FP2 Chapter 3 – Further Complex Numbers Dr J Frost (jfrost@tiffin.kingston.sch.uk) Last modified: 11th December 2015 OVERVIEW This chapter is divided up into two main sections. You are pretty much guaranteed one exam question on each of the two halves: 1 Different forms of a complex number: • Cartesian π§ = π₯ + ππ¦ • Modulus-Argument Form π§ = π cos π + π sin π and • NEW! Exponential form π§ = ππ ππ Multiplying, dividing and raising to a power of complex numbers (the last using De Moivre’s Theorem) Uses of De Moivre’s Theorem (trig identities, roots) Loci with Complex Numbers e.g. “Draw π§ − 2π = 9 on an Argand diagram.” 2 2 -1 RECAP :: Modulus-Argument Form Im[z] (π₯, π¦) If π = π + ππ (and suppose in this case π§ is in the first quadrant), what was: π Re[z] modulus ? argument ? Then In terms of π and π: π = π ππ¨π¬?π½ π = π π¬π’π§?π½ π = π ππ¨π¬ π½ +? π π¬π’π§ π½ π= π§ = ππ +? ππ π = arg π§ = π ? π πππ§−π When −π < π < π it is known as the principal? argument. This is known as the modulus-argument form of π§. ? RECAP :: Modulus-Argument Form π + ππ −1 π π 1 ? 1 1+π ?2 − 3+π 2 1−π 3 2 ? −1 − π ?2 ? ? π½ Mod-arg form π π§ = cos π? + π sin π ? π π π π§ = cos ? + π sin ? 2 2 2 π π π π§ = 2 cos? + π sin ? 4 4 4 5π 5π 5π ? π§ = 2 cos ? + π sin 6 6 6 π π π −? π§ = 2 cos − ? + π sin − 3 3 3 3π 3π 3π π§ = 2 cos − ? + π sin − −? 4 4 4 Exponential Form We’ve seen the Cartesian form a complex number π§ = π₯ + π¦π and the modulus-argument form π§ = π(cos π + π sin π). But wait, there’s a third form! In the later chapter on Taylor expansions, you’ll see that you that you can write functions as an infinitely long polynomial: cos π₯ = 1 π₯2 − 2! π₯4 + 4! π₯6 − 6! +β― π₯3 π₯5 sin π₯ = π₯ − + −β― 3! 5! π₯2 π₯3 π₯4 π₯5 π₯6 π₯ π = 1+π₯+ + + + + + 2! 3! 4! 5! 6! It looks like the cos π₯ and sin π₯ somehow add to give π π₯ . The one problem is that the signs don’t quite match up. But π changes sign as we raise it to higher powers. ππ 2 ππ 3 ππ 4 π = 1 + ππ + + + +β― 2! 3! 4! π 2 ππ 3 π 4 = 1 + ππ − − + +β― 2! 3! 4! π2 π4 π6 π3 π5 = 1− + − +β― +π π− + −β― 2! 4! 6! 3! 5! = cos π + π sin π ? iπ ? ? Holy smokes Batman! ? Exponential Form ! Exponential form π§ = ππ ππ You need to be able to convert to and from exponential form. π + ππ −1 Mod-arg form π§ = cos π?+ π sin π ? π§ =? π ππ ? −0.98π π§ = 13π 2 − 3π −1 + π Exp Form π π 2 cos + π sin 10 10 3π 3π 2 cos ?+ π sin 4 4 3π 3π 2 cos ? + π sin 5 5 π§= ππ 2π 10 ? π§= 3ππ 2π 4 π§= 23ππ 2π 5 πππ + π = π This is Euler’s identity. It relates the five most fundamental constants in maths! To get Cartesian form, put in modulusargument form first. Notice this is not a principal argument. A Final Example 1 2 Use π ππ = cos π + π sin π to show that cos π = (π ππ + π −ππ ) π ππ = cos π + π sin π π −ππ = cos −π + π sin −π = cos π − π sin π ? ∴ π ππ + π −ππ = 2 cos π 1 ∴ cos π = (π ππ + π −ππ ) 2 Exercise 3A Multiplying and Dividing Complex Numbers FP1 recap: If π§1 = π1 cos π1 + π sin π1 and π§2 = π2 cos π2 + π sin π2 Then: π§1 π§2 = ππ ππ (ππ¨π¬ π½π + π½π + π?π¬π’π§(π½π + π½π )) π§1 ππ = (ππ¨π¬ π½π − π½π + π π¬π’π§(π½π − π½π )) π§2 ππ ? i.e. If you multiply two complex numbers, you multiply the moduli and add the arguments, and if you divide them, you divide the moduli and subtract the arguments. Similarly if π§1 = π1 π ππ1 and π§2 = π2 π ππ2 Then: ?π π§1 π§2 = ππ ππ ππ π½π +π½ π§1 ππ π π½ −π½ = π π ?π π§2 ππ Examples 5π 5π π π 3 cos + π sin × 4 cos + π sin 12 12 12 12 π π = ππ ππ¨π¬ + π πππ π? π = ππ π + π π = πππ 2 cos π π 2π 2π + π sin × 3 cos − π sin 15 15 5 5 π π 2π 2π = 2 cos + π sin × 3 cos − + π sin − 15 15 5 5 π π = π ππ¨π¬ − + π πππ − π ? π =π−π ππ π π 2 cos 12 + π sin 12 π ππ ππ = ππ¨π¬ − + π π¬π’π§ − 5π 5π π π π 2 cos + π sin ? 6 6 π π =− − π π π Test Your Understanding FP2 June 2013 Q2 π§ = 5 3 − 5π Find (a) |π§| (b) arg(π§) in terms of π π π π€ = 2 cos + π sin 4 4 Find (c) π€ π§ (d) arg (1) (2) ? ? (1) π€ π§ ? (2) ? Exercise 3B De Moivre’s Theorem We saw that: π§1 π§2 = ππ ππ (ππ¨π¬ π½π + π½π + π π¬π’π§(π½π + π½π )) Can you think therefore what π§ π is going to be? ππ = ππ ππ¨π¬?ππ½ + π π¬π’π§ ππ½ This is known as De Moivre’s Theorem. FP2 June 2013 Q4 Prove by induction that π§ π = π π cos ππ + π sin ππ ? They so went there... Secret Alternative Way ? De Moivre’s Theorem for Exponential Form If π§ = ππ ππ then ππ = π ππ½ ππ = ππ?ππππ½ Examples Simplify 9π 9π 5 cos +π sin 17 17 2π 2π 3 cos 17 − π sin 17 ππ ππ π πππ + π πππ ππ ππ = ππ ππ πππ − + π πππ − ππ ππ = ππ¨π¬ ππ + π π¬π’π§ ππ = ππ¨π¬ π + π π¬π’π§ π = −π Express 1 + 3 π 7 πππ πππ + π π¬π’π§ ππ ππ = π ππ ππ ? πππ − ππ + π πππ − ππ ππ¨π¬ in the form π₯ + ππ¦ where π₯, π¦ ∈ β. π π π + π π = π(ππ¨π¬ + π π¬π’π§ ) π π π ππ ππ Therefore π + π π = ππ ππ¨π¬? + π π¬π’π§ π π π π = πππ ππ¨π¬ + π π¬π’π§ = ππ + ππ π π π π Exercise 3C Applications of de Moivre #1: Trig identities Express cos 3π in terms of powers of cos π Is there someway we could use de Moivre’s theorem to get a cos 3π term? ππ¨π¬ π½ + π π¬π’π§ π½ π ?= ππ¨π¬ ππ½ + π π¬π’π§ ππ½ Could we use the LHS to get another expression? Use a Binomial expansion! cos π + π sin π 3 = π ππ¨π¬ π π½ + π ππ¨π¬ π π½ π π¬π’π§ π½ + π ππ¨π¬ π½ π π¬π’π§ π½ π + π π¬π’π§ π½ ? = ππ¨π¬ π π½ + ππ ππ¨π¬ π π½ π¬π’π§ π½ − π ππ¨π¬ π½ π¬π’π§π π½ − π π¬π’π§π π½ How could we then compare the two expressions? Equating real parts: ππ¨π¬ ππ½ = ππ¨π¬ π π½ − π ππ¨π¬ π½ π¬π’π§π π½ ? =β― = π ππ¨π¬ π π½ − π ππ¨π¬ π½ π Applications of de Moivre #1: Trig identities Express (a) cos 6π in terms of cos π. (b) sin 6π ,π sin π ≠ ππ, in terms of cos π. cos π + π sin π 6 = cos 6π + π sin 6π = cos 6 π + 6 cos 5 π π sin π + β― = cos6 π + 6π cos 5 π sin π − 15 cos 4 π sin2 π − 20π cos 3 π sin3 π + 15 cos 2 π sin4 π + 6π cos π sin5 π − sin6 π Equating real parts: cos 6π = cos 6 π − 15 cos4 π sin2 π + 15 cos 2 π sin4 π − sin6 π = 32 cos6 π − 48 cos 4 π + 18 cos2 π − 1 Equating imaginary parts: (and simplifying) sin 6π = 32 cos 5 π − 32 cos 3 π + 6 cos π Test Your Understanding FP2 June 2011 Q7 (a) Use de Moivre’s theorem to show that sin 5π = 16 sin5 π − 20 sin3 π + 5 sin π (5) Hence, given also that sin 3π = 3 sin π − 4 sin3 π (b) Find all the solutions of sin 5π = 5 sin 3π in the interval 0 ≤ π < 2π. Give your answers to 3 decimal places. (6) ? ? Finding identities for sinπ π and cos π π The technique we’ve seen allows us to write say cos 3π in terms of powers of cos π (e.g. cos 3 π). Is it possible to do the opposite, to say express cos 3 π in terms of a linear combination of cos 3π and cos π (with no powers)? 1 1 If π§ = cos π + π sin π, what is π§ + π§ and π§ − π§ ? π = π−π = ππ¨π¬ π½ + π π¬π’π§ π½ −π = ππ¨π¬ −π½ + π π¬π’π§ −π½ = ππ¨π¬ π½ − π π¬π’π§ π½ π π π ∴ π + π = β― = π ππ¨π¬ π½ π−π =? ππ π¬π’π§ π½ 1 And what is π§ π + π ? π§ Just using de Moivre’s theorem again: π? π π + π = π ππ¨π¬ ππ½ π ! Results you need to use (but you don’t need to memorise) 1 1 π§ + = 2 cos π π§ π + π = 2 cos ππ π§ π§ 1 1 π π§ − = 2π sin π π§ − π = 2π sin ππ π§ π§ Finding identities for sinπ π and cos π π Express cos5 π in the form π cos 5π + π cos 3π + π cos π 5 2 cos π 5 1 = π§ + Starting π§ point? 10 5 1 + 3+ 5 π§ π§ π§ 1 1 1 32 cos5 π = π§ 5 + 5 + 5 π§?3 + 3 + 10 π§ + π§ π§ π§ = 2 cos 5π + 5(2 cos 3π) + 10(2 cos π) ? 1 5 5 Therefore cos 5 π = 16 cos 5π + 16 cos 3π + 8 cos π ? 32 cos5 π = π§ 5 + 5π§ 3 + 10π§ + ? 1 π§ + = 2 cos π π§ 1 π§ − = 2π sin π π§ 1 + π = 2 cos ππ π§ 1 π π§ − π = 2π sin ππ π§ π§π Notice how the powers descend by 2 each time. Using identities below. Finding identities for sinπ π and cos π π 1 4 3 4 Prove that sin3 π = − sin 3π + sin π 3 1 Starting 2π sin π 3 = π§ − π§ point? 3 1 = − 3π§?+ − 3 π§ π§ 1 1 3 = π§ − 3? − 3 π§ − π§ π§ = 2π sin 3π − 3(2π sin π) −8π sin3 π π§3 ? Bro Speed Tip: Remember that terms in such an expansion oscillate between positive and negative. Again using identities. Therefore dividing by −8π: 1 3 sin3 π = − sin 3π + ? 4 sin π 4 1 π§ + = 2 cos π π§ 1 π§ − = 2π sin π π§ 1 + π = 2 cos ππ π§ 1 π π§ − π = 2π sin ππ π§ π§π Test Your Understanding (a) Express sin4 π in the form π cos 4π + π cos 2π + π (b) Hence find the exact value of π 2 0 sin4 π ππ 4 2π sin π 4 1 = Starting π§− π§ point? 4 1 16 sin π = π§ − 4π§ + ? 6− 2+ 4 π§ π§ 1 1 = π§ 4 + 4 −?4 π§ 2 + 2 + 6 π§ π§ = 2 cos 4π − 4 ? 2 cos 2π + 6 1 1 3 ∴ sin4 π = cos 4π − cos 2π?+ 8 2 8 4 π 2 0 sin4 π ππ = 4 2 1 1 3 sin 4π − sin 2π?+ π 32 4 8 1 π§ + = 2 cos π π§ 1 π§ − = 2π sin π π§ π 2 0 = 3π 16 1 + π = 2 cos ππ π§ 1 π π§ − π = 2π sin ππ π§ π§π Exercise 3D Applications of de Moivre #2: Roots π§ π = π π cos ππ + π sin ππ We have so far used de Moivre’s theorem when π was an integer. It also works however when π is a rational number! (proof not required) Solve π§ 3 = 1 Plot these roots on an Argand diagram. The modulus-argument form of 1 is 1(πππ 0 + π π ππ 0) thus: π§ 3 = 1 cos 0 + π sin 0 = 1(cos 0 + 2ππ + π sin 0 + 2ππ 1 3 π§ 2π 3 4π cos 3 2π 3 4π π sin 3 + π sin 2π 3 2π 3 = cos 2ππ + π sin 2ππ 2ππ 2ππ = cos +? π sin 3 3 π = 0, π§1 = cos 0 + π sin 0 = 1 π = 1, π§2 = cos π§2 ? 1 = −2 + π 1 π = 2, π§3 = + = −2 − π These are known as the ‘cube roots of unity’. 3 2 3 2 π§1 2π 3 π§3 1 is always a root – the rest are spread out equally distributed. Can write three roots as 1, π, π2 . Note that 1 + π + π2 = 0 Applications of de Moivre #2: Roots Solve π§ 4 = 2 + 2 3 π π π + π sin 3 3 π π = 4 cos + 2ππ + π sin( + 2ππ) 3 3 π 2ππ π 2ππ π§ = 2 cos + + π sin + 12 4 12 4 π§ 4 = 4 cos π π + π sin 12 12 7π 7π cos + π sin 12 12 13π 13π cos + π sin 12 12 π = 0, π§ = 2 cos π = 1, π§ = 2 π = 2, π§ = 2 = 2 cos − ?11π 12 + π sin − 11π 12 π = 3, π§ = β― We need principal roots. Alternative is to use π = 0, 1, −1, −2 instead of 0, 1, 2, 3, which will give your principal roots directly. Test Your Understanding FP2 June 2012 Q3 a) Express the complex number −2 + 2 3 π in the form π cos π + π sin π , − π < π ≤ π. (3) b) Solve the equation π§ 4 = −2 + 2 3 i giving the roots in the form π cos π + π sin π , −π < π ≤ π. (5) ? ? You can avoid the previous large amount of working by just dividing your angle by the power (4), and adding/subtracting increments of 2π again divided by the power. Exercise 3E Chapter 3 Part II – Locus of Points Dr J Frost (jfrost@tiffin.kingston.sch.uk) Loci You have already encountered loci in FP1 as a set of points which satisfy some restriction. For example, the definition of a parabola: A set of points equidistant from a line (the directrix) and a point (the focus). We often want to find an equation which defines the set of points or the opposite. π¦ 3 -3 3 -3 π₯ π§ = 3 means that the modulus of the complex number has to be 3. What points does this give us on the Argand diagram? Click to Brosketch > On quick reminder… π§ = π + ππ = ππ +? ππ π§ − 3 = π + ππ − π = π − π + ππ =? π−π π + ππ Loci of form π§ − π§1 = π π¦ What does π§ − π§1 = π mean? The difference between some point π§ from a fixed point π§?1 is of length π. i.e. The point π§ is a distance π from π§1 . π π§1 π₯ ! π§ − π§1 = π is represented by a circle centre (π₯1 , π¦1 ) with radius π, where π§1 = π₯1 + ππ¦1 π¦ Sketch the locus of points represented by π§ − 5 − 3π = 3 5,3 π₯ π§ − 5 − 3π = |π§?− 5 + 3π | Find the Cartesian equation of this locus. π§ − 5 − 3π = 3 π₯ + ππ¦ − 5 − 3π = 3 π₯−5 +π π¦−3 =3 Group by real/imaginary. ? Find modulus and square both sides. π₯−5 2 + π¦−3 2 =9 Loci of form π§ − π§1 = π§ − π§2 What does π§ − π§1 = π§ − π§2 mean? π¦ The distance of π§ from π§1 is the same as the distance from?π§2 . π§2 π§1 π₯ ! π§ − π§1 = |π§ − π§2 | is represented is represented by a perpendicular bisector of the line segment joining the points π§1 to π§2 . Sketch the locus of points represented by π§ = |π§ − 6π|. Write its equation. π¦ (0,6) Click to Brosketch > (The equation is obviously π¦ = 3, but let’s do it the same algebraic way) π₯ + ππ¦ = Equation π₯+π π¦−6 ? π₯ 2 + π¦ 2 = π₯ 2 + π¦ − 6 2 Note that both sides were squared. π¦=3 (0,0) π₯ Test Your Understanding Find the Cartesian equation of the locus of π§ if π§ − 3 = |π§ + π|, and sketch the locus of π§ on an Argand diagram. π¦ π₯ + ππ¦ − 3 = π₯ + ππ¦ + π π₯ − 3 + ππ¦ = π₯ + π π¦ + 1 π₯ − 3 2 + π¦2 = π₯2 + π¦ + 1 2 π¦ = −3π₯ +4 Equation ? 4 Argand Diagram ? (0, −1) 4 3 (3,0) Loci of form π§ − π§1 = π π§ − π§2 Find the Cartesian equation of the locus of π§ if π§ − 6 = 2|π§ + 6 − 9π|, and sketch the locus of π§ on an Argand diagram. π₯ + ππ¦ − 6 = 2 π₯ + ππ¦ + 6 − 9π π₯ − 6 + ππ¦ = 2 π₯ + 6 + π π¦ − 9 π₯ − 6 2 + π¦ 2 = 2Equation π₯+6 2+ ? π¦−9 π₯ − 6 2 + π¦2 = 4 π₯ + 6 2 + π¦ − 9 2 … π₯ + 10 2 + π¦ − 12 2 = 100 π¦ −10,12 12 π΅ Argand 2 You can see that at any point on the circle, you’ll be twice the distance from π΄ as you will be for π΅ (this is Diagram ? not required on your sketch). It’s a little surprising we got a circle, so an algebraic method is definitely needed here. π₯ π΄ Test Your Understanding FP2 June 2012 Q8a The point π represents a complex number π§ on an Argand diagram such that π§ − 6π = 2|π§ − 3| (a) Show that, as π§ varies, the locus of π is a circle, stating the radius and the coordinates of the centre of this circle. ? arg π§ − π§1 = π π¦ Click to Brosketch > π 6 arg π§ = ? At any point, the angle made π relative to the origin is 6 π 6 π₯ π arg π§ = 6 π arg π₯ + ππ¦ = 6 π¦ π 1 = tan = π₯ 6 3 1 π¦= π₯, π₯ > 0, π¦ > 0 3 This bit is important. The locus is referred to as a ‘half line’. Equation ? π§ + 3 + 2π can be rewritten as π§ − −3 − 2π 3π 4 (−3, −2) π¦ Click to Brosketch > π₯ One way of thinking about it: If we were to add 3 + 2π to π§ − (−3 − 2π), we’d then have had 3π arg π§ = 4 , which would be centred at the origin. arg π§ + 3 + 2π = arg π₯ + ππ¦ + 3 + 2π = 3π 4 arg π₯ + 3 + π π¦ + 2 = 3π 4 3π 4 Equation ? π¦+2 3π = tan = −1 π₯+3 4 π¦ = −π₯ − 5, π₯ < −3, π¦ β» 2 ? Harder πππ ones arg π§−6 π§−2 = π 4 ? π 4 Suppose πΏ1 is half-line such that arg π§ − 6 = π, and L2 the half-line such that arg π§ − 2 = π. π It follows π − π = 4 arg π§ − 6 − arg π§ − 2 = π¦ πΏ1 πΏ2 π 4 π(π₯, π¦) π 2 π Where is the 4 in this diagram? Exterior angle of triangle is sum of other two ? on right. interior, so where indicated As the two angles π and π vary (as π§ varies), but π the 4 is constant, π forms an arc, given that angles in same segment are equal. π 4 π π π₯ 6 2 How could we find the centre of the circle? Angle at centre will be double. Resulting triangle will need to be isosceles as radius of circle is constant. ? Equation: π − π Recall that arg π§1 π§2 = arg π§1 + arg π§2 π§1 arg = arg π§1 − arg π§2 π§2 π ? + π−π π = π, π > π Another Example π§ π If arg = , sketch the locus of π(π₯, π¦) which is π§−4π 2 represented by π§ on an Argand diagram. π§ = arg π§ − arg π§ − 4π = π − π π§ − 4π π where π − π = arg 2 π Since the angle at π is always 2 , it follows that?the line segment (0,0) to (4,0) must be the diameter, so the equation of the circle is π₯2 + π¦ − 2 2 = 4 π¦ 4 π 2 π₯ Bro Exam Note: No questions have appeared of the π§−π form arg π§−π = π since the new syllabus came into play, but it IS explicitly referenced in the specification. Another Example Given that the complex number π§ satisfies the equation π§ − 12 − 5π = 3, find the minimum value of |π§| and the maximum. Since |π§| is the distance to the origin, this is minimised and maximised on the line which connects the origin to the centre of the circle. π¦ (12,5) 13 3 ? π₯ Minimum = 10 Maximum = 16 Exercise 3F Regions How would you describe each of the following in words? Therefore draw each of the regions on an Argand diagram. π§ − 4 − 2π ≤ 2 π§ − 4 < |π§ − 6| “The distance from 4 + 2π is less than 2”. π¦ ? (4,2) 4 π 0 ≤ arg π§ − 2 − 2π ≤ 4 π ? 4 (2,2) π₯ “The distance from 4 is less than from 6.” ? π₯=5 4 6 π§ − 4 − 2π ≤ 2, π§ − 4 < |π§ − 6| π and 0 ≤ arg π§ − 2 − 2π ≤ 4 An intersection of the three other regions. (I couldn’t be ? bothered to draw this. Sorry) Test Your Understanding P6 June 2003 Q4(i)(b) Shade the region for which π π π§ − 1 ≤ 1 and ≤ arg π§ + 1 ≤ 12 2 ? Transformations If π§ is used to represent the complex number which forms a locus of points, what do you think will happen for each of the following π€: π€ = π§ + π + ππ π€ = ππ§ π€ = ππ§ + π + ππ π Translation by ? π Enlargement by scale factor π centre 0,0 ? Enlargement followed by translation. ? Example: If π represents the complex number π§ where π§ = 2, describe 1 the locus of the image of π under the transformation π€ = 2 π§ + π 1 π§ 2 1 π€−π = π§ 2 1 π€−π = π§ 2 1 π€−π = 2 =1 2 π€−π = ? STEP 1: Make π§ (or multiple of π§) the subject) STEP 2: Apply |..| to both sides (because we know what |π§| is) STEP 3: Use π§1 π§2 = π§1 |π§2 | if necessary. STEP 4: Substitute using original equation. π + ππ π§ π€ More Difficult Example Show that effect on the circle π§ = 1 of the transformation π€ = π−π€ ? π€ − 5π π−π€ |π§| = π€ −?5π 5ππ§+π π§+1 STEP 1: Make π§ (or multiple of π§) the subject) π§= STEP 2: Apply |..| to both sides (because we know what |π§| is) π−π€ = ? π€ − 5π STEP 3: Use π§1 π§2 = π§1 |π§2 | or π§1 π§2 = π§1 π§2 necessary. π−π€ 1= π€? − 5π STEP 4: Substitute using original equation. π€ − 5π = π − π€ π€ − 5π =?|π€ − π| STEP 5: Subsequent manipulation to put in form we recognise the loci of. π¦ (0,5) π¦=3 Sketch ? (0,1) π₯ if Test Your Understanding FP2 June 2009 Q6 A transformation π from the π§-plane to the π€-plane is given by π§ π€= , π§ ≠ −π π§+π The circle with equation π§ = 3 is mapped by π onto the curve πΆ. (a) Show that πΆ is a circle and find its centre and radius. (8) (b) The region π§ < 3 in the π§-plane is mapped by π onto the region π in the π€-plane. Shaded the region π on an Argand diagram. (2) (Hint: be careful and don’t guess! Try a value of π§ where you know that π§ < 3, and then transform it to see where it ends up.) ? ? π€ = π§2 For the transformation π€ = π§ 2 , where π§ = π₯ + ππ¦ and π€ = π’ + ππ£, find the locus of π€ when π§ lies on a circle with equation π₯ 2 + π¦ 2 = 16 π§ =4 ? STEP 0: Represent Cartesian equation as equation involving π§. (note: π§ 2 is also fine, so don’t do this step) STEP 1: Make π§ (or multiple of π§) the subject π€ = π§2 π€ = |π§ 2 | ? STEP 2: Apply |..| to both sides (because we know what |π§| is) ? STEP 3: Use π§1 π§2 = π§1 |π§2 | or π€ = π§ π§ π€ = 16 ? 0,0 radius 4. i.e. A circle with centre π§1 π§2 = π§1 π§2 necessary. STEP 4: Substitute using original equation. if A slightly different one Bro Tip: Sometimes we need to represent π§ as π₯ + π¦π and π€ as π’ + ππ£ so we can reason about real and imaginary parts. ππ§−2 A transformation π on the π§-plane to the π€-plane is given by π€ = 1−π§ , π§ ≠ 1. Show that as π§ lies on the real axis in the π§-plane, then π€ lies on the line π in the π€plane. Sketch π on an Argand diagram. π€+2 ? π€+π π€+2 π₯= ? π€+π π§= STEP 1: Make π§ (or multiple of π§) the subject. We’re told π§ lies on the real-axis, so what can we replace it with? (given that π§ = π₯ + ππ¦) π’ + ππ£ + 2 π’ + 2 + ππ£ = π’ + ππ£ + π π’ + π π£ + 1 π’ + 2 + ππ£ π’ − π π£ + 1 = × π’+π π£+1 π’−π π£+1 π’ π’+2 +π£ π£+1 π’π£ − π’ + 2 π£ + 1 = + π π’2 + π£ + 1 2 π’2 + π£ + 1 2 It would be nice to compare real and imaginary components on each side. First replace π€ = π’ + ππ£ and do the complex number division on the RHS (as per FP1). π₯= ? π’π£ − π’ + 2 π£ + 1 0= π’2 + π£ + 1 2 0 = π’π£ − π’?+ 2 π£ + 1 1 π£ =− π’−1 2 Now compare imaginary components (since we get rid of π₯) π£ Sketch ? −2 −1 π’ Exercise 3H Summary Euler’s relation and de Moivre’s theorem/applications You’ve been asked to prove these before in exams. Note: both cos ππ and sin ππ to powers and vice versa. (this is a screenshot from the official specification) Loci/Transformations |π§ − π| = π was a circle centred at π with radius π. π§ − π = π|π§ − π| was a circle but required an algebraic approach. arg π§ − π = π½ was a half-line starting at π. π§−π arg = π½ required using circle π§−π theorems. It will probably be the case that π π π½ = or . 4 2
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