C. Complex Numbers. x 6x + 2 = 0. This equation was known to have three real roots, given by simple combinations of the expressions

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1 C. Complex Numbers. Complex arithmetic. Most people thik that complex umbers arose from attempts to solve quadratic equatios, but actually it was i coectio with cubic equatios they first appeared. Everyoe kew that certai quadratic equatios, like x 2 + = 0, or x 2 + 2x + 5 = 0, had o solutios. The problem was with certai cubic equatios, for example x 6x + 2 = 0. This equatio was kow to have three real roots, give by simple combiatios of the expressios () A = + 7, B = 7; oe of the roots for istace is A + B: it may ot look like a real umber, but it turs out to be oe. What was to be made of the expressios A ad B? They were viewed as some sort of imagiary umbers which had o meaig i themselves, but which were useful as itermediate steps i calculatios that would ultimately lead to the real umbers you were lookig for (such as A + B). This poit of view persisted for several hudred years. But as more ad more applicatios for these imagiary umbers were foud, they gradually bega to be accepted as valid umbers i their ow right, eve though they did ot measure the legth of ay lie segmet. Nowadays we are fairly geerous i the use of the word umber : umbers of oe sort or aother do t have to measure aythig, but to merit the ame they must belog to a system i which some type of additio, subtractio, multiplicatio, ad divisio is possible, ad where these operatios obey those laws of arithmetic oe lears i elemetary school ad has usually forgotte by high school the commutative, associative, ad distributive laws. To describe the complex umbers, we use a formal symbol i represetig ; the a complex umber is a expressio of the form (2) a + ib, a, b real umbers. If a = 0 or b = 0, they are omitted (uless both are 0); thus we write a + i0 = a, 0 + ib = ib, 0 + i0 = 0. The defiitio of equality betwee two complex umbers is () a + ib = c + id a = c, b = d. This shows that the umbers a ad b are uiquely determied oce the complex umber a + ib is give; we call them respectively the real ad imagiary parts of a + ib. (It would be more logical to call ib the imagiary part, but this would be less coveiet.) I symbols, (4) a = Re (a + ib), b = Im (a + ib)

2 2 8.0 NOTES Additio ad multiplicatio of complex umbers are defied i the familiar way, makig use of the fact that i 2 = : (5a) Additio (a + ib) + (c + id) = (a + c) + i(b + d) (5b) Multiplicatio (a + ib)(c + id) = (ac bd) + i(ad + bc) Divisio is a little more complicated; what is importat is ot so much the fial formula but rather the procedure which produces it; assumig c + id = 0, it is: a + ib ac + bd bc ad (5c) Divisio = a + ib c id = + i c + id c + id c id c 2 + d 2 c 2 + d 2 This divisio procedure made use of complex cojugatio: if z = a + ib, we defie the complex cojugate of z to be the complex umber (6) z = a ib (ote that zz = a 2 + b 2 ). The size of a complex umber is measured by its absolute value, or modulus, defied by (7) z = a + ib = a 2 + b 2 ; (thus : zz = z 2 ). Remarks. Oe ca legitimately object to defiig complex umbers simply as formal expressios a + ib, o the grouds that formal expressio is too vague a cocept: eve if people ca hadle it, computers caot. For the latter s sake, we therefore defie a complex umber to be simply a ordered pair (a, b) of real umbers. With this defiitio, the arithmetic laws are the defied i terms of ordered pairs; i particular, multiplicatio is defied by (a, b)(c, d) = (ac bd, bc + ad). The disadvatage of this approach is that this defiitio of multiplicatio seems to make little sese. This does t bother computers, who do what they are told, but people do better at multiplicatio by beig told to calculate as usual, but to use the relatio i 2 = to get rid of all the higher powers of i wheever they occur. Of course, eve if you start with the defiitio usig ordered pairs, you ca still itroduce the special symbol i to represet the ordered pair (0, ), agree to the abbreviatio (a, 0) = a, ad thus write (a, b) = (a, 0) + (0, )(b, 0) = a + ib. 2. Polar represetatio. Complex umbers are represeted geometrically by poits i the plae: the umber a+ib is represeted by the poit (a, b) i Cartesia coordiates. Whe the poits of the plae are thought of as represetig complex umbers i this way, the plae is called the complex plae. By switchig to polar coordiates, we ca write ay o-zero complex umber i a alterative form. Lettig as usual x = r cos α, y = r si α, we get the polar form for a o-zero complex umber: assumig x + iy = 0, (8) x + iy = r(cos α + i si α).

3 C. COMPLEX NUMBERS Whe the complex umber is writte i polar form, we see from (7) that r = x + iy, (absolute value, modulus) We call α the polar agle or the argumet of x + iy. I symbols, oe sometimes sees α = arg (x + iy) (polar agle, argumet). The absolute value is uiquely determied by x + iy, but the polar agle is ot, sice it ca be icreased by ay iteger multiple of 2λ. (The complex umber 0 has o polar agle.) To make α uique, oe ca specify 0 α < 2λ pricipal value of the polar agle. This so-called pricipal value of the agle is sometimes idicated by writig Arg (x + iy). For example, Arg ( ) = λ, arg ( ) = ±λ, ±λ, ±5λ,.... Chagig betwee Cartesia ad polar represetatio of a complex umber is the same as chagig betwee Cartesia ad polar coordiates. Example. Give the polar form for: i, + i, i, + i. Solutio. i = i cos π 2 + i = 2 (cos 2π + i si 2 + i = 2 (cos π + i si π 4 4 ) π ) i = 2 (cos π 4 + i si π ) 4 The abbreviatio cis α is sometimes used for cos α + i si α; for studets of sciece ad egieerig, however, it is importat to get used to the expoetial form for this expressio: (9) e iθ = cos α + i si α Euler s formula. Equatio (9) should be regarded as the defiitio of the expoetial of a imagiary power. A good justificatio for it however is foud i the ifiite series t t 2 t e t = ! 2!! If we substitute iα for t i the series, ad collect the real ad imagiary parts of the sum (rememberig that ad so o, we get i 2 =, i = i, i 4 =, i 5 = i,..., iθ e = ) ) α 2 α 4 α α i α ! 4!! 5! = cos α + i si α, i view of the ifiite series represetatios for cos α ad si α. Sice we oly kow that the series expasio for e t is valid whe t is a real umber, the above argumet is oly suggestive it is ot a proof of (9). What it shows is that Euler s formula (9) is formally compatible with the series expasios for the expoetial, sie, ad cosie fuctios. Usig the complex expoetial, the polar represetatio (8) is writte as (0) x + iy = r e iθ

4 4 8.0 NOTES The most importat reaso for polar represetatio is that multiplicatio of complex umbers is particularly simple whe they are writte i polar form. Ideed, by usig (9) ad the trigoometric additio formulas, it is ot hard to show that iθ iθ i(θ+θ ) e e = e. This gives aother justificatio for defiitio (9) it makes the complex expoetial follow the same expoetial additio law as the real expoetial. Thus we ca multiply two complex umbers i polar form by () r e iθ r e iθ = r r e i(θ+θ ) ; multiplicatio rule to multiply two complex umbers, you multiply the absolute values ad add the agles. By repeated applicatio of this, we get the rule (sometimes called DeMoivre s formula for raisig a complex umber to a positive iteger power: usig the otatio of (0), ( ) iθ (2) r e iθ = r e ; i particular, (cos α + i si α) = cos α + i si α. Example 2. Express ( + i) 6 i the form a + bi. Solutio. We chage to the polar form, use (2), the chage back to Cartesia form: ( + i) 6 = ( 2 e iπ/4 ) 6 = ( 2) 6 e i 6π/4 = 8 e i π/2 = 8i. The aswer may be checked by applyig the biomial theorem to ( + i) 6 ad collectig the real ad imagiary parts. Divisio of complex umbers writte i polar form is doe by the rule (check it by crossmultiplyig ad usig the multiplicatio rule): r e iθ r = e i (θ θ ) ; divisio rule r e iθ r to divide by a complex umber, divide by its absolute value ad subtract its agle. Combiig pure oscillatios of the same frequecy. The equatio which does this is widely used i physics ad egieerig; it ca be expressed usig complex umbers: () A cos λt + B si λt = C cos (λt + ζ), where A + Bi = Ce iλ ; i other words, C = A 2 + B 2, ζ = ta B/A. To prove (), we have ( ) A cos λt + B si λt = Re (A + Bi) (cos λt + i si λt). Complex expoetials iλt = Re (C e iλ e ) = Re (C e λt+λ ) = C cos (λt + ζ). Because of the importace of complex expoetials i differetial equatios, ad i sciece ad egieerig geerally, we go a little further with them. Euler s formula (9) defies the expoetial to a pure imagiary power. The defiitio of a expoetial to a arbitrary complex power is: a ib (4) e a+ib = e e = e (cos b + i si b). We stress that the equatio (4) is a defiitio, ot a self-evidet truth, sice up to ow o meaig has bee assiged to the left-had side. From (4) we see that a (5) Re (e a+ib ) = e cos b, Im (e a+ib ) = e a si b. a

5 C. COMPLEX NUMBERS 5 The complex expoetial obeys the usual law of expoets: z z (6) e z+z = e e, as is easily see by combiig (4) ad (). The complex expoetial is expressed i terms of the sie ad cosie by Euler s formula (9). Coversely, the si ad cos fuctios ca be expressed i terms of complex expoetials. There are two importat ways of doig this, both of which you should lear: (7) cos x = Re (e ix ), si x = Im (e ix ) ; ix ), 2 (8) cos x = i (eix e ix 2 (e ix + e si x = ). The equatios i (8) follow easily from Euler s formula (9); their derivatio is left for the exercises. Here are some examples of their use. Example. Express cos x i terms of the fuctios cos x, for suitable. Solutio. We use (8) ad the biomial theorem, the (8) agai: cos x = ix ) 8 (e ix + e ix = + e ix 8 (e ix + e ix + e ) = cos x + cos x. 4 4 As a prelimiary to the ext example, we ote that a fuctio like ad its derivative ad itegral with respect to x is defied to be (9) D(u + iv) = Du + idv, (u + iv) dx = u dx + i v dx. From this it follows by a calculatio that (20) D(e (a+ib)x = (a + ib)e (a+ib)x, ad therefore e (a+ib)x dx = e (a+ib)x. a + ib Example 4. Calculate e x cos 2x dx by usig complex expoetials. Solutio. The usual method is a tricky use of two successive itegratio by parts. Usig complex expoetials istead, the calculatio is straightforward. We have ( ) (+2i)x e x cos 2x = Re e, by (4) or (5); therefore ) e x cos 2x dx = Re e (+2i)x dx, by (9). Calculatig the itegral, e (+2i)x dx = e (+2i)x by (20); + 2i ) 2 ( ) = i e x cos 2x + i e x si 2x, 5 5 ix e = cos x + i si x is a complex-valued fuctio of the real variable x. Such a fuctio may be writte as u(x) + i v(x), u, v real-valued usig (4) ad complex divisio (5c). Accordig to the secod lie above, we wat the real part of this last expressio. Multiply usig (5b) ad take the real part; you get e x cos 2x + 2 e x si 2x. 5 5

6 6 8.0 NOTES I this differetial equatios course, we will make free use of complex expoetials i solvig differetial equatios, ad i doig formal calculatios like the oes above. This is stadard practice i sciece ad egieerig, ad you eed to get used to it. 4. Fidig -th roots. To solve liear differetial equatios with costat coefficiets, you eed to be able fid the real ad complex roots of polyomial equatios. Though a lot of this is doe today with calculators ad computers, oe still has to kow how to do a importat special case by had: fidig the roots of z = α, where α is a complex umber, i.e., fidig the -th roots of α. Polar represetatio will be a big help i this. Let s begi with a special case: the -th roots of uity: the solutios to z =. To solve this equatio, we use polar represetatio for both sides, settig z = re iθ o the left, ad usig all possible polar agles o the right; usig the expoetial law to multiply, the above equatio the becomes r e iθ = e (2kπi), k = 0, ±, ±2,.... Equatig the absolute values ad the polar agles of the two sides gives r =, α = 2kλ, k = 0, ±, ±2,..., from which we coclude that 2kλ ( ) r =, α =, k = 0,,...,. I the above, we get oly the value r =, sice r must be real ad o-egative. We do t eed ay iteger values of k other tha 0,..., sice they would ot produce a complex umber differet from the above umbers. That is, if we add a, a iteger multiple of, to k, we get the same complex umber: α = 2(k + a)λ = α + 2aλ; ad e iθ = e iθ, sice e 2aπi = (e 2πi ) a =. We coclude from ( ) therefore that (2) the -th roots of are the umbers e 2kπi/, k = 0,...,. 2 π i e πi e This shows there are complex -th roots of uity. They all lie o the uit circle i the complex plae, sice they have absolute value ; they are evely spaced aroud the uit circle, startig with ; the agle betwee two cosecutive oes is 2λ/. These facts 4 π i π i are illustrated o the right for the case = 6. e e 5

7 C. COMPLEX NUMBERS 7 From (2), we get aother otatio for the roots of uity (ω is the Greek letter zeta ): (22) the -th roots of are, ω, ω 2,..., ω, where ω = e 2πi/. We ow geeralize the above to fid the -th roots of a arbitrary complex umber w. We begi by writig w i polar form: iθ w = r e ; α = Arg w, 0 α < 2λ, i.e., α is the pricipal value of the polar agle of w. The the same reasoig as we used above shows that if z is a -th root of w, the iθ i(θ+2kπ)/ (2) z = w = r e, so z = r e, k = 0,,...,. Comparig this with (22), we see that these roots ca be writte i the suggestive form (24) w = z r e iθ/ 0, z 0 ω, z 0 ω 2,..., z 0 ω, where z 0 =. As a check, we see that all of the complex umbers i (24) satisfy z = w : (z 0 ω i 0 ω i ) = z = z 0 i, sice ω =, by (22); = w, by the defiitio (24) of z 0 ad (2). Example 5. Fid i Cartesia form all values of a) b) 4 i. Solutio. a) Accordig to (22), the cube roots of are, ω, ad ω 2, where 2λ 2λ ω = e 2πi/ = cos + i si = + i 2 2 2πi/ ω 2 = e = cos 2λ + i si 2λ = i. 2 2 The greek letter ω ( omega ) is traditioally used for this cube root. Note that for the polar agle of ω 2 we used 2λ/ rather tha the equivalet agle 4λ/, i order to take advatage of the idetities cos( x) = cos x, si( x) = si x. Note that ω 2 = ω. Aother way to do this problem would be to draw the positio of ω 2 ad ω o the uit circle, ad use geometry to figure out their coordiates. b) To fid 4 i, we ca use (24). We kow that 4 =, i,, i (either by drawig the uit circle picture, or by usig (22)). Therefore by (24), we get 4 λ i = z 0, z 0 i, z 0, z 0 i, where z 0 = e πi/8 = cos λ + i si ; 8 8 λ λ = a + ib, b + ia, a ib, b ia, where z 0 = a + ib = cos + i si. 8 8

8 8 8.0 NOTES Example 6. Solve the equatio x 6 2x + 2 = 0. Solutio. Treatig this as a quadratic equatio i x, we solve the quadratic by usig the quadratic formula, the two roots are + i ad i (check this!), so the roots of the origial equatio satisfy either x = + i, or x = i. This reduces the problem to fidig the cube roots of the two complex umbers ± i. We begi by writig them i polar form: πi/4 πi/4 + i = 2 e, i = 2 e. (Oce agai, ote the use of the egative polar agle for i, which is more coveiet for calculatios.) The three cube roots of the first of these are (by (2)), 6 2 e πi/2 = 6 2 cos ) λ λ + i si 2 2 ) 6 2 e πi/4 = 6 2 cos λ λ + i si, sice 4 4 ) 6 2 e 7πi/2 = 6 2 cos 7λ 7λ i si, sice 2 2 ) + i + i The secod cube root ca also be writte as 6 2 =. 2 2 λ 2 + 2λ = λ 4 ; λ 2 2λ = 7λ 2. This gives three of the cube roots. The other three are the cube roots of i, which may be foud by replacig i by i everywhere above (i.e., takig the complex cojugate). The cube roots ca also accordig to (24) be described as πi/2 πi/2 z, z ω, z ω 2 ad z 2, z 2 ω, z 2 ω 2, where z = 6 2 e, z 2 = 6 2 e. Exercises: Sectio 2E