay real umber to the iterval ( π/2, π/2) I must have (make a figure if you are cofused here) arcta y x, x > 0, arcta y x + π, x < 0, y 0, arcta y arg z = x π, x < 0, y < 0, π 2, x = 0, y > 0, π 2, x = 0, y < 0, idetermiate, x = y = 0. We of course have (see Fig. 1) arg z = arg z (is it true for ay z?). To go back from polar coordiates to rectagular, oe of course uses x = r cos θ, y = r si θ, 5

6 ad hece z = x + iy = r(cos θ + i si θ) = z (cos Arg z + i si Arg z). The last equality is the polar form of a complex umber z. Fially I am ready to see the geometric meaig of complex multiplicatio. complex umbers z 1 = r 1 (cos θ 1 + i si θ 1 ), z 2 = r 2 (cos θ 2 + i si θ 2 ). I have i other words z 1 z 2 = r 1 r 2 ( (cos θ1 cos θ 2 si θ 1 si θ 2 ) + i(si θ 1 cos θ 2 + si θ 2 cos θ 1 ) ) = r 1 r 2 (cos(θ 1 + θ 2 ) + i si(θ 1 + θ 2 )), z 1 z 2 = z 1 z 2, Arg z 1 z 2 = Arg z 1 + Arg z 2, Let z 1, z 2 be two where the last equality meas that the set of all possible values Arg z 1 z 2 is obtaied by formig all possible sums from the sets Arg z 1 ad Arg z 2. Oes agai i words: to multiply two complex umbers meas geometrically to obtai the vector with the modulus with is the product of two moduli, ad with the argumet which is the sum of the correspodig argumets (make a figure for, say, z = 1 ad w = i). Sice z 1 = z z 2 = w z = wz 1, ad the modulus of z 1 = 1/ z ad argumet Arg z 1 = Arg z, where the last equality uderstood modulo 2π, the z 1 = z 1 z 2, Arg z 1 = Arg z 1 Arg z 2. z Powers ad roots Sice for ay atural z 2 z = z... z ad usig the formula for complex multiplicatio i polar form derived above, I have If r = 1 the I obtai De Moivre s theorem z = ( r(cos θ + i si θ) ) = r (cos θ + i si θ). (cos θ + i si θ) = cos θ + i si θ. If I defie z 0 = 1 ad z = 1 z the De Moivre s theorem becomes true for ay iteger Z. I will use this formula to solve the equatio z = 1, i other words I wat to determie all possible -th roots of uity 1. Sice 1 = 1 ad Arg 1 = 2πk for k Z, I have, assumig that z = r(cos θ + i si θ), r (cos θ + i si θ) = cos 2πk + i si 2πk, 6

7 which implies that r = 1 = r = 1, ad θ = 2πk, k Z. How may really differet θ did I get? Note that for k = 0, 1,..., 1 all θ will be betwee 0 ad 2π, but for k = I get θ = 2π, which correspods to the same polar agle as θ = 0. Therefore I obtaied Propositio 2.2. Equatio z = 1 has always distict roots ẑ k = 1 = cos 2πk which are called the -th roots of uity. I exactly the same way I ca solve the equatio 2πk + i si, k = 0, 1,..., 1, z = w for a give complex w = ρ(cos φ + i si φ). Fill i the missed details to covice yourself that i this case ( z k = w = ρ cos φ + 2πk + i si φ + 2πk ), k = 0,..., 1, where φ is ay argumet of w. Usig the formula for the multiplicatio of complex umbers it follows that I ca write z k = w = ( ρ cos arg w + si arg w ) ẑ k, k = 0,..., 1. Problem 2.5. Show that i geeral the equalities arg zw = arg z + arg w, arg z w = arg z arg w are icorrect. Problem 2.6. Show that the sum of all the -th roots of uity is always zero ( 1). What geometric fact does it express? Problem 2.7. I the first lecture I have derived Cardao s formula to solve a depressed cubic polyomial. I told you that this formula allows to fid oe real root of it, ad two others ca be foud from the correspodig quadratic equatio. Actually, usig the roots of uity the formula that I derived ca be modified to give all three (i geeral complex) roots. Ca you fill i the ecessary details? 2. Examples with solutios Let us see how the material I discussed so far i this lecture ca be used to solve cocrete examples. 7

8 Example 2.3. What is the polar form of z = 1 i 3? I have z = ( 3) 2 = 2, ta θ = ta arg z = 3 1 = 3 = arg z = 2π 3, hece 1 i ( ( 3 = 2 cos 2π 3 Example 2.. What is the modulus ad argumet of z = si π 8 i cos π 8? ) ( + i si 2π )). 3 The modulus is easy sice z = si 2 π 8 + cos2 π 8 = 1. For the argumet we have hece arg z = π + arcta si π 8 cos π 8 Example 2.5. Compute ( 1 + i 3) 60. First = π + arcta cot π ( π 8 = π + arcta ta 2 π ) = 5π 8 8, Arg z = 5π i 3 = 2 + 2πk, k Z. ( cos 2π 3 + i si 2π 3 Hece by De Moivre s formula ( 1 + i ( 3) 60 = 2 60 cos 2π 3 + i si 2π ) 60 ( = 2 60 cos 60 2π i si 602π 3 Example 2.6. Fid a formula for si 3θ. By De Moivre s formula (cos θ + i si θ) 3 = cos 3θ + i si 3θ. Raisig to the third power the left had side I get (cos θ + i si θ) 3 = cos 3 θ + 3i cos 2 θ si θ 3 cos θ si 2 θ i si 3 θ ). ) = ad recallig that two complex umbers are equal if ad oly if their real ad imagiary parts are equal simultaeously, I coclude si 3θ = 3 cos 2 θ si θ si 3 θ = 3 si θ si 3 θ. Example 2.7. Compute all 1 i. I have 1 i = ( ( 2 cos π ) ( + i si π )), hece by the formula I have I get 1 i = 8 2 ( π/ + 2πk cos + i si ) π/ + 2πk, k = 0, 1, 2, 3. 8