MATH1035: Workbook Four M. Daws, 2009

MATH1035: Workbook Four M. Daws, 2009 Roots of uity A importat result which ca be proved by iductio is: De Moivre s theorem atural umber case: Let θ R ad N. The cosθ + i siθ = cosθ + i siθ. Proof: The result is obviously true whe = 1. We proceed by iductio, so suppose the claim is true for. The cosθ + i siθ +1 = cosθ + i siθ cosθ + i siθ by the iductio hypothesis, = cosθ cosθ siθ siθ + i cosθ siθ + siθ cosθ = cos + 1θ + i si + 1θ. So the result is true by mathematical iductio. Usig the complex expoetial otatio, De Moivre s theorem says simply that e iθ = e iθ for N, θ R. This looks obvious! However, it s ot quite: we have to do a little work. Remember that we have previously proved that e iθ e iϕ = e iθ+ϕ ideed, the proof was rather similar to the above, usig trigoometric double agle formulae: this is ot a coicidece! So, if e iθ = e iθ which is true if = 1 the e iθ +1 = e iθ e iθ = e i+1θ. So by iductio, the expoetial form of De Moivre s theorem is true this was so simple, you almost did t eed to use iductio. We ca prove a better result: De Moivre s theorem: Let θ R ad Z. The cosθ + i siθ = cosθ + i siθ. Proof: Whe 1 we have already proved this. Whe = 0, the as z 0 = 1 for ay o-zero 1 z C, ad cos0 + i si0 = 1, we have that the result holds whe = 0. Whe < 0, let m = N, so that cosθ + i siθ = cosθ + i siθ m = 1 cosθ + i siθ m 1 = by De Moivre for m, cosmθ + i simθ cosmθ i simθ = cos 2 mθ + si 2 = cosmθ i simθ mθ = cosθ + i siθ, which completes the proof. 1 Why is this true? To be really precise, this is actually a defiitio. 1

Agai, this is very easy i the expoetial form, we have already show that for ay θ R, we have 1/e iθ = e iθ. I ll leave it as a exercise to writeup the above proof usig expoetial otatio. De Moivre s theorem allows us to solve equatios of the form z = 1, where Z. Notice that for θ R, if θ = 2kπ for some k Z, the cosθ + i siθ = cosθ + i siθ = cos2kπ + i si2kπ = 1. Thus, whe 0, some solutios to z = 1 are 2kπ 2kπ z = cos + i si for k = 0, 1,, 1. Notice that there is o poit cosiderig k =, for the 2kπ/ = 2π ad the cos2π = cos0 ad si2π = si0, so we just get the same aswer as for k = 0. Similarly, if we try ay other k Z, we ll get the same result as for some 2 umber i the set {0, 1,, 1}. Claim: Let N, 0. The the solutios to z = 1 are z = cos 2kπ + i si 2kπ for k = 0, 1,, 1. Proof: We showed that these are solutios, so we eed to show that they are the oly solutios. Let z = 1. Thus z = z = 1 why? Agai, techically, this is a proof by iductio, but it s so simple I ll skip it. So z = 1, as z is a positive real, ad the oly solutio to t = 1 for t R with t > 0, is t = 1. We showed before that if z = 1, the z = cosθ + i siθ for some θ R. Ideed, we ca choose θ i the rage [0, 2π. So by De Moivre, 1 = z = cosθ + i siθ. So siθ = 0, which meas that θ is a iteger multiple of π. Similarly, cosθ = 1, which meas that θ is a iteger multiple of 2π. So θ = 2kπ for some k Z, that is, θ = 2kπ/. As 0 θ < 2π, we have 0 2kπ < 2π = 0 2kπ < 2π = 0 k <, usig that > 0. So k is i the set {0, 1,, 1} as required. If < 0, the z = 1 if ad oly if 1 = 1/z = z, ad the we ca apply the above theorem for. We call solutios to z = 1 the th roots of uity. This is a slightly old-fashioed phrase: uity meas 1, ad the we take the th root of 1. Questio: What are the complex solutios to z 2 = 1 + i? Aswer: Write z i polar form as z = rcosθ + i siθ, so by De Moivre, z 2 = r 2 cos2θ + i si2θ. Hece z 2 = 1 + i is equivalet to r 2 = 1 + i ad 2θ beig the argumet of 1 + i. Now, 1 + i = 1 + 1 = 2 r = 2 1/4. The Pricipal Argumet of 1 + i is π/4, ad so the argumet of 1 + i is π/4 or 2π + π/4, ad so forth. Hece θ = π/8 or π + π/8 you ca check that ay other choice would just be oe of these added to a multiple of 2π. So z 2 = 1 + i z = 2 1/4 cosπ/8 + i siπ/8 or z = 2 1/4 cosπ + π/8 + i siπ + π/8. Of course, the secod solutio is just the egative of the first, as we were solvig a quadratic equatio i this specific case. 2 You might like to thik: which umber? 2

Quatifiers: ad You will have see phrases like there exists or for all quite a bit by ow. Just as we write to mea implies, we write to mea for all : remember this as upside dow A for all. We write to mea there exists : remember this as backwards E for there exists. These are called quatifiers. We ca ow traslate phrases i Eglish ito symbols. A example: x R [x 2 0] meas For all real umbers x, we have that x 2 0. We use the brackets simply to make the result easier to read. Just traslatig, you might have come up with For all x R, x 2 0, but this is still hard to read, so we traslate a bit more to the versio I have. Notice that this statemet is true. Some further examples: a R [a 2 = 2] which meas There is a real umber a, with a 2 = 2. We ca make this eve clearer: The square-root of 2 is a real umber. x C y C [xy = yx] meas: For all complex umbers x ad y, we have xy = yx. We ca say this is a simpler way: Multiplicatio of complex umbers is commutative. θ R Z [cosθ + i siθ = cosθ + i siθ]. This is just De Moivre s theorem! Thigs get a bit more complicated whe we combie both ad. For example, r R s R t C [rs = t 2 ]. This true statemet meas that for all r, s R, there exists t C with t 2 = rs. You could also say: For ay two real umbers r, s R, there is a square-root of rs i C. Fix a fuctio f : X Y. The cosider: y Y x X [fx = y] This meas: For each y Y there is some x X with fx = y. This is the defiitio of what it meas for f to be surjective! Notice that this statemet could be true or false, depedig upo what f is. Cosider ow y Z x Z [x < y]. So we traslate: There exists a iteger y, for all itegers x, we have that x is less tha y. If we thik about this for a little while, we see that this is equivalet to There exists a iteger y such that all itegers are less tha y. This is false of course: as y + 1 is ot less tha y. Let us thik about this a bit further: we caot fid a sigle couter-example: istead, we have to fid a ew couter-example for each y. But this is easy, for give ay y Z, we ca fid x Z with x y; just let x = y + 1 or x = y or x = y + 100 or whatever. If we swap the order aroud, ad cosider x Z y Z [x < y], the we do get a true statemet. If we traslate, we get For all itegers x, there exists a iteger y with y bigger tha x. This is true, because if I give you x, the you ca certaily fid some y with x < y, for example, y = x + 1 would work. Notice how I phrased this as a game or challege : I thik this is a good way to thik about such problems. 3

Agai, let f : X Y be a fuctio, ad cosider: x X y X [x y = fx fy]. This meas: For ay choice of x ad y i X, if x y the fx fy. This is the defiitio of what it meas for f to be ijective: differet elemets of X get mapped by f to differet elemets of Y. We saw before that this is equivalet to: x X y X [fx = fy = x = y]. This 2d coditio is easier to check. Notice that I used the implies symbol, but whe I traslated, I used the if... the... form, which I thik is easier to uderstad. We ca go the other way, ad covert mathematics statemets which are i Eglish ito statemets i symbols. We ca also work with geeral sets. Some examples: Let A = {1, 2, 3, 4, 5, 6}. The there is a atural umber greater tha all elemets of A ca be traslated to x N y A[x > y]. We ca traslate Every atural umber is either eve or odd as x N N[x = 2 or x = 2 1]. How did I do this? Well, a atural umber is eve if it s of the form 2 for some N; a atural is odd if it s of the form 2 1. Let B = 1, 123 R. The the square of ay umber x i B is greater tha x is x B[x 2 > x]. Fially, a example of a false statemet. There is a greatest real umber smaller tha 1. Let s defie a set C =, 1, so C is the collectio of all real umbers less tha 1. The our statemet is that C has a greatest elemet. This ca ow be writte as x C y C[x y]. Fially, why is this false? Equivalece relatios We wat to abstract the idea of equality, which will ultimately lead us to a rigourous defiitio of the ratioal umbers. A relatio o a set X is a very geeral idea, which we itroduce by givig some examples: 1. Numbers x ad y are related if ad oly if x = y. 2. A iteger x is related to a iteger y if ad oly if x divides y. 3. A perso p is related to a perso q if ad oly if p is a brother or sister of q. 4. Complex umbers w ad z are related if ad oly if w z < 1. We deote a relatio by R, ad for x, y X, we write xry if x is related to y. For example, we ca re-write the above as: 1. O R or C or... xry if ad oly if x = y. 2. O Z, defie xry if ad oly if x divides y. 3. O the set of people, defie prq if ad oly if p is a brother or sister of q. 4. O C, defie wrz if ad oly if w z < 1. 4

There are some commo properties of relatios. Let R be a relatio o X. The: R is reflexive if xrx for all x X. R is symmetric if xry implies that yrx. R is trasitive if, wheever xry ad yrz, the xrz as well. We ca write these usig quatifiers: R is reflexive if x X[xRx]. R is symmetric if x X y X[xRy = yrx]. R is trasitive if x X y X z X[xRy ad yrz = xrz]. Let us cosider our examples: 1. xry x = y. This is reflexive, as xrx x = x which is always true. This is symmetric, as if xry the x = y so also y = x so yrx. This is trasitive, as if xry ad yrz, the x = y ad y = z, so also x = y = z so x = z so xrz. 2. O Z, xry x divides y. This is reflexive, because for ay x R, certaily x divides x, so xrx holds. This is ot symmetric: a couter-example is that 5R10, because 5 divides 10, but 10 R5, because 10 does ot divide 5. As usual, a sigle couter-example is eough! This is trasitive, for if xry ad yrz, the x divides y ad y divides z. That meas I ca fid k, l Z with y/x = k ad z/y = l. But the z/x = ly/x = lk which is a iteger, showig that x divides z, that is, xrz. 3. prq if ad oly if p is a brother or sister of q. This is ot reflexive, because a perso is ot a siblig of themselves! This is symmetric: if prq the p is a sister or brother of q, so that q is a sister or brother of p, so qrp. This is also trasitive, for if prq ad qrs, the p ad q are sibligs, ad also q ad s are sibligs, so p ad s are also sibligs, so prs. 4. O C, defie wrz if ad oly if w z < 1. This is reflexive, as for ay w C, we have that w w = 0 < 1, so wrw. This is also clearly symmetric: if wrz the w z < 1 so also z w < 1 so zrw. It is ot trasitive: a couter-example is foud by settig x = 0, y = 4i/5 ad z = 4i/5 4/5. The x y = 0 4i/5 = 4/5 ad y z = 4i/5 4i/5 + 4/5 = 4/5 so xry ad yrz. However, x z = 0 4i/5 + 4/5 = 32/25 > 1 so x Rz. Defiitio: A equivalece relatio is a relatio which is reflexive, symmetric ad trasitive. We ted to write istead of R for a equivalece relatio. Let s do a log worked example. Claim: O C, z w if ad oly if e z = e w defies a equivalece relatio. 5

Proof: This is reflexive, as e z = e z for ay z, so z z. This is symmetric, for z w = e z = e w = e w = e z = w z. Similarly, this is trasitive, for if z w ad w x the e z = e w ad e w = e x so e z = e x, so z x. Suppose we fix z = a + bi C. Which w C satisfy z w? Well, let w = c + di so z w e z = e w e a e bi = e c e di e a = e c ad e bi = e di. Takig the logarithm, we see that e a = e c if ad oly if a = c. However, e bi = e di e di bi = e 0 = 1 cosd b = 1 ad sid b = 0. This is equivalet to d b beig a iteger multiple of 2π. That is, there exists k Z with d b = 2πk. We have show: Claim: For z = a + bi C fixed, we have that w = c + di is such that z w if ad oly if c = a ad d = b + 2πk for some k Z. However, if c = a ad d = b + 2πk the w = c + di = a + bi + 2πki = z + 2πki. So really we have: Claim: For z C fixed, we have that z w if ad oly if w = z + 2πki for some k Z. Or, i symbols, z w k Z[w = z + 2πki]. Ideas for the tutorial Here are some ideas for thigs which might be iterestig to discuss; aswers will be put o the VLE. yit might be a good idea to draw the solutios to z 5 = 1 o the Argad diagram. Do similar, or maybe slightly more complicated, examples. ya typical questio usig De Moivre is: Express si θ 3 usig siθ ad si3θ. Hit: We have two ways of workig out cos θ + i si θ 3, amely, De Moivre, ad just expadig out. The look at the imagiary parts. ythik about the fial example i the quatifiers sectio: amely, that There is a great real umber smaller tha 1 is false. yhere is a proof that a relatio R is reflexive if it s both symmetric ad trasitive. Ideed, suppose xry, so as R is symmetric, we have that yrx. Thus xry ad yrx, so as trasitive, also xrx. Thus R is reflexive. Why is this ot true? Thik: fid a couter-example! What, precisely was my mistake? yyou probably kow that a quadratic equatio has at most 2 roots or solutios. A cubic has a most three roots. A polyomial of degree meaig the highest power ivolved is has a most roots. If we believe this, the the argumet o the middle of page 2 is made a lot easier! However, why does a degree polyomial have at most roots? 6

Problem Set 4 Due i at the lecture o Wedesday 17 November. I the followig, I ll ofte write short proof. You should ow be startig to do some work i rough, ad the be writig up eat fial aswers to had i. By short proof, I mea a icely writte, short ad to-the-poit proof: similar to the way I wrote proofs out above. It s a learig process to decide how much detail you eed to give i a proof: ask i tutorials if you are usure if your proofs are too short or too log. 1. For the complex umber questios, there are electroic resources o the VLE which will help you. a Write the followig, i the simplest way possible, i the form a + bi. i. cos3π/28 + i si3π/28 7 ii. cos2π/3 i si2π/3 11 iii. 3 + i 201. Hit: Covert to the form rcosθ + i siθ. b Fid the solutios, with z C, to the followig equatios see the example at the ed of page 2. i. z 5 = 1 ii. z 8 = 1 iii. z 6 2 + i = 0. Hit: Covert 2 + i ito polar form. 2. a Covert the followig ito words. Try, if possible, to fid the icest form, ot just For all..., there exists.... I each case, decide if the result is true ad if so, write a short proof: some thigs are so obvious they do t eed a proof! or false ad if so, give a couter-example. i. x R y R [x 2 = y]. ii. y R x R [x 2 = y]. iii. y N x N [x 2 = y]. iv. x R [x 2 + 3x + 2 = 0]. v. x N [x 2 + 3x + 2 = 0]. vi. x N y C [yy = x]. b Covert the followig ito symbols. i. There exist two prime umbers whose sum is eve. You might wat to defie a set, P = {prime umbers}. ii. For each atural umber, there is a prime umber bigger tha. iii. There exists a ratioal umber greater tha 5. iv. For ay real umbers a ad b, there exists a solutio i C to the equatio x 2 + ax + b = 0. c Let A = {1, 2, 3} ad B = [0,. For each of the followig, prove that the statemet is true, or give a couter-example. You do t have to traslate the symbols ito words, but obviously this might help you! i. x A y Q [y > x]. ii. x B y Q [y > x]. iii. y Z x A [y > x]. iv. y Z x B [y > x]. v. x A [x 2 x]. 3. For the followig relatios, decide if they are reflexive or ot, symmetric or ot, ad trasitive or ot. You do ot eed to give a proof, but do give a couter-example if you thik the property does ot hold. 7

a O R, let xry if ad oly if x y N. b O R, let xry if ad oly if Z[y = 10 x]. c O Q Q, let x 1, x 2 Ry 1, y 2 if ad oly if x 1 = y 2. 4. The followig are equivalece relatios. I each case, give a short proof that we do have a equivalece relatio. Fix some x, ad the fid a simple coditio o y which tells us whe x y. For example, above we showed that z w e z = e w is a equivalece relatio. The z w is equivalet to w = z + 2πki for some k Z. a O C, defie x y if ad oly if Rex = Rey. b O Z, defie x y if ad oly if 10 divides x y. Agai, it might help to re-write this as x y k Z[x y = 10k]. c Let Q = {x 1, x 2 Z Z : x 2 0} i words, Q is the set of ordered pairs i Z Z with a o-zero secod co-ordiate. O Q, defie x 1, x 2 y 1, y 2 if ad oly if x 1 y 2 = x 2 y 1. Optioal questios for further practise 1. Fid the solutios, with z C, to the followig equatios. a z 2 + 2z + 4 = 0 b z 8 + 2z 4 + 4 = 0 2. Covert the followig ito words. Are they true or false? a x R y R[x + y = 1]. b y R x R[x + y = 1]. c x N y N[x + y = 1]. 3. The followig are equivalece relatios. Prove this; the fix x ad fid a simple descriptio of those y with x y. a Let X ad Y be sets, ad f : X Y a fuctio. For x, y X defie x y if ad oly if fx = fy. b O C, defie x y if ad oly if x = y. c For x R, let x be the biggest iteger less tha or equal to x. So 1 = 1, π = 3, 17 = 4 ad 0.6 = 1. O R, we defie x y if ad oly if x = y. Hit: It might help to prove that for ay x R, we have x x < x + 1. 8