Unit 3 Specialist Maths

Unit 3 Specialist Maths

succeeding in the vce, 017 extract from the master class teaching materials Our Master Classes form a component of a highly specialised weekly program, which is designed to ensure that students reach their full potential (including the elite A and A+ scores). These classes incorporate the content and teaching philosophies of many of the top schools in Victoria, ensuring students are prepared to a standard that is seldom achieved by only attending school. These classes are guaranteed to motivate students and greatly improve VCE scores! For additional information regarding the Master Classes, please do not hesitate to contact us on (03) 9663 3311 or visit our website at www.tsfx.com.au. essential for all year 11 and 1 students! important notes Our policy at TSFX is to provide students with the most detailed and comprehensive set of notes that will maximise student performance and reduce study time. These materials, therefore, include a wide range of questions and applications, all of which cannot be addressed within the available lecture time i.e. Due to time constraints; it is possible that some of the materials included in this booklet will not be addressed during the course of these lectures. Where applicable, fully worked solutions to the questions in this booklet will be handed to students on the last day of each subject lecture. Although great care is taken to ensure that these materials are mistake free, an error may appear from time to time. If you believe that there is an error in these notes, please let us know asap (admin@tsfx.com.au). Errors, as well as clarifications and important updates, will be posted at www.tsfx.com.au/vce-updates The views and opinions expressed in this booklet and corresponding lecture are those of the authors/lecturers and do not necessarily reflect the official policy or position of TSFX. TSFX - voted number one for excellence and quality in VCE programs. copyright notice These materials are the copyright property of The School For Excellence and have been produced for the exclusive use of students attending this program. Reproduction of the whole or part of this document constitutes an infringement in copyright and can result in legal action. No part of this publication can be reproduced, copied, scanned, stored in a retrieval system, communicated, transmitted or disseminated, in any form or by any means, without the prior written consent of The School For Excellence (TSFX). The use of recording devices is STRICTLY PROHIBITED. Recording devices interfere with the microphones and send loud, high-pitched sounds throughout the theatre. Furthermore, recording without the lecturer s permission is ILLEGAL. Students caught recording will be asked to leave the theatre, and will have all lecture materials confiscated. it is illegal to use any kind of recording device during this lecture

COMPLEX NUMBERS THE IMAGINARY NUMBER Real numbers (R) Irrational numbers (Q ) Rational numbers (Q), e,, non-recurring decimals p (Numbers in the form ; p, q Z, q 0) q Integers (Z) (whole numbers) {, 3,, 1, 0, 1,, 3,..} Natural numbers (N) (Counting numbers 1 3, ) You would already know how to solve quadratic equations via many different methods, e.g. using the cross method of factorisation, completing the square and the quadratic formula method. Irrespective of the choice of method, you would have found some equations that could not be solved across the Real Number System, as we could not find the square root of a negative number. For example: x 5 0 x 5 An Italian mathematician, Rafael Bombelli, proposed the expression 40 (5 15)(5 15), which was valid if it was possible to take the square root of a negative number. He defined an imaginary number (denoted as i) that has the property i 1. i 1 The School For Excellence 017 Succeeding in the VCE Unit 3 Specialist Maths Page 1

This number is used in many fields of mathematics, including algebra, where it may be used to find solutions for all algebraic expressions. So now: x x x 5 0 5 5 1 x 5 1 x 5i (This is the solution under the complex number system.) QUESTION 1 Solve x 6x 5 0. Solution b x b 4ac a 6 6 4 1 5 6 36 100 6 64 6 64 1 6 8i x 3 4i The School For Excellence 017 Succeeding in the VCE Unit 3 Specialist Maths Page

QUESTION Solve the following expressions by first completing the square. (a) x 4x 13 0 (b) x 5x 7 0 (c) 3 3 The School For Excellence 017 Succeeding in the VCE Unit 3 Specialist Maths Page 3

QUESTION 3 Solve the following expressions over the complex number field using the quadratic formula. (a) 3x 7x 10 (b) 3 5 7 0 The School For Excellence 017 Succeeding in the VCE Unit 3 Specialist Maths Page 4

OPERATIONS INVOLVING IMAGINARY NUMBERS Any expression in the form bi (where b 0) is called a pure imaginary number. The rules of arithmetic and algebra for real numbers also apply to pure imaginary numbers i.e. imaginary numbers may be added, subtracted, multiplied and divided in the same way as real numbers. For example: (a) (b) (c) 3i 7i 10i 7i i 5i 3 i i 4i 8i 8( i ) i 8( 1) i 8i 4 4 (d) (i) 16( i) 16( i ) 16( 1) 16 (e) ( a ac) i ai( a c) a c ai ai Note: Every integral power of i can be expressed as one of i, i, 1, 1. For example: i i 3 1 i i ( 1) i i i 4 ( i ) ( 1) 1 i 5 ( i 4 ) i (1) i i This pattern continues in a periodic fashion, i.e. 4n i 1 for n 4n1 i i i i 4n 4n3 1 i The School For Excellence 017 Succeeding in the VCE Unit 3 Specialist Maths Page 5

PROPERTIES OF COMPLEX NUMBERS Expressions of the form 1 6i are referred to as complex numbers. The general form of a complex number is x yi. x represents the real part of, called Re. y represents the imaginary part of, called Im. The letter C is used to denote the set of complex numbers: { : x yi, x, y R}. If there is no y value then is a real number, and therefore, real numbers are a subset of complex numbers i.e. R C. EQUALITY Two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal. For example: If a bi and w c di then w if (if and only if) a c and b d. If QUESTION 4 ( x 1) 7i and w 5 ( y 4) i find the value of x and y if w. Solution Since w then x 1 5 and 7 y 4. Therefore: x 6 and y 3. x 3 For to be equal to w, then x and y both must equal 3. The School For Excellence 017 Succeeding in the VCE Unit 3 Specialist Maths Page 6

ADDITION AND SUBTRACTION To add/subtract complex numbers, we add or subtract like terms. We add the imaginary components together and the real components together. Note: ( a bi) ( c di) ( a c) ( b d ) i The sum or difference of two complex numbers is itself a complex number. If 1 a bi and c di then 1 1. If 1 a bi and c di and e fi 3 then 1 3 1 3. If QUESTION 5 3 5i and w 7i find: (a) w w ( 3 ) (5 7) i 5 i (b) w w ( 3 ) (5 7) i 1 1i The School For Excellence 017 Succeeding in the VCE Unit 3 Specialist Maths Page 7

MULTIPLICATION The philosophy and procedures involved in the multiplication of complex numbers is identical to that employed with algebraic expressions. When a complex number is multiplied by a real number (scalar), we multiply each component of the complex number by the scalar quantity (k). i.e. k( a bi) ka kbi. For example: (1 3i) (1) (3i) 6i. The product of two complex numbers is obtained by multiplying each component of the brackets by one another. For example: If a bi and w c di then. w ac adi bci bdi ( ac bd) ( ad bc) i Important Notes: If 1 a bi and c di then 1.. 1. If 1 a bi and c di and e fi 3 then 1.. 3 1.. 3. The product of two complex numbers may result in either a real or non-real number. If QUESTION 6 3 4i and w 3i, find. w. Solution. w (3 4i)( 3i) 6 9i 8i 1i 6 i 1 1 18 i The School For Excellence 017 Succeeding in the VCE Unit 3 Specialist Maths Page 8

COMPLEX CONJUGATES The complex conjugate of a bi is defined as a bi, where ( bar ) is the complex conjugate of (i.e. to find the conjugate you change the sign of the imaginary part). For example: The complex conjugate of 5 i is written as 5 i. The complex conjugate is similar to the conjugates in surds. Recall that when a surd is multiplied by its conjugate, a real number is formed. For example: ( 3 )(3 ) 9 3 3 9 7. Similarly, when a complex number is multiplied by its complex conjugate, a real number is also formed. If 3 i then 3 i. (3 i)(3 i) 9 6i 6i 4i 9 4 13 In general:. a b MAGNITUDE The magnitude of the complex number by : a ib is called the modulus and is denoted a b That is, Re Im In general,. a b. The School For Excellence 017 Succeeding in the VCE Unit 3 Specialist Maths Page 9

THE MULTIPLICATIVE INVERSE The multiplicative inverse (or inverse under multiplication) of is denoted as If a bi then the multiplicative inverse is given by 1 1 1. a bi 1. Rationalising produces: 1 1 a bi a b For example: If 4 5i then 1 1 1 (4 5i) 4 5i = 4 5 4 5i (4 5i) (4 5i) 41 41 41 i. Note: When a complex number and its multiplicative inverse are multiplied, the result is 1 always equal to 1, that is:. 1. DIVISION OF COMPLEX NUMBERS The quotient of two complex numbers is obtained in the form and bottom by the complex conjugate. a bi by multiplying the top For example: 5 5 (3 i ) 15 i5 15 i5 15 5 i 3 i (3 i ) (3 i ) 9 11 11 11 Handy tip: 1 i i since 1 1 i i i i i i i i 1 QUESTION 7 3 i Express in the form of 5 3i Solution a bi. 3 i 5 3i 3 i 5 3i 5 3i 5 3i 15 9i 10i 6i 5 15i 15i 9i 15 19i 6 5 9 1 9 19i 34 9 34 19 i 34 The School For Excellence 017 Succeeding in the VCE Unit 3 Specialist Maths Page 10

Given QUESTION 8 3 i, w 5 i and a i find, in Cartesian form: (a) w a (b). a (c) a (d) (e) 1 1 w. a (f) Im4 3w (g) Rea iim (h) 4i3iw Re( a) (i) 8 iw (j) 10 i Re a Solution The School For Excellence 017 Succeeding in the VCE Unit 3 Specialist Maths Page 11

The School For Excellence 017 Succeeding in the VCE Unit 3 Specialist Maths Page 1

QUESTION 9 If 3 i then ( )(1 ) is equal to A 5 9i B 9 5i C 3 7i D 5 9i E 5 5i Solution Use technology to answer this question: Answer is A The School For Excellence 017 Succeeding in the VCE Unit 3 Specialist Maths Page 13

QUESTION 10 If a bi, a, b 0 and a, b R, then is equal to A B C D E 1 a bai b b ai b 1 a a 1 The School For Excellence 017 Succeeding in the VCE Unit 3 Specialist Maths Page 14

THE COMPLEX NUMBER PLANE As complex numbers do not possess the order properties of real numbers, it is not possible to represent them as points on a number line. However, as complex numbers have two dimensions the real part and the imaginary part they can be represented on a plane. This plane is commonly referred to as an Argand diagram or the -plane, and consists of two perpendicular axes: The vertical axis which represents the imaginary component of the complex number and it is labelled Im(). The horiontal axis which represents the real component of the complex number and it is labelled Re(). Each point on an Argand diagram represents a complex number. The complex number a bi is located at the point ( a, b), and is referred to as the Cartesian form of the complex number. The complex number i is located at the point (, ). The complex number 3 i is located at the point ( 3, 1). Note: The complex numbers are points in space. Do not connect lines to the origin. The School For Excellence 017 Succeeding in the VCE Unit 3 Specialist Maths Page 15

QUESTION 11 Given 3 i and w 1 3i plot the following results on an Argand diagram. (a) (b) w (c) w (d) 3 (e). w (f) wi (g) i (h) 3 i What is the effect of multiplying expressions by 3 i, i, i? Solution Im() 10 8 6 4-4 6 8 10 Re() - i rotation of anticlockwise i rotation of anticlockwise 3 i rotation of 3 anticlockwise n i rotation of n anticlockwise we can see that every time we multiply by i, the point is rotated anticlockwise. The School For Excellence 017 Succeeding in the VCE Unit 3 Specialist Maths Page 16

GEOMETRICAL INTERPRETATION OF SUBTRACTION The distance between the two complex numbers 1 and is equal to 1. This provides a simple geometrical interpretation for subtraction. Proof: Let 1 x1 iy1 and x iy. Then x iy ) ( x iy ) ( x x ) i( y ). 1 ( 1 1 1 y1 Then 1 ( x x1 ) i( y y1) ( x x1 ) ( y y1). Now recall that points x, ) and ) ( 1 y1 x1) ( y 1) ( x, y. ( x y is the formula for the distance between the two The School For Excellence 017 Succeeding in the VCE Unit 3 Specialist Maths Page 17

POLAR FORM Instead of using the Cartesian coordinates ( x, y), a point can be represented by using the polar coordinates [ r, ]. Where: r is the distance from the origin (positive). is the angle moved from the horiontal. Using Pythagoras Theorem: r x y As sin y r and cos x : r Then sin and x rcos y r By definition: x yi (Cartesian form) Therefore: rcos risin r(cos isin ) rcis (Polar form) Note: cis is a common abbreviation for cos i sin. r represents the distance between the origin and the point P, and is denoted by (modulus of ). r x y Im( ) is the argument of, where tan y Re( ) x, and represents the angle formed in moving in an anticlockwise direction from the positive Re() axis to the point P. The argument of can be measured in two ways. arg = n, n Z. This gives many possible values (4 quadrant theory) as any multiple of can be added to. OR Arg = (expressed with an upper case A) where. The Argument of is known as the Principal Value (or Argument) of and is expressed as an angle in the interval (-, ]. There is only one answer in the interval. Note: VCAA expects that answers are simplified to Principal Argument. The School For Excellence 017 Succeeding in the VCE Unit 3 Specialist Maths Page 18

The argument of ero is not defined. Remember Im( ) 0 then tan, which is undefined! Re( ) 0 Im( ) tan y Re( ) x, so if 0 0 i If rcis, then rcis( ). Proof: rcis( ) r cos( ) ir sin( ) r cos( ) ir sin( ) r cos( ) ir sin( ) r cos( ) ir sin( ) Using the symmetry properties cos( ) cos( ) and sin( ) sin( ). rcis ( ) Note: The CAS calculator does not display polar form as rcis( ), but instead uses Euler i form re. EXAMPLE Find: (a) cis in Cartesian Form. 4 (b) 1 3i in Polar Form. This allows us to understand one of the famous mathematical equations that links four fundamental mathematical constants: The School For Excellence 017 Succeeding in the VCE Unit 3 Specialist Maths Page 19

i e 1 The School For Excellence 017 Succeeding in the VCE Unit 3 Specialist Maths Page 0

TECHNOLOGY TIP Using the Define function on CAS to make shortcuts for Complex Number functions Type the following rule into the CAS Calculator to convert from Cartesian to Polar Form. Define ctp a, b a b arg a bi for ClassPad. Define ctp a, b a b angle a bi for TI-nspire. Note: The right side of the rule is a 1X matrix and the Complex Number i must be used. CAS must be set to radian mode. Type the following rule into the CAS Calculator to convert from Polar to Cartesian Form. Define ptc r, r cos sin i EXAMPLE Find: (a) 3 cis in Cartesian Form. 4 (b) i in Polar Form. Solution (a) i (b) cis 3 The School For Excellence 017 Succeeding in the VCE Unit 3 Specialist Maths Page 1

CONVERTING CARTESIAN FORMS INTO POLAR FORM Step 1: Calculate the value of r. Use r x y. Step : Draw the complex number on an Argand diagram (to make sure you know which quadrant it is in). Step 3: Use tan OPP y ADJ x to find the sie of the angle. Step 4: Determine the Argument of. Take into account which quadrant it is in. = where. Remember that Arg Step 5: Substitute r and into rcis. QUESTION 1 Convert 1 3i into polar form. Solution Calculate the value of r. Use r x y : Since x yi, then x 1 and y 3. r x y ( 1) 4 ( 3) r Draw the complex number on an Argand diagram: Use tan OPP y ADJ x to find the sie of the angle: tan 3 3 Substitute r and into rcis : cis 3 The School For Excellence 017 Succeeding in the VCE Unit 3 Specialist Maths Page

QUESTION 13A Convert the following expressions into polar form: (a) i (b) 3 i (c) 3 i The School For Excellence 017 Succeeding in the VCE Unit 3 Specialist Maths Page 3

QUESTION 13B Convert the following expressions into polar form: (a) 3 4i (b) 5 6i The School For Excellence 017 Succeeding in the VCE Unit 3 Specialist Maths Page 4

QUESTION 14 If = x + (x + 1)i, (a) Find the real values of x for which =. (b) Find the value of x for which Arg () = 3. Solution The School For Excellence 017 Succeeding in the VCE Unit 3 Specialist Maths Page 5

QUESTION 15 If 3i and w3 4i, then w equals A B C D E 13 5 13 5 (3 + 4i) 6 + 17i 13 5 (3 4i) 13 5 (3 + 4i) Solution Use technology to answer this question: Answer is E The School For Excellence 017 Succeeding in the VCE Unit 3 Specialist Maths Page 6

QUESTION 16 Arg ((1 + i)(i 3)) is equal to A B C D E 1 11 1 1 5 1 6 Solution The School For Excellence 017 Succeeding in the VCE Unit 3 Specialist Maths Page 7

CONVERTING THE POLAR FORM INTO CARTESIAN FORM METHOD: Step 1: Expand the given expression using rcis rcos risin. Step : Determine the quadrant in which the given angle lies. Step 3: Solve the expression taking into account the quadrant in which the given angle lies. QUESTION 17 Convert each of the following into Cartesian form. (a) Solution (a) 5 4 cis (b) 7cis 3 3 5 cis 3 Expand the given expression using rcis rcos risin : 5 5 5 cis cos i sin 3 3 3 Determine the quadrant in which the given angle lies: As 5, the given angle lies in the 4 th quadrant, Re() > 0, Im () < 0 3 Solve the expression taking into account the quadrant in which the given angle lies: 5 5 cos i sin 3 3 cos i sin 3 3 1 3 i 1 3i The School For Excellence 017 Succeeding in the VCE Unit 3 Specialist Maths Page 8

(b) 4 7cis 3 Expand the given expression using 4 4 4 7cis 7cos 7isin 3 3 3 rcis r cos r sini : Determine the quadrant in which the given angle lies: 4 As, the given angle lies in the nd quadrant, Re() < 0 and Im() > 0 3 Solve the expression taking into account the quadrant in which the given angle lies: 4 4 7cos 7i sin 3 3 7 cos 7i sin 3 3 1 3 7 7 i 7 (1 3i) 7 7 3i The School For Excellence 017 Succeeding in the VCE Unit 3 Specialist Maths Page 9

QUESTION 18 Convert the following expressions into Cartesian form: (a) 3 cis 4 (b) 5 5cis 6 The School For Excellence 017 Succeeding in the VCE Unit 3 Specialist Maths Page 30

QUESTION 19 The Cartesian form of A 1 3 i 7 3cis 6 is: B C D E 1 3 i 3 1 i 3 3 i 3 1 i Solution Use technology to answer this question: Note: rcis = i re on CAS Note that you must use Euler form to input the value into your calculator. Answer is D The School For Excellence 017 Succeeding in the VCE Unit 3 Specialist Maths Page 31

MULTIPLICATION AND DIVISION IN POLAR FORM If the complex numbers are already written in polar form, you can multiply and divide these complex expressions in a quick and efficient manner. If rcis and rcis 1 1 1 then:. r. rcis( ) 1 1 1 1 r1 r cis ( 1 ) Remember to express the angle in your final answer as a measure between. If (a) (b) QUESTION 0 5 5cis and w cis find: 3 6. w w Solution (a) 5 10 11. w 5 cis 10cis 10cis 3 6 6 6 6 10cis 6 (b) w 5 5 cis 3 6 5 3 cis 5 cis 5 10 5 9 cis cis 6 6 6 Always state the principal value in the final answer. The School For Excellence 017 Succeeding in the VCE Unit 3 Specialist Maths Page 3

If (a) QUESTION 1 3 cis and w 3cis find: 4 6. w (b) w The School For Excellence 017 Succeeding in the VCE Unit 3 Specialist Maths Page 33

QUESTION Let (a) 1 1 i and 1 i 3. Find 1 in exact Cartesian form. (b) Express in polar form. (c) Given that 1 cos isin, find the polar form of 4 4 1. (d) Hence find the exact value of cos. 1 The School For Excellence 017 Succeeding in the VCE Unit 3 Specialist Maths Page 34

QUESTION 3 5 If 1 1 i and cis 6 then 1 is equal to A B C D E cis 5 4 4 cis 13 1 cis 11 1 4 cis 3 5 cis 3 Solution Using technology: The School For Excellence 017 Succeeding in the VCE Unit 3 Specialist Maths Page 35

Let r cis ) and r cis ). 1 1 ( 1 GEOMETRICAL INTERPRETATION OF MULTIPLICATION AND DIVISION ( Multiplying 1 by, that is, the product 1, can be interpreted geometrically as the rotation of 1 about the origin in an anti-clockwise sense through an angle, together with a scaling of the modulus by a factor of r. In particular, multiplying 1 by i cis, that is, the product i 1, can be interpreted geometrically as the rotation of 1 about the origin in an anti-clockwise sense through an angle of radians. This is something we discovered earlier. Dividing 1 by, that is, the quotient 1, can be interpreted geometrically as the rotation of 1 about the origin in a clockwise sense through an angle together with 1 1 a scaling of the modulus by a factor of r. In particular, dividing 1 by i cis, that is, the quotient 1, can be interpreted i geometrically as the rotation of 1 about the origin in a clockwise sense through an angle of radians. Note: 1 1 i i i i i 1 The School For Excellence 017 Succeeding in the VCE Unit 3 Specialist Maths Page 36

QUESTION 4* The diagram below shows two distinct points P and Q in the complex plane that represent the complex numbers and w respectively. The origin is denoted by O. The triangle OPQ is isosceles and POQ is a right angle. Q Imaginary P O Real Prove that w 0. Solution The School For Excellence 017 Succeeding in the VCE Unit 3 Specialist Maths Page 37

DE MOIVRE S THEOREM If an expression can be written in polar form, we can find solutions for expressions written to high powers, fractional powers and negative powers using De Moivre s Theorem. De Moivre s Theorem states that if rcis then: n r n cis( n) Remember to express the angle in your final answer as a measure between. QUESTION 5 If 5cis, find: 3 (a) 4 1 (b) (c) Solution (a) 5 4 cis4 3 8 65cis 3 65cis 3 4 1 (b) (c) The School For Excellence 017 Succeeding in the VCE Unit 3 Specialist Maths Page 38

QUESTION 6 4 (1 i) Simplify ( 3 i) 5 and express in Cartesian form. Solution The School For Excellence 017 Succeeding in the VCE Unit 3 Specialist Maths Page 39

Using technology to check your answer: The School For Excellence 017 Succeeding in the VCE Unit 3 Specialist Maths Page 40

QUESTION 7 (a) Simplify the following expressions, giving your answer in Cartesian form. (i) (1 3 ) 3 i The School For Excellence 017 Succeeding in the VCE Unit 3 Specialist Maths Page 41

(ii) 5 cis 3 4 ( 3 i) 3 The School For Excellence 017 Succeeding in the VCE Unit 3 Specialist Maths Page 4

(iii) (1 i) 4 ( 3 + i) 5 1 (b) If acis and 1 3i find the value of a. 3 The School For Excellence 017 Succeeding in the VCE Unit 3 Specialist Maths Page 43

b (c) If acis and w cis and cis 6 4 w 1 find the value of a and b. b (d) Given that acis and w cis find the value of a and b if 3 4 3 w 108cis. The School For Excellence 017 Succeeding in the VCE Unit 3 Specialist Maths Page 44

QUESTION 8* Given 4 3i and 7 i: 1 (a) Find 1 in exact Cartesian form. (b) Hence find 1 in exact Polar form. (c) Write both 1 and in exact Polar form. (d) Hence find the exact value of tan 3 1 tan 4 7. 1 1 The School For Excellence 017 Succeeding in the VCE Unit 3 Specialist Maths Page 45

QUESTION 9* m Find 3 i 3 i Solution m : 0. m The School For Excellence 017 Succeeding in the VCE Unit 3 Specialist Maths Page 46

QUESTION 30* (cos sin ) and w n a, where a and n are both positive integers. Let w r i (a) Show that sin( n ) 0. (b) Show that n w a. (c) Show that w and w satisfy the equation n a Solution. The School For Excellence 017 Succeeding in the VCE Unit 3 Specialist Maths Page 47

QUESTION 31* (cos isin ), show that: (a) By considering the real and imaginary parts of 5 (i) sin(5 ) 16sin 5 0sin 5sin (ii) cos(5 ) 16cos 5 0cos 5cos Solution i 5 i i i (cos sin ) (cos sin ) (cos sin ) (cos sin ) i i i (cos sin cos sin )(cos sin cos sin )(cos sin ) 4 3 4 3 i i 3 3 [icos sin icossin 4cos sin ] (cos isin ) [cos cos sin cos sin ] [sin cos sin sin cos ] 4 4 i 3 i 3 i [cos 5 10cos 3 sin 5sin 4 cos ] 4 3 5 i[5cos sin 10sin cos sin ] cos 6cos sin sin 4 cos sin 4 sin cos (cos sin ) [cos 5 10cos 3 (1 cos ) 5(1 cos ) cos ] i 3 5 [5(1 sin ) sin 10sin (1 sin ) sin ] 16cos 5 0cos 3 5cos i 16sin 5 0sin 3 5sin Also: 5 (cos isin ) cis(5 ) cos(5 ) isin(5 ) Now equate real and imaginary parts of each expression for (cos isin). 5 The School For Excellence 017 Succeeding in the VCE Unit 3 Specialist Maths Page 48

5 3 (b) Hence show that a solution to 16x 0x 5x 0 is x sin. 5 (c) Hence show that the exact value of sin 1 is equal to 10 5. 5 4 The School For Excellence 017 Succeeding in the VCE Unit 3 Specialist Maths Page 49

5 (d) Find an exact solution to 16x 0x 5x 1 of the form a( b c), where a, b and c are integers. 3 The School For Excellence 017 Succeeding in the VCE Unit 3 Specialist Maths Page 50

SOLVING EQUATIONS IN THE FORM Z n = a USING DE MOIVRE S THEOREM Equations of the form n a can be solved using De Moivre s Theorem. The solution to such equations gives the nth roots of a. METHOD: Given n a : Step 1: Express the left hand side of the equation in the form r n cisn. Let rcis n n r cisn Step : Express the right hand of the equation in polar form. Express a in the form n rcisn rcis 1 1 rcis, say a rcis. 1 1 Step 3: Equate the modulus and arguments and solve for r and the basic value of. r n r 1 r n r1 n k, k 0, 1,.... 1 Step 4: Divide by the value of n (the power in n a ). Step 5: Add (or subtract) 1 ( 1 k ) n to the value of, ( n 1) times. n Step 6: Express the answers in the form rcis with as the principal angle. The School For Excellence 017 Succeeding in the VCE Unit 3 Specialist Maths Page 51

QUESTION 3 Solve the following equations using De Moivre s Theorem. Give your answer in Polar form. (a) 4 16 3 (b) 8 (c) 3 Solution 1 i (a) 4 16 n Express the left hand side of the equation in the form rcisn : 4 4 r cis 4 Express the right hand of the equation in polar form: 16 16cis 0 Equate the modulus and arguments and solve for r and the basic value of : r 4 16 4 0 r 0 Divide by the value of n (the power in n a ): 4 Add to the value of, ( n 1) times: n 0 3 Express the answers in the form rcis with as the principal angle: cis0 cis cis cis Note that 3 cis cis. The School For Excellence 017 Succeeding in the VCE Unit 3 Specialist Maths Page 5

3 (b) 8 (c) 3 1 i The School For Excellence 017 Succeeding in the VCE Unit 3 Specialist Maths Page 53

FINDING n th ROOTS OF A COMPLEX NUMBER AN ALTERNATIVE APPROACH In general, there are n distinct values for the n th root of a complex number. This is a consequence of the Fundamental Theorem of Algebra. When the roots are represented on an Argand diagram, they lie on the circumference n of a circle with radius r 1/ and centre at the origin. The roots are evenly spaced around this circle, each root being separated by an angle. n The n th roots of REAL numbers occur in conjugate pairs. The n th roots of non-real numbers do not occur in conjugate pairs. This is a consequence of the Conjugate Root Theorem. To find the n th root of a complex number in polar form: Step 1: Express in the form rcis( m ), where m 0, 1,, and Arg(). Step : Apply De Moivre s Theorem: m 0, 1,, 1/ n r 1/ n cis m, where n th Step 3: To obtain the n distinct n roots 1,, 3, n, substitute n consecutive values of m. For example, m 0,1,, 3, n 1: m = 0: m = 1: 1 r r 1/ n cis θ n 1/ n cis θ n m = : 3 m = n 1: r n 1/ n cis θ 4 n r 1/ n θ ( n 1) cis n Step 4: Where necessary, convert the polar form of the roots into Cartesian form. Spacing of solutions on the Argand plane: nth roots of any number will be apart. n Square roots apart (180 ). Cube roots apart (10 ). 3 Fourth roots apart (90 ). What this means is that if you are able to find one 4 The School For Excellence 017 Succeeding in the VCE Unit 3 Specialist Maths Page 54

solution, then by geometry you are able to find all the others. The School For Excellence 017 Succeeding in the VCE Unit 3 Specialist Maths Page 55

QUESTION 33 Plot the solutions to each of the following on the complex plane. (a) (b) Find the fourth roots of, give answers Cartesian form. Find the fifth roots of 3, give answers Polar form. Solution (a) Im() Re() (b) Im() Re() The School For Excellence 017 Succeeding in the VCE Unit 3 Specialist Maths Page 56

QUESTION 34A (a) Express 4 4 i in exact polar form. (b) Show that one of the cube roots of 4 4 i is u cis 4. (c) Find the remaining two cube roots of 4 4 i in exact polar form. (d) Express u cis in exact Cartesian form. 4 (e) Carefully plot the three cube roots of 4 4 i on the following Argand diagram. Solution The School For Excellence 017 Succeeding in the VCE Unit 3 Specialist Maths Page 57

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QUESTION 34B (VCAA 01) 3 Consider the equation 1 0, C. (a) Given that cis is a root of the equation, find the other two roots in the form 3 a ib, where a, b R. (b) Plot all of the roots clearly on the Argand diagram below. The School For Excellence 017 Succeeding in the VCE Unit 3 Specialist Maths Page 59

The School For Excellence 017 Succeeding in the VCE Unit 3 Specialist Maths Page 60

THE N TH ROOTS OF UNITY The n th roots of unity have the polar form cis m, where m 0,1,, 3, n 1. n ( m 0 ) is always an n th root of unity. The n th roots of unity are the solutions to the polynomial equation 1 0. Since a real solution to this equation is always 1, 1 is always a real factor of the n equation. By performing polynomial long division, it follows that 1 can be 3 n1 factorised as ( 1)(1 ). It follows that all non-real roots of unity 1 3 satisfy the polynomial equation n 1 0. n k If is a root of unity and k is an integer, then is also a root of unity. If n is even, then 1 is also an n th root of unity. Furthermore, if n is even then n k, where k is an integer, and so: n k k k k k1 3 k 1 1 1 ( 1) 1 1 1 1 is always a solution of 1 0, 1 k k 1 0. It follows that 1 is a factor of either 1 1 3 k 1. Since 1 k will always be a factor of or FINDING SQUARE ROOTS IN EXACT CARTESIAN FORM If we are required to find the square roots or cube roots of a given complex number then we can apply De Moivre s Theorem to find the solutions in polar form. For example, if asked find the cube roots of i, then we can express the equation as 3 i. The three solutions of would represent the cube roots of i. Similarly, if the square roots are required, then the equation can be expressed as where c represents the complex number. c When dealing with square roots, we can also find the solutions in Cartesian form by letting a ib so that a ib. The School For Excellence 017 Succeeding in the VCE Unit 3 Specialist Maths Page 61

QUESTION 35 Find the square roots of Solution 3 4i in exact Cartesian form. Let the required square roots have the form Then: a ib where a and b are real numbers. 3 4i ( a ib) 3 4i a abib 3 4i Equate real and imaginary parts of each side: Real part: a b 3 (1) Imaginary part: ab 4 ab () From equation (), b b 3 a. Substituting into equation (1) it follows that: b b 4 b 4 3 4 b 3b b 4 3b 4 0 ( b 4)( b 1) 0 b 1 0 or b 4 0 But b 4 0 has no real solution for b. Therefore b 1 a. Therefore the square roots of 3 4i are i and i. The School For Excellence 017 Succeeding in the VCE Unit 3 Specialist Maths Page 6

QUESTION 36 Find the square roots of Solution 5 1i in exact Cartesian form. Use technology to check your answers: The School For Excellence 017 Succeeding in the VCE Unit 3 Specialist Maths Page 63

QUESTION 37 The square roots of 4cis 3 are: A cis and cis 3 3 B 4cis and 4cis 3 3 C cis and cis 3 3 D 4 4 16cis and 16cis 3 3 E cis and cis 3 3 Solution Answer is C The School For Excellence 017 Succeeding in the VCE Unit 3 Specialist Maths Page 64

POLYNOMIALS OVER C THE FUNDAMENTAL THEOREM OF ALGEBRA Every polynomial p () of degree n can be factorised into n complex linear factors (these factors are not necessarily non-repeated). Hence, using the Null Factor Law, the polynomial equation p ( ) 0 of degree n has n complex solutions (these solutions are not necessarily all distinct). The polynomial p ( ) 3 1 has three complex linear factors. 3 3 Hence, 1 1 0 has three complex solutions. THE FACTOR THEOREM If p () is a polynomial and p ( ) 0, then is a factor of p (). THE CONJUGATE ROOT THEOREM If a polynomial equation p ( ) 0 has REAL coefficients and a ib is a solution, then a ib is also a solution; that is, eroes (roots) of a polynomial with REAL coefficients occur in conjugate pairs. If a polynomial p () has REAL coefficients and ( ) is a linear factor, then ( ) is also a linear factor. 3 3 For example, solutions to the equation 1 1 0 occur in conjugate pairs. Note: The conjugate root theorem IS NOT VALID if p () has NON-REAL coefficients. 3 3 For example, there are no conjugate pair solutions to the equation i i 0. The School For Excellence 017 Succeeding in the VCE Unit 3 Specialist Maths Page 65