1Number ONLINE PAGE PROOFS. systems: real and complex. 1.1 Kick off with CAS

1Numer systems: real and complex 1.1 Kick off with CAS 1. Review of set notation 1.3 Properties of surds 1. The set of complex numers 1.5 Multiplication and division of complex numers 1.6 Representing complex numers on an Argand diagram 1.7 Factorising quadratic expressions and solving quadratic equations over the complex numer field 1.8 Review

1.1 Kick off with CAS Exploring imaginary numers 1 Enter 1 into your CAS technology. If the calculator output mode is set to real numers, then the result for 1 will e Error: Non-real result. If the calculator output mode is set to rectangular or polar, then the result for 1 will e i. When exploring numers within the real numer field only, 1 is undefined. Within the field of complex numers, 1 = i. Therefore i = 1. With the calculator output mode set to rectangular or polar, simplify the following: a i 3 i c i 5 d i 6 e i 7 f i 8. 3 Complete the following: a i n =, for n = 1,, 3,,... i n+1 =, for n = 1,, 3,,... With the calculator output mode set to rectangular or polar, expand the following: a ( 3i)(3 i) (5 i)( + i) c (3 + i)(5 i) d ( 3i)( + 3i) e (i 1)(1 i) f (x yi)(x + yi). 5 With the calculator output mode set to rectangular or polar, factorise the following over the complex numer field: a x 10z + 5 x + z + 6 c 3x 7z + 7 d x + 5z + e 3x z 1. Please refer to the Resources ta in the Prelims section of your ebookplus for a comprehensive step-y-step guide on how to use your CAS technology.

think a 1. Units 1 & AOS Topic 1 Concept 1 Set notation Concept summary Practice questions WoRKEd ExAMPLE 1 Review of set notation Sets contain elements. In this topic the elements are numers. For example, the following are six elements: 1,, 3,, 5, 6. ξ is the universal set the set of all elements under consideration. So, in this example, ξ = {1,, 3,, 5, 6}. is the empty or null set. This set contains no elements. = {}. An upper case letter, such as A, represents a suset of ξ. In our example, A = {1, 3, 5} and B = {1,, 3, }. is read as is an element of. For example, 3 A. is read as is not an element of. For example, A. is read as is a suset of. For example, {1, 3} A. is read as is a superset of. For example, A {1, 3}. Related symols, such as, and, are also used. A is the complement of A. This set contains all the elements not in A that are in ξ. For example, given ξ = {1,, 3,, 5, 6}, if A = {1, 3, 5}, then A = {,, 6}. A B is the union of A and B. This set contains all the elements in sets A and B. For the example aove, A B = {1,, 3,, 5}. A B is the intersection of A and B. This set contains all the elements in oth A and B. For the example aove, A B = {1, 3}. C\D is read as C slash D. This set contains all the ξ A B elements in C that are not in D. If C = {1,, 5, 6} and D = {, 5}, then C\D = {1, 6}. This notation 1 is particularly useful in modifying a given set to 5 exclude a small numer of elements. 3 A Venn diagram may e used to illustrate 6 set notation. ξ = {,, 6, 8, 10, 1}, C = {, 8, 1} and D = {, 6, 10, 1}. a Illustrate these sets on a Venn diagram. Then state: C d C D f C D Draw a Venn diagram and enter the elements in the appropriate region. c C D e (C D) g C \{}. WritE/DrAW a ξ C D 8 1 6 10 MATHS QUEST 11 SPECIALIST MATHEMATICS VCE Units 1 and

c d e f The set C is the complement of C and contains all elements not in the set C. The set C D is the union of C and D and contains all elements in sets C and D. The set C D is the intersection of C and D and contains elements common to sets C and D. The set (C D) is the complement of the union of sets C and D. It contains elements not in the union of sets C and D. In this case, there are no elements not in the union of sets C and D. The set C D is the intersection of C and D. It contains elements common to the sets C and D. There are no common elements to C and D. g The set C \{} is the set C without the element. It contains all the elements of the set C ut not. Units 1 & AOS Topic 1 Concept Sets of numers Concept summary Practice questions WoRKEd ExAMPLE think Natural numers Numers were invented to quantify ojects in the environment. Hunter gatherers used counting or natural numers to communicate how many of a particular animal were seen on a hunting trip. The set of natural numers is given as N = {1,, 3,...}. Natural numers are positive whole numers. C = {, 6, 10} c C D = {,, 6, 8, 10, 1} d C D = {1} e (C D) = f C D = g C \{} = {6, 10} Proof of divisiility y mathematical induction Worked example demonstrates the proof of divisiility y mathematical induction. Prove that 6 n + is divisile y 5 for all n N. WritE 1 Let n = 1. 6 1 + = 10, which is divisile y 5 Assume that it is true for n = k. Assume that when n = k, 6 k + is divisile y 5 3 Consider n = k + 1. When n = k + 1, 6 k + 1 + = (6 k 6 1 ) + Need to add and sutract a numer that is a multiple of 6. (6 k 6 1 ) + = (6 k 6 1 ) + + 5 Factorise the expression so that 6 k + is inside the racket. = 6 1 (6 k + ) + = 6(6 k + ) 0 Topic 1 NUMER SySTEMS: REAL ANd CoMPLEx 5

6 From step, 6 k + is divisile y 5. Since oth 6 k + and 0 are divisile y 5, 6 k+1 + is divisile y 5. 7 State your conclusion. Since 6 n + is divisile y 5 for n = 1 and it is assumed to e true for n = k, which implies that it is true for n = k + 1, the statement is true for all whole numers, n. Prime and composite numers A prime numer is a positive natural numer that has exactly two factors: itself and one. Composite numers are positive natural numers that have more than two factors, including at least one prime factor. Is there an infinite numer of prime numers? Let us assume that there is a finite numer of prime numers, say p 1, p, p 3,..., p n. Let M = ( p 1 p p 3... p n ) + 1. M could either e a prime numer or a composite numer. If M is a prime numer, then p n cannot e the last prime numer. If M is a composite numer, then none of the primes p 1, p, p 3,..., p n can divide into it, since there will always e a remainder of 1. Hence a prime factor exists that is not in p 1, p, p 3,..., p n. Therefore, we can conclude that there are infinitely many prime numers. Integers and rational numers The systematic consideration of the concept of numer in algera, and the numers required to solve equations of the form x + = 0 and 3x + 1 = 0, resulted in the invention of integers and rational numers. The set of integers is given y Z = {..., 3,, 1, 0, +1, +, +3,...}. They are positive and negative whole numers, including zero. Z is the set of negative integers: Z = {..., 3,, 1}. Z + is the set of positive integers: Z + = {1,, 3,...}. Therefore, Z = Z {0} Z +. The set of rational numers is given y Q. These are numers of the form p q, where p Z and q Z \{0}. Whole numers are also rational numers. Consistent with the definition of Q, Z Q. Q is the set of negative rational numers. Q + is the set of positive rational numers. Therefore, Q = Q {0} Q +. Rational numers in their simplest form with denominators such as, 8, 16, 6 produce terminating decimals. Some examples include: 1 = 0.5, 3 8 = 0.375, 7 89 13 = 0.375, = 0.71, = 1.91 875 16 15 6 6 MATHS QUEST 11 SPECIALIST MATHEMATICS VCE Units 1 and

WoRKEd ExAMPLE 3 think a Rational numers in their simplest form with denominators such as 3, 6, 7, 9, 11, 13, 1, 15, 17 produce non-terminating recurring or repeating decimals. Some examples include: 1 3 = 0.333... = 0.3, 1 5 = 0.1666... = 0.16, = 0.16 66... = 0.16 6 1 17 99 = 0.171 717... = 0.17, 3 17 = 0.8 571 8 571... = 0.8 571, = 1.307 69 7 13 Using a calculator, express the following rational numers in decimal form. a 5 16 7 Since the denominator is 16, expect a terminating decimal. 1 Since the denominator is 7, expect a non-terminating, repeating decimal. Indicate the repeating sequence using dot notation. WritE a 5 16 = 0.315 = 0.571 8 571... 7 = 0.571 8 7 Irrational numers Irrational numers are given y I. They are numers that can e placed on a numer line and may e expressed as non-terminating, non-recurring decimals. For example:, 3, 5 + 1, 1 3, 5 3, π. Irrational numers cannot e written in the form p, where p Z and q Z \{0}. q Many irrational numers in decimal form, such as and π, have digits that have no pattern. For these numers, it is impossile to predict the next digit from the preceding digits. However, other irrational numers can e constructed with a pattern; for example: 0.101 100 111 000 111 100 00... and 0.010 110 111 011 11... There are two important susets of the set of irrational numers: the set of algeraic numers and the set of transcendental numers. Algeraic numers are those that are the solution of an algeraic polynomial equation of the form: a n x n + a n 1 x n 1 +... + a x + a 1 x + a 0, where a 0, a 1, a,..., a n 1, a n Z. For example, algeraic numers include 3 1 3 from one of the solutions of x 3 3 = 0 and 3 from x 8 = 0. Transcendental numers occur in the evaluation of some functions, such as trigonometric functions. For example, sin (3.1 ) and π are transcendental numers. The functions that produce these numers are often called transcendental functions. Topic 1 NUMER SySTEMS: REAL ANd CoMPLEx 7

Why is 7 an irrational numer? Assume that 7 is a rational numer and can e written in the form p, a fraction in q simplest form, where p Z and q Z \ 0. 7 = p q 7 = p q p = 7q Therefore, p is divisile y 7, which means that p is divisile y 7. Let p = 7k, k Z (7k) = 7q 9k = 7q q = 7k Therefore, q is divisile y 7, which means that q is divisile y 7. As p and q have a common factor of 7, this contradicts the fact that p is a fraction q written in simplest form. The assumption that 7 is rational is incorrect, hence 7 is an irrational numer. Real numers Finally, the set of real numers is given as R. R includes all numers that can e put on a numer line, where R = Q I. The Venn diagram shows the relationships etween R, Q, I, Z and N. ξ = R (real numers) Q (rational numers) Z (integers) N (natural numers) I (irrational numers) Asolute value of a real numer The asolute value a of a real numer a is the distance from a to zero. 8 = 8 11 = 11 = 8 MATHS QUEST 11 SPECIALIST MATHEMATICS VCE Units 1 and

WoRKEd ExAMPLE For each of the numers elow, using R, Q, I, Z and N, state all the sets for which they are a memer. think a 5 17 3 c 3 d 7.179 e.153 f 17.135... g 1.011 011 101 111... h 3 1 5 i 17 1 WritE a 5 is an integer. a 5 is a negative integer (Z ). It is also a rational numer (Q) and a real numer (R). 17 is a rational numer, as it can e written 3 as a fraction. 17 is a rational numer (Q) and a real 3 numer (R). c 3 is an irrational numer. c 3 is an irrational numer (I) and a real numer (R). d e f 7.179 is a rational numer, as it is a recurring decimal..153 is a rational numer, as it is a terminating decimal. 17.135... is an irrational numer as there is no indication that there is a recurring pattern. d 7.179 is a rational numer (Q) and a real numer (R). e.153 is a rational numer (Q) and a real numer (R). f 17.135... is an irrational numer (I) and a real numer (R). g 1.011 011 101 111... is an irrational numer. g 1.011 011 101 111... is an irrational numer (I) and a real numer (R). h 3 1 5 can e simplified to and is therefore a natural numer. h 3 1 5 is a natural numer (N). It is also an integer (Z), a rational numer (Q) and a real numer (R). i 17 1 is an irrational numer. i 17 1 is an irrational numer (I) and a real numer (R). WoRKEd ExAMPLE 5 think Express each of the following in the form a, where a Z and Z \{0}. a 0.6 0.3 WritE a 1 Write 0.6 in expanded form. a 0.6 = 0.666 666... [1] Multiply [1] y 10. 10 0.6 = 6.666 66... [] Topic 1 NUMER SySTEMS: REAL ANd CoMPLEx 9

3 Sutract [1] from []. 9 0.6 = 6 0.6 = 6 9 State the simplest answer. = 3 1 Write 0.3 in the expanded form. 0.3 = 0.3 33... [1] Multiply [1] y 100. 100 0.3 = 3.3 33... [] 3 Sutract [1] from []. 99 0.3 = 3 State the simplest answer. 0.3 = 3 99 WoRKEd ExAMPLE 6 think The asic properties of numer are assumed to e true if a counterexample cannot e found. For example, the statement the product of two integers is an integer is accepted as true ecause a counterexample has not een found, ut the statement the quotient of two integers is an integer is false ecause a counterexample is not an integer. 3 Determine counterexamples for the following. a The product of two irrational numers is irrational. The sum of two irrational numers is irrational. a Take a simple irrational numer such as. Multiply y an irrational numer, say. State your answer. 1 Take two irrational numers such as 0.1011 001 110 00... and 0.010 011 000 111.... Add these numers. WritE a Because =, which is a rational numer, the statement the product of two irrational numers is irrational is shown to e false. 0.101 100 111 000... + 0.010 011 000 111... = 0.111 111 111 111... The digits form a pattern so the sum is a rational numer. State your answer. Because 0.111 111 111 111... is a rational numer, the statement the sum of two irrational numers is irrational has een shown to e false. Standard form or scientific notation Very large or very small numers are conveniently expressed in standard form, a 10, where a R, 1 a < 10 and Z. For example, 1 3 111 = 1.3 111 10 6 and 0.000 000 000 05 =.5 10 11. decimal places and signifi cant fi gures The numerical answer to a calculation may e required to e given correct to a set numer of decimal places, and this is done through a process of rounding. To determine the numer of decimal places contained in a numer, count the numer 10 MATHS QUEST 11 SPECIALIST MATHEMATICS VCE Units 1 and

WoRKEd ExAMPLE 7 of digits after the decimal point. For example, 0.35 has decimal places. For numers expressed to a given numer of decimal places, rememer to round up if the next digit is 5 or more. For example, rounded to decimal places,.3 ecomes.3 and.36 ecomes.. To determine the numer of significant figures contained in a numer, count the numer of digits from the first non-zero digit. For example, 0.035 contains significant figures. Any zeros at the end of a numer after a decimal point are considered to e significant. For example, 1.0 has 3 significant figures. The trailing zeros at the end of a numer are not considered to e significant. For example, 000 has significant figures. For numers expressed to a given numer of significant figures, rememer to round. For example, rounded to significant figures,.3 ecomes. and.36 also ecomes.. Some examples are shown in the following tale. Numer significant figures 3 significant figures decimal places 3 decimal places 71 860.37 8 70 000 7 000 71 860. 71 860.38 1.38 9 1. 1. 1. 1.39 1.006 8 1.0 1.01 1.01 1.007 0.016 78 0.017 0.016 8 0.0 0.017 0.001 556 0.0016 0.001 56 0.00 0.00 0.199 1 0.0 0.199 0.0 0.199 Calculate the following products and quotients without using a calculator, expressing your answer in scientific notation correct to 1 significant figure. a 8 10 3 10 10 7 1017 8 10 10 think a 1 Multiply the terms y using the properties of indices: a n a m = a n + m. Write the answer in standard form, correct to 1 significant figure. 1 Multiply the terms y using the properties of indices: a n a m = a n m. WritE a 8 10 3 10 10 = 10 1 Write the answer in standard form, correct to 1 significant figure. 10 1 =. 10 10 1 =. 10 15 = 10 15 7 1017 = 0.875 107 10 8 10 0.875 10 7 = 0.9 10 7 or = 9 10 6 Topic 1 NUMER SySTEMS: REAL ANd CoMPLEx 11

Susets of the set of real numers Notation There are different forms of notation for representing susets. 1. Set notation For example {x: x Z, 1 x 5}, which is read as the set of numers x such that x is an element of the set of integers and x is greater than or equal to 1 and less than or equal to 5. If x R, it is not necessary to include the nature of x. For example, {x: x } represents the set of real numers greater than or equal to. Each of the two sets aove may e represented on a numer line as follows. 1 0 1 3 5 6 7 1 0 1 3 5 6 If x Q, the graph on the numer line appears to look like the corresponding graph for x R ecause the numer line appears to e continuous (although all irrational numers are missing). For example, {x: x Q, x } would appear to e identical to the graph of {x: x } shown aove. If individual numers are excluded from a given set, indicate this on a numer line y an open circle. If individual numers are included in a given set, indicate this on the numer line y a closed circle. For example, {x: x }\{3} is represented on a numer line elow. 3 1 0 1 3 A given set can e stated in more than one way using set notation. For example, {1,, 3,, 5} can e written as {x: x Z, 0 < x < 6}, {x: 1 x 5} or {x: x Z +, x 5}.. Interval notation Interval notation uses rackets, either square rackets [ 1, 6] or curved rackets (1, 6) to descrie a range of numers etween two numers. Square rackets include the numers at the end of the interval; curved rackets exclude them. For example: (3, 8) represents all real numers etween 3 and 8, excluding oth 3 and 8. [, 6] represents all real numers etween and 6, including oth and 6. [, 10) represents all real numers etween and 10, including and excluding 10. Example sets are illustrated on the following numer lines. x x x z a x [a, ] = {x : a x } a x [a, ) = {x : a x < } 1 MATHS QUEST 11 SPECIALIST MATHEMATICS VCE Units 1 and

a x (a, ) = {x : a < x < } a x (, a] = {x : x a} WoRKEd ExAMPLE 8 think Note: Curved rackets are always used for ±. a (a, ) = {x : x > a} x List the following sets and then express each set using set notation. Illustrate each set on a numer line. a {Integers etween 3 and } {Integers less than } a 1 This set involves integers. List the set of integers. Express the set using set notation. Draw a numer line showing arrowheads on each end. Ensure that the numers from 3 to are shown using an appropriate scale. Since the set of integers is to e represented, do not join the dots. 1 This set involves integers. List the set of integers. Express the set using set notation. Draw a numer line showing arrowheads on each end. Ensure that the numers from and elow are shown, using an appropriate scale. Since the set of integers is to e represented, do not join the dots, ut show an arrow on the left side of 3. WoRKEd ExAMPLE 9 WritE/DrAW Use set notation to represent the following sets. a {Rational numers greater than 7} a {, 1, 0, 1,, 3} = {x: x Z, 3 < x < } 3 1 0 1 3 5 6 x {...,, 1, 0, 1} = {x: x Z, x < } 3 1 0 1 3 x z {Integers etween and including oth 100 and 300, except for 00} c {Positive integers less than 9 and greater than 50} d {Real numers that are less than 7 and greater than } e {Positive real numers that are less than or greater than 7} Topic 1 NUMER SySTEMS: REAL ANd CoMPLEx 13

think a c d e think a The numers in this set elong to the set of rational numers, Q. The numers in this set elong to the set of integers, Z. Exclude 00. The numers in this set elong to the set of positive integers, Z +. Express the set of positive integers less than 9 and greater than 50 as the union of two sets. The numers in this set elong to the set of real numers, R, that are less than 7 and greater than. The numers in this set elong to the set of positive real numers, R +, that are less than or greater than 7. WoRKEd ExAMPLE 10 ExErCisE 1. PrACtisE Use interval notation to represent the following sets. a {x: < x 3} {x: x } c {x: 3 < x 5} {x: x < 7} d {x: 3 < x 5} {x: x < 7} x R. Only the end point 3 is included; therefore, use a square racket. x R. Negative infinity is always preceded y a round racket when using interval notation. Review of set notation WritE a {x: x Q, x > 7} WritE a (, 3] (, ] c x R. Only the inner end points are included. c (3, 5] [, 7) = (3, 7) d x R. Only the inner end points are included. d (3, 5] [, 7) = [, 5] 1 WE1 If ξ = {1,, 3,, 5, 6}, A = {1, } and B = {, 3}, show these on a Venn diagram, and then state the following sets. a A A B c A B d A\{} If ξ = {, 8, 1, 16, 0,, 8, 3, 36}, A = {, 8, 0} and B = {0,, 8, 3, 36}, show these on a Venn diagram, and then state the following sets. a B A B c A B d (A B) 3 WE Prove that n 1 is divisile y 3 for all n N. Prove that n 3 + n is divisile y 3 for all n N. {x: x Z, 100 x 300}\{00} c {x: x Z +, x < 9} {x: x Z +, x > 50} d {x: x < 7} {x: x > } or, more simply, {x: < x < 7} e {x: 0 < x < } {x: x > 7} or R + \{x: x 7} 5 WE3 Use a calculator to express the following rational numers in decimal form. a 13 15 6 1 MATHS QUEST 11 SPECIALIST MATHEMATICS VCE Units 1 and

6 Use a calculator to express the following rational numers in decimal form. a 17 16 1 13 7 WE For each numer, using Z, N, Q, I and R, state all the sets of which it is a memer., 16 8, 1 16, 3, 6 3, 161 7 8 Specify to which sets (Z, N, Q, I and R) each of the following numers elong. 5 1 5, π, 1.7,.567,.135 18 97...,.3 33 333. 9 WE5 Express each of the following in the form a, where a Z and Z \ 0. a 0.1.13 10 Express the following in their simplest rational form. a 0. 1.13 c 0.13 d.113 11 WE6 Find a counterexample, if possile, for the following statements. If a counterexample is found, the statement is false (F). If a counterexample is not found, accept the statement to e true (T). a The product of two integers is an integer. The division of an integer y an integer is a rational numer. 1 Find a counterexample, if possile, for the following statements. If a counterexample is found, the statement is false (F). If a counterexample is not found, accept the statement to e true (T). a The difference of two irrational numers is irrational. The sum of an irrational numer and a rational numer is irrational. 13 WE7 Calculate the following products and quotients without using a calculator, expressing your answer in scientific notation to 1 significant figure. a 1.5 10 16 10 1 1. 10 3 10 10 c 3. 10 5 10 15 1 Calculate the following products and quotients without using a calculator, expressing your answer in scientific notation to 1 significant figure. a 7 10 1 9 10 8 8 1017 c.5 101 8 10 7 10 10 5 10 8 15 WE8 List the following sets and then express each set using set notation. Then illustrate each set on a numer line. a {Integers etween 6 and 1} {Integers from 3 to } c {Integers greater than 6 and less than or equal to } d {Positive integers less than 5} 16 List the following sets and then express each set using set notation. Then illustrate each set on a numer line. a {Integers less than 5} {Integers greater than } c {Negative integers greater than 5} 17 WE9 Use set notation to represent the following sets. a {Rational numers greater than 5} {Rational numers greater than 5 and less than or equal to 0} c {Positive rational numers less than 0} d {Integers etween 5 and 0, except for 8 and 9} e {Positive integers less than 100, except for integers etween 0 and 50} Topic 1 Numer systems: real and complex 15

Consolidate 18 Use set notation to represent the following sets. a {Real numers from to 5, including } {Real numers that are less than 5 and greater than 3} c {Real numers that are less than 3 and greater than 7} d {Positive real numers that are less than 3 and greater than 7} 19 WE10 Use interval notation to represent the following sets, then illustrate the sets on a numer line. a {x: 3 x 1} {x: x < } c {x: < x < 1} d {x: x } 0 Use interval notation to represent the following sets, then illustrate the sets on a numer line. a {x: x < 5} {x: x < 6} {x: x < 5} {x: x < 6} c {x: x < 5} {x: < x 6} d {x: x > 5} {x: < x 6} 1 Copy the Venn diagram at right and then shade the region represented y each of the following sets. ξ A B a A A B c A B d (A B) \ (A B) e A B f A B g (A B) Complete the following tale. Numer 167.1066 7.6699 8.000 56 0.999 87 0.076 768 0.000 17 95 3 significant figures significant figures 3 Simplify the following. a 7 decimal places 3 5 6 + 6 The smallest suset of R in which 3 7 elongs is: A Z + B Z C Q + D Q E I 8 3 decimal places 5 The smallest suset of R in which elongs is:.567 A Z + B Z C Q + D Q E I 6 If ξ = {1,, 3,, 5, 6, 7, 8}, A = {1,, 3, } and B = {5}, then A \B is: A {1,, 3,, 5} B {5, 6, 7, 8} C D {6, 7, 8} E {1,, 3,, 5, 6, 7, 8} 7 3.010 and 9 57 to significant figures are: A 3.01 and 9 50 B 3.010 and 9 50 C 3.01 and 9 60 D 3.010 and 9 60 E 3.010 and 9 57 9 16 MATHS QUEST 11 SPECIALIST MATHEMATICS VCE Units 1 and

MAstEr 1.3 Units 1 & AOS Topic 1 Concept 3 Surds Concept summary Practice questions WoRKEd ExAMPLE 11 think 8 0.3, 0.3 and 0.3 33 333... respectively elong to which of the following sets? A Z, Z, I B Q, Q, I C Q, I, I D Z, Q, I E Q, Q, Q 9 Which of the following sets is an incorrect representation of the set {all integers from 1 to 5}? A {1,, 3,, 5} B {x: x Z, 1 x 5} C {x: x Z, 1 x < 6} D Z + \{x: x Z, x 6} E [1, 5] 30 For the set illustrated on the given x numer line, which of the following 5 3 1 0 1 3 5 cannot e true? A ( 5, 5] B {x: 5 < x 5} C {x: x Q, 5 < x 5} D {Real numers from 5 to 5, not including 5} E [ 5, 5] 31 Calculate the following products and quotients using a calculator, expressing your answer in scientific notation to 3 significant figures. a 1.57 10 1 3.6677 10 9 8.583 10 5 9.57 10 7 5.7789 1017 c d.578 101 (8.775 10 7 + 7.3 10 6 ).6999 10 10 5.878 10 13 3 a Using your calculator, investigate the percentage of prime numers: i etween 1 and 9 ii etween 10 and 99 iii etween 100 and 999. What conclusion can you make aout the percentage of prime numers etween 10 n and 10 n+1 1 as n? Properties of surds A surd is an irrational numer of the form n a, where a > 0 and n Z +. In this section we will focus on the surds of the form a, where a Q. For example, 1 is a surd, ut 36 = 6 is a rational numer and not a surd. Simplifying surds cannot e simplified ecause it does not have a perfect square factor, ut 8 can e simplified since 8 = = = =. A surd is not simplified until all perfect square factors are removed, so the simplified version of 3 is not 8 ut. Simplify the following surds. a 38 3 05 c 1 8 a 1 Express 38 as a product of two factors where one factor is the largest possile perfect square. WritE a 38 = 6 6 Express 6 6 as the product of two surds. = 6 6 3 Simplify the square root from the perfect square (that is, 6 = 8). 1 Express 05 as a product of two factors, one of which is the largest possile perfect square. = 8 6 3 05 = 3 81 5 175 Topic 1 NUMER SySTEMS: REAL ANd CoMPLEx 17

Express 81 5 as a product of two surds. = 3 81 5 3 Simplify 81. = 3 9 5 Multiply the whole numers outside the root. = 7 5 c 1 Express 175 as a product of two factors where one factor is the largest possile perfect square. WoRKEd ExAMPLE 1 think Addition and sutraction of surds Only like surds may e added or sutracted. Like surds, in their simplest form, have the same numer under the square root sign. For example, 5 3 + 7 3 = (5 + 7) 3 = 1 3 and 5 3 7 3 = (5 7) 3 = 3. Simplify each of the following expressions involving surds. Assume that a and are positive real numers. a 3 6 + 17 6 6 5 3 + 1 5 + 3 8 c 1 100a 3 + a 36a 5 a WritE c 1 8 175 = 1 8 5 7 Express 5 7 as a product of surds. = 1 8 5 7 3 Simplify 5. = 1 8 5 7 Multiply the numers outside the square root. = 5 8 Units 1 & AOS Topic 1 Concept Addition and sutraction of surds Concept summary Practice questions a All three terms contain the same surd a 3 6 + 17 6 6 = (3 + 17 ) 6 ( 6), so group like terms and simplify. = 18 6 1 Simplify the surds where possile. 5 3 + 1 5 + 3 8 = 5 3 + 3 5 + 3 = 5 3 + 3 5 + 3 = 5 3 + 3 5 + 6 Collect the like terms. = 9 3 + c 1 Simplify the surds where possile. c 1 100a 3 + a 36a 5 a = 1 10 a a + a 6 a 5 a = 1 10 a a + a 6 a 5 a = 5a a + 6a a 10a Add the like terms. = 11a a 10a 7 18 MATHS QUEST 11 SPECIALIST MATHEMATICS VCE Units 1 and

Units 1 & AOS Topic 1 Concept 5 Multiplication and division of surds Concept summary Practice questions WoRKEd ExAMPLE 13 think Multiplication of surds Using the property a = a, where a, R +, 6 = 1 = 3 = 3. Using the distriutive property a( + c) = a + ac, ( 3 + 6) = 3 + 6 = 6 + 1 = 6 + 3. Using an extension of the distriutive property, ( 3 + 1) ( 3 ) = 3 3 3 + 3 = 3 3 = 1 3. When appropriate, the expansion of a perfect square may e used; that is, (a + ) = a + a + and (a ) = a a +. For example, ( 3 ) = 3 3 + = 5 6. defi nition of the conjugate The conjugate of a + is a. The conjugate of 3 5 is 3 + 5. The product of a conjugate pair is rational if the numers under the square root are rational. For example, ( 3 + ) ( 3 ) = 3 3 3 + 3 = 3 6 + 6 = 1. This is a special case of the difference of perfect squares expansion, (a + )(a ) = a. Multiply the following surds, expressing answers in simplest form. a 6 1 6 3 5 WritE 70 1 a 1 Write the expression and simplify 1. a 6 1 6 = 6 3 6 = 6 3 6 = 1 3 6 Multiply the coefficients and multiply the surds. = 18 3 Simplify the product surd. = 9 = 3 = 7 1 Multiply the coefficients and multiply the surds. 3 5 70 1 = 3 0 Simplify the product surd. = 3 0 3 Simplify y dividing oth 10 and 0 y 10 (cross-cancel). 10 10 = 3 5 1 70 10 700 100 7 = 3 0 10 7 = 3 7 or 3 7 Topic 1 NUMER SySTEMS: REAL ANd CoMPLEx 19

WoRKEd ExAMPLE 1 Expand and simplify the following where possile. a 7( 18 3) 3( 10 5 3) c ( 5 + 3 6)( 3 ) think division of surds a = WritE a 1 Write the expression. a 7( 18 3) Simplify 18. = 7(3 3) 3 Expand the racket. = 7 3 + 7 3 Simplify. = 3 1 3 7 1 Write the expression. 3( 10 5 3) i Expand the rackets. ii Be sure to multiply through with the negative. a, where a, R+. For example, = 3 10 3 5 3 = 30 + 10 9 3 Simplify. = 30 + 10 3 = 30 + 30 c 1 Write the expression. c ( 5 + 3 6) ( 3 ) Expand the rackets. = 5 3 + 5 + 3 6 3 + 3 6 3 Simplify. = 15 10 + 6 18 3 1 = 15 10 + 6 3 3 3 = 15 10 + 18 6 3 6 = 6 = 3. a Using the property = a = a, where a and are rational, we can express answers with rational denominators. For example, 6 = 6 6 6 = 1 = 3 = 3 6 6 3 Using the property of conjugates, inomial surds in the denominator may e rationalised. For example, 7 7 + = 7 7 + 7 7 1 1 + 11 3 1 = = 7 7 5 7 By multiplying the original surd y, we are multiplying y 1, so the 7 numer is unchanged ut is finally expressed in its rational denominator form. 0 MATHS QUEST 11 SPECIALIST MATHEMATICS VCE Units 1 and

WoRKEd ExAMPLE 15 think Express the following in their simplest form with a rational denominator. a 9 88 6 99 6 13 c WritE 1 6 3 + 1 3 6 + 3 a 1 Rewrite the surds, using a = a. a 9 88 6 99 = 9 6 Simplify the fraction under the root. = 9 6 3 Simplify the surds. = 9 6 3 Multiply the whole numers in the numerator and those in the denominator and simplify. 1 Write the fraction. 6 13 Multiply oth the numerator and the denominator y the surd 13. = 88 99 8 9 6 = 13 13 13 = 78 13 c 1 Write the first fraction. c 1 = 6 3 Multiply the numerator and the denominator y the conjugate of the denominator. 1 = 6 3 6 + 3 6 + 3 3 Expand the denominator. = 6 + 3 () 6 3 Simplify the denominator. = 6 + 3 1 1 5 Write the second fraction. 3 6 + 3 6 Multiply the numerator and the denominator y the conjugate of the denominator. = 1 3 6 + 3 3 6 3 3 6 3 7 Expand the denominator. = 3 6 3 3 6 3 8 Simplify the denominator. = 3 6 3 Topic 1 NUMER SySTEMS: REAL ANd CoMPLEx 1

9 Add the two fractions together. Find the lowest common denominator first. Exercise 1.3 PRactise Properties of surds 1 WE11 Simplify the following surds. a 56 c 15 d 98 e 8 Simplify the following surds. 18 a 300 7 80 c 3 WE1 Simplify the following expressions. d 18 5 a 7 + 3 5 6 3 + 5 7 6 7 c 3 5 6 3 + 5 5 8 5 d 18 1 + 75 + 7 Simplify the following expressions. e 3 50 10 a 50 7 + 80 + 5 3 1 5 18 + 7 + 5 98 c 3 3 + 5 3 5 d 7 3 3 + 5 8 5 8 8 5 5 3 5 WE13 Express the following surds in their simplest form. a 6 15 3 5 7 6 Simplify the following surds. 0 15 a 7 3 1 3 7 WE1 Expand, giving your answers in their simplest form. a 3( 5 ) 3(3 3 + ) c ( 5 3) ( 5 ) d ( 18 1) ( 3 ) 8 Expand and simplify. a ( 5 + 7) ( 1 + 3 18) c ( 5 3) ( 5 + 3) d (5 5 10) (5 5 + 10) 9 WE15 Express the following surds in their simplest form with a rational denominator. a 18 3 d 3 7 5 3 3 e 8 3 1 6 + 3 1 = 6 + 3 1 = 6 + 3 10 Add the numerators. = 7 6 11 Simplify where appropriate. = 6 6 + 3 6 3 + 3 6 3 + 3 6 3 c f 5 3 1 5 3 MATHS QUEST 11 SPECIALIST MATHEMATICS VCE Units 1 and

Consolidate 10 Rationalise the denominators. a d 3 3 1 8 1 + 8 5 + 3 e 5 3 5 3 c 5 + 3 5 3 f 18 3 8 5 11 Express the following surds in their simplest form with a rational denominator. 1 a 3 + 1 1 + 3 3 3 1 + 3 3 c 3 5 3 3 5 1 + 3 3 e + 3 3 5 3 6 + 3 3 d 5 + 3 3 3 5 3 3 5 3 5 + 3 3 f 5 + 3 3 3 5 3 3 5 + 3 5 + 3 3 1 Given that x = 3, find each of the following, giving the answer in surd form with a rational denominator. a x + 1 x 1 c x x x x x + d x + x e x x 1 f x x 9 x + 3 g Using your answers to e and f, state if 3 x x 1 = 0 and x x 9 = 0. is a solution of 13 Show that 5 3 is a solution of one of the following equations: x 13x + 10 = 0 or x 10x + 13 = 0. 1 Show that + 1 is a solution of oth of the following equations: x x + 1 = 0 and x ( + 3)x + + 3 = 0. 15 Expressed in its simplest form, 3 5 75 3 7 1 8 equals: A 3 B 3 C 7 3 D 0 E 3 3 1a 16 Expressed in its simplest form, 3 equals: 7a A a B a C a D a 0 E a 17 Expressed in its simplest form, (3 3 8)( 3 3 8) equals: A 11 3 6 B 10 3 6 C 78 17 D 18 E 18 3 6 18 Expressed in its simplest form, A 5 3 D 5 6 15 1 6 1 equals: B 5 6 E 5 6 C 5 3 Topic 1 Numer systems: real and complex 3

19 Expressed in its simplest form, 5 + 3 5 3 equals: A D 13 + 3 15 1 + 3 15 B E 1 15 13 + 15 C 18 3 15 Master 1. Units 1 & AOS Topic 1 Concept 6 Complex numers Concept summary Practice questions 0 Expressed in its simplest form, 3 5 5 3 5 + 5 3 5 + 5 3 5 5 equals: A 3 5 B 30 5 C 30 5 D 5 30 5 E 3 5 1 An ice-cream cone with measurements as shown is completely filled with ice-cream, and has a hemisphere of ice-cream on top. a Determine the height of the ice-cream cone in simplest surd form. Determine the volume of the ice-cream in the cone. c Determine the volume of the ice-cream in the hemisphere. d Hence, find the total volume of ice-cream. A gold ar with dimensions of 5 0, 3 1 and 6 cm is to e melted down into a cylinder of height 10 cm. a Find the volume of the gold, expressing the answer in the simplest surd form and specifying the appropriate unit. Find the radius of the cylinder, expressing the answer in the simplest surd form and specifying the appropriate unit. c If the height of the cylinder was 3 0 cm, what would e the new radius? Express your answer in the simplest surd form. The set of complex numers 7 cm The need to invent further numers ecame clear when equations such as x = 1 and x = 9 were considered. Clearly there are no real solutions, so imaginary numers were invented, using the symol i, where i = 1. The equation x = 1 has two solutions, x = i and x = i. As 9 = 9 1 = 9 1 = 3 i = 3i, x = 9 has the solutions x = ±3i. 175 cm Quadratic equations such as x x + 5 = 0 were investigated further. Using the general formula for the solution of a quadratic equation, that is, if ax + x + c = 0, then x = ± ac ± 16 0, x = = ± = ± i = ± i. If the a discriminant, ac, is negative, the equation has no real solutions, ut it does have two complex solutions. A complex numer is any numer of the form x + yi, where x, y R. C is the set of complex numers where C = {x + yi : x, y R}. MATHS QUEST 11 SPECIALIST MATHEMATICS VCE Units 1 and

Just as x is commonly used in algera to represent a real numer, z is commonly used to represent a complex numer, where z = x + yi. If x = 0, z = yi is a pure imaginary numer. If y = 0, z = x is a real numer, so that R C. This is represented on the Venn diagram at right. ξ Q Z N R I C WoRKEd ExAMPLE 16 Units 1 & AOS Topic 1 Concept 7 Operations with complex numers Concept summary Practice questions Notation If z = a + i, the real component of z is Re(z) = a, and the imaginary component of z is Im(z) =. For example, if z = 3i, Re(z) = and Im(z) = 3 (not 3i). Similarly, Re( + 6i) = and Im( + 6i) = 6. Equality of complex numers If a + i = c + di, then a = c and = d. For two complex numers z 1 and z to e equal, oth their real and imaginary components must e equal. If (x + ) + (y )i = (x + y) + xi, find x and y. think 1 Let the real parts e equal and the imaginary parts e equal to form two equations. WritE Re: x + = x + y [1] Im: y = x [] Rearrange the linear simultaneous equations. x y = [3] x + y = [] 3 Add equations [3] and [] to solve for x. x = x = 1 Sustitute x = 1 into equation [] to find the y = 1 value of y. y = 3 5 State the answer. x = 1, y = 3 Multiplication of a complex numer y a real constant If z = a + i, then kz = k(a + i) = ka + ki. For example, if z = + 3i, then 3z = 3( + 3i) = 6 9i. Adding and sutracting complex numers If z 1 = a + i and z = c + di, then z 1 + z = (a + c) + ( + d)i and z 1 z = (a c) + ( d)i. Modulus of a complex numer The magnitude (or modulus or asolute value) of a complex numer z = x + yi is denoted y z and z = x + y. Topic 1 NUMER SySTEMS: REAL ANd CoMPLEx 5

WoRKEd ExAMPLE 17 If z 1 = 3i and z = 3 + i, find: a z 1 + z 3z 1 z c z 1 d z. think a Use the definition for addition of complex numers: z 1 + z = (a + c) + ( + d)i First multiply each complex numer y the constant and then use the definition for sutraction of complex numers to answer the question. ExErCisE 1. PrACtisE ConsoliDAtE The set of complex numers 1 WE16 Solve to find x and y in the following. a (x + 1) + (y 1)i = + 3i c (x + i) + (3 yi) = x + 3i Solve the following. a (x + 3yi) + (x + yi) = 3 + i (x + i) + ( + yi) = x + 3yi c (x 3i) + ( 3 + y)i = y xi 3 WE17 If z 1 = i and z = 3 + i, find: WritE c Use z = x + y. c z 1 = + 3 (x + ) (3 + yi) = + 5i d (x + i) + (y + xi) = 7 + i a z 1 + z z 1 + 3z c z 1 d z. If z 1 = 3 i and z = 3i, evaluate the following. a z 1 3z z 1 + z c z 1 + 3 z 5 Express the following in terms of i. a 16 7 c + 0 6 Simplify. a 10 + 10 1 8 7 Simplify the following numers. a 5 9 + c 11 81 8 State the values of Re(z) and Im(z) for the following. a 3 + i + i c ( 1) + ( + 1)i 9 State the values of Re(z) and Im(z) for the following. a z 1 + z = ( 3) + ( 3 + )i = 1 + i 3z 1 z = 3( 3i) ( 3 + i) = 6 9i ( 1 + 16i) = 18 5i = 13 d Use z = x + y. d z = ( 3) + = 5 z = 5 a 8 0 6 c 13i 6 MATHS QUEST 11 SPECIALIST MATHEMATICS VCE Units 1 and

MAstEr 1.5 WoRKEd ExAMPLE 18 think 10 Find the following components. a Re( + 3i + 3( i)) Re( 3 + i + ( 3 3i)) c Im(( 3i) 3( i)) d Im( 3 i + ( 3 6i)) 11 If z 1 = i and z = 3 i, then Re(z 1 3z ) equals: A 13 B 13 C 5 D 5 E 1 If z 1 = i and z = 3 i, then Im(z 1 3z ) equals: A i B C D 8 E 8i 13 If ( + xi) + ( 3i) = x + 3yi, then the respective values of x and y are: A 6, 1 B 3, 6 C 6, 3 D 6, 3 E 1, 6 1 Simplify (1 i) + (3 + 6i) (10 + 10i). 15 Find the error in the student s work shown elow. (7 3i) + ( + i) 9 i 9 ( 1) 9 + 1 10 16 Simplify (13 5i) + (7 5i), giving the answer in simplest form of a + i. 17 Find the imaginary part of 1 i + (10 5i). Multiplication and division of complex numers Multiplication of complex numers If z 1 = a + i and z = c + di where a,, c, d R, then z 1 z = (a + i)(c + di) = ac + adi + ci + di = (ac d) + (ad + c)i Note that this is an application of the distriutive property. Simplify: a i( 3i) ( 3i)( 3 + i). WritE a Expand the rackets. a i( 3i) = i 6i Expand the rackets as for inomial expansion and simplify. = 6 + i ( 3i)( 3 + i) = 6 + 8i + 9i 1i = 6 + 17i Topic 1 NUMER SySTEMS: REAL ANd CoMPLEx 7

WoRKEd ExAMPLE 19 think The conjugate of a complex numer If z = a + i, then its conjugate, z, is z = a i. The sum of a complex numer and its conjugate z + z = a + i + a i = a, which is a real numer. The product of a complex numer and its conjugate: If z 1 = + 3i and z = 5i, find: zz = (a + i)(a i) = a (i) = a +, which is a real numer. a z 1 + z z 1 + z c z 1 z d z 1 z. a 1 Determine the conjugate of each complex numer using the definition: if z = a + i, then its conjugate, z = a i. WritE a z 1 = + 3i z 1 = 3i z = 5i z = + 5i Evaluate z 1 + z. z 1 + z = ( 3i) + ( + 5i) = + i 1 To evaluate z 1 + z, first evaluate z 1 + z. z 1 + z = ( + 3i) + ( 5i) = i Evaluate z 1 + z. z 1 + z = + i c Evaluate z 1 z using inomial expansion. c z 1 z = ( 3i)( + 5i) = 8 + 10i + 1i 15i = 7 + i d Evaluate z 1 z first and then evaluate z 1 z. d z 1 z = ( + 3i)( 5i) = 8 10i 1i 15i = 7 i z 1 z = 7 + i division of complex numers If z 1 = a + i and z = c + di where a,, c, d R, then using the conjugate: z 1 = a + i z c + di = a + i c + di c di c di 8 MATHS QUEST 11 SPECIALIST MATHEMATICS VCE Units 1 and

= ac adi + ci di c (di) (ac + d) (ad c)i = c + d = ac + d ad c c + d c + d i WoRKEd ExAMPLE 0 think Express each of the following in the form a + i. a i c (3 i) 1 3 i 3i d 3i + i WritE a Divide each term of the numerator y. a i = 1 i Multiply the numerator and denominator y i 3 i i and then divide each term of the numerator y 3. 3i i = 3i 3 Write the answer in the required form a + i. = 3 i c 1 Express (3 i) 1 as a reciprocal. c (3 i) 1 = 1 3 i Multiply the numerator and denominator y the = 1 conjugate of the denominator. 3 i 3 + i 3 + i = 3 + i 9 (i) = 3 + i 9 i = 3 + i 13 3 Write the answer in the required form a + i. = 3 13 + 13 i d 1 Multiply the numerator and denominator y the conjugate of the denominator. d 3i = 3i i + i + i i = i 6i + 3i i = 1 8i 5 Write the answer in the required form a + i. = 1 5 8 5 i Topic 1 NUMER SySTEMS: REAL ANd CoMPLEx 9

Exercise 1.5 PRactise Consolidate Multiplication and division of complex numers 1 WE18 Simplify the following, giving each answer in its simplest a + i form. a i( + 3i) ( 3i)(1 + i) c ( i)(1 3i) d ( 3i) e (6 + 7i)(6 7i) Expand and simplify: a i(3 i) (3 + 5i)(3 3i) c (a + i)(a i). 3 WE19 Give the conjugate of the following complex numers. a 3 + i + 3i c i d 8i If z 1 = 3i and z = 3 i, evaluate the following, giving each answer in its simplest a + i form. a z 1 z 1 z c z 1 z d (z 1 ) e i z f (z 1 + z ) 5 WE0 Express each of the following in the form a + i. a 3 i 3 + i c 1 + i 5i 3 i + i ( + i) d e (3 + i) 1 f (3 + i) 1 + i 6 Simplify each of the following if z 1 = 3i and z = 3 i. a (z 1 ) 1 z 1 z 1 z d e z 1 + 1 f z 1 z z z 1 z 1 z 7 If z 1 = a + i and z = c + di, show that: a z 1 z = z 1 z z 1 + z = z 1 + z c z 1 = z 1. z z 8 Find x and y in each of the following. a (x + yi)( + i) = 3 + 6i 9 Solve for z. a ( + 3i)z = i c x + yi 1 + i = 1 + i z 1 z ( 3i)z = 3 i 10 For each of the following, state z and find z 1, then state z 1 in terms of z. a z = + 5i z = a + i 11 Expressed in a + i form, ( 3 3i) (3 3 i) equals: A 13 3i B 1 13 3i C (6 3 + 6) 13 3i D (6 3 6) 13 3i E 1 5 3i 1 Expressed in a + i form, 3 3i 3 3 i equals: A 31 5 3 31 i B 1 31 5 3 31 i C 31 13 3 31 i D 3 5 3 31 i E 1 3 5 3 3 i 1 30 MATHS QUEST 11 SPECIALIST MATHEMATICS VCE Units 1 and

13 Expressed in a + i form, (1 + i) + (1 + i) equals: A 5 i B 9 7 i C 0 D 9 9 i E 3 i MAstEr 1.6 Units 1 & AOS Topic 1 Concept 8 Argand diagrams Concept summary Practice questions WoRKEd ExAMPLE 1 1 If (x + yi)(3 i) = + 5i, then the respective values of x and y are: A 3 and 13 13 B 1 3 and 13 13 C 7 3 and 13 13 D 1 3 and E 3 and 5 5 5 5 15 Evaluate the following, giving each answer in the form a + i. 1 a 3i + 1 (3 i) 1 c + 3i + + 3i ( i) + 3i 16 Evaluate the following, giving each answer in the form a + i. 3 i a 3 + i + 3 + i i 3 i 3 + i 3 + i i 17 If z = 1 + i, simplify z 3 + z + z + 1, giving answer in the form a + i. 18 Write + 5i in the form a + i. 3 + i Representing complex numers on an Argand diagram Real numers can e represented on a numer line, ut complex numers, with their real and imaginary components, require a plane. The Argand diagram or Argand plane has a horizontal axis, Re(z), and a vertical axis, Im(z). The complex numer z = a + i is represented y the point (a, ). Because of the similarities with the Cartesian plane, a + i is referred to as the Cartesian or rectangular form. The complex numers + 3i,, 3i and i are shown on the Argand plane at right. Im(z) a Express the following in their simplest form: i 0, i, i, i 3, i, i 5. + 3i 0 Re(z) 3i i Use the pattern in these results to find the simplest form for i 8, i 1 and i 63. c Illustrate the points from part a on an Argand plane, and state their distance from the origin and the angle of rotation aout the origin to rotate from one power of i to the next. think a Use i = 1 and index law rules to simplify each term. WritE/DrAW a i 0 = 1 i = 1 i = i i = 1 i 1 = i i 3 = i i = i i 5 = i i = i Topic 1 NUMER SySTEMS: REAL ANd CoMPLEx 31

The pattern repeats as shown in part a. i 8 = (i ) c 1 Rule up a pair of laelled, scaled axes for the Argand plane. Place each of the points from part a onto the plane and lael them. Determine the distance of each point from the origin. 3 State the angle of rotation aout the origin to rotate from one power of i to the next. Exercise 1.6 PRactise Consolidate = 1 i 1 = (i 5 ) i Representing complex numers on an Argand diagram 1 WE1 Give the following in their simplest form. a i 7 i 37 c i d i 15 Give the following in their simplest form. c = i i 63 = (i 15 ) i 3 = i 3 = i Im(z) i 0, i 1 i i, i 5 0 1 1 1 Re(z) a (i) 6 ( i) 8 c (i) 9 d ( i) 9 3 Plot the following points on an Argand plane. a + 3i 3i c + 3i d 3i e 3i f Write down the complex numer represented y Im(z) the points A to F on the Argand diagram at right. D 5 If z 1 = + 5i and z = 5 7i, find the simplest F A algeraic expression for each of the following and E represent them on an Argand diagram. a z 1 + z B 0 Re(z) z 1 z C 6 Assume z 1 = 3 + 5i and z = z 1 to answer the following. a Sketch z 1 and z on an Argand diagram. State the angle of rotation aout the origin to rotate from z 1 to z. 7 If z = 3i, show on an Argand diagram: a z z c z. i 3 All points are 1 unit from the origin. The angle of rotation aout the origin to rotate from one power of i to the next is 90 in an anticlockwise direction. 3 MATHS QUEST 11 SPECIALIST MATHEMATICS VCE Units 1 and

Master 1.7 Units 1 & AOS Topic 1 Concept 9 Factorising quadratic expressions over the complex numer field Concept summary Practice questions 8 Represent the following on an Argand diagram if z = i. a zi z c z 1 9 a If z = 3 + i, state z and calculate z 1. Plot z, z, and z 1 on an Argand plane. c What transformation plots z onto z 1? d What is the relationship etween the origin and the points representing z and z 1? 10 a If z = + 3i, calculate iz, i z, i 3 z and i z. Plot z, iz, i z, i 3 z and i z on an Argand plane. c State a transformation that will plot the point i n z onto i n + 1 z for n Z +. 11 a If z = 1 + i, calculate z, z 1, z 0, z, z 3 and z. Plot z, z 1, z 0, z, z, z 3 and z on an Argand plane. c State the rotation and the change in distance from the origin required to plot the point z n onto z n + 1 for n Z. d State the rotation and the distance from the origin required to plot the point z 0 onto z n for n Z +. 1 State, using the results from question 11, the following powers of z in their simplest Cartesian form. a z 5 z 3 c z 10 d z 17 e z 13 13 If z = x + yi then mod z = z = x + y, which is the distance of z from the origin. What does z = 1 represent? 1 Descrie the graph of the following if z = x + yi. a z 1 = 1 z + 1 = c z + 3i = 3. Factorising quadratic expressions and solving quadratic equations over the complex numer field In the introduction to complex numers, a quadratic equation of the form ax + x + c = 0, where a R\{0},, c R, was solved using the quadratic formula x = ± ac. The expression under the square root sign is called the a discriminant and is represented y Δ, where Δ = ac. The discriminant can e used to determine the nature of the solutions. It can also e used to determine possile methods for factorising a quadratic expression. The following tale gives the method for factorising a quadratic expression and the nature of the solutions for given values of Δ, where a,, c Q\{0}. Values of the discriminant Factorising methods Nature of solution(s) Δ = 0 A perfect square. State Two equal rational solutions the answer. Δ is a perfect square. Factorise over Q or Two rational solutions complete the square. Δ > 0 and is not a Complete the square. Two irrational solutions perfect square. Δ < 0 Complete the square. Two complex solutions Topic 1 Numer systems: real and complex 33

WoRKEd ExAMPLE think a Factorising quadratic expressions over the complex numer field Factorising over R implies that all the coefficients must e real numers and factorising over C implies that all the coefficients must e complex numers. As factors over C are required in this section, the variale lael will e z. In Worked example elow, the factors for the expressions in parts a and are factors over oth R and C, ut the factors for the expression in part c are factors over C only. It is still correct to say that z + 3 does not factorise over R. If c = 0 and a, R\{0}, then factorise az + z y taking out the common factor z(az + ). If = 0 and a, c R\{0}, then factorise az + c using completing the squares to factorise. Factorise each of the following quadratic expressions over C. a z + 6z z 6 c z + 3 Factorise z + 6z y taking out the highest common factor. 1 Factorise z 6 y taking out the highest common factor. Factorise further using the difference of two squares. c 1 Factorise z + 3 y taking out the common factor of. Factorise further using the difference of two squares. Let 3 = 3 i 3 Rationalise the denominators y multiplying the relevant terms y. WoRKEd ExAMPLE 3 WritE a z + 6z = z(z + 3) z 6 = (z 3) c z + 3 = z + 3 = (z 3) (z + 3) = z 3 i = z 3 i z + 3 i = z 6 i z + 6 i 3 Factorise each of the following quadratic expressions over C. a z 6z + 9 z z 60 c z 6z 6 d z 3z 5 think a 1 Calculate the value of the discriminant to determine the nature of the factors. WritE a Δ = ac = ( 6) (1)(9) = 0 3 MATHS QUEST 11 SPECIALIST MATHEMATICS VCE Units 1 and

Since Δ = 0, the expression is a perfect square. 1 Calculate the value of the discriminant to determine the nature of the factors. z 6z + 9 = (z 3) Δ = ac = ( ) (1)( 60) = 56 Since Δ = 56, which is a perfect square, the factors will e rational. c 1 Calculate the value of the discriminant to determine the nature of the factors. Since Δ = 8, which is not a perfect square ut is positive, use completing the square to find two factors over R. d 1 Calculate the value of the discriminant to determine the nature of the factors. Since Δ = 31, which is not a perfect square ut is negative, use completing the square to find two factors over C. c d z z 60 = (z 10)(z + 6) Δ = ac = ( 6) ()( 6) = 8 z 6z 6 = (z 3z 3) = z 3z + 3 = z 3 1 = z 3 1 Δ = ac = ( 3) ( )( 5) = 31 Solving quadratic equations over the complex numer field Two methods can e used to solve quadratic equations over the complex numer field: 1. Factorise first and use the null factor property to state solutions.. Use the formula for the solution of a quadratic equation. The null factor property states that if a = 0, then a = 0 or = 0 or a = = 0. From Worked example 3 d, 3 3 z 3 + 1 z 3z 5 = z + 3 z + 5 = z + 3 z + 3 = z + 3 = z + 3 + 31 16 31 16 i + 5 9 16 = z + 3 31 i z + 3 + 31 i Topic 1 Numer systems: real and complex 35

Units 1 & AOS Topic 1 Concept 10 Solving quadratic equations over the complex numer field Concept summary Practice questions WoRKEd ExAMPLE think a 1 Use the quadratic formula z = ± z 3z 5 = z + 3 31 i z + 3 + 31 i, so the solutions of z 3z 5 = 0 are from z + 3 31 i = 0 and z + 3 + 31 i = 0. The solutions are z = 3 ± 31 i. If az + z + c = 0, where a C \{0},, c C, the formula for the solution of the quadratic equation is z = ± ac. a Solve the following using the formula for the solution of a quadratic equation. a z + z + 5 = 0 iz + z 5i = 0 ac to solve over C, a where a =, =, c = 5. WritE ± 16 0 a z = ± = ± 6 = ± 6i = Express the answer in the form a + i. = 1 ± 6 i 1 Use the quadratic formula z = ± ac to solve over C, a where a = i, =, c = 5i. z = ± 16 10i i ± 16 0 = i ± = i ± 6 = i i i ( ± 6i)i = Express the answer in the form a + i. = i ± 6 = 6 + i or 6 + i 36 MATHS QUEST 11 SPECIALIST MATHEMATICS VCE Units 1 and

Exercise 1.7 PRactise Consolidate Factorising quadratic expressions and solving quadratic equations over the complex numer field 1 WE Factorise the following quadratic expressions over C. a z 6 z 3 c 3z + 6 Factorise the following quadratic expressions over C. d z + 1 a z z 6z z c z z d z 3z 3 WE3 Factorise the following quadratic expressions over C using the completion of the square method. a z + z + 1 z + 10z + 16 c z + 5z 3 d z + z 3 e z + 8z + 16 Factorise the following quadratic expressions over C using the completion of the square method. a z + z + 3 z 5z + c z + 8z + 8 d z + 5z + e z + z 1 5 WE Factorise the following quadratic expressions over C, and then solve the given quadratic equations. a 3z = 0 z + 5 = 0 c z 7z = 0 d z 6z + 5 = 0 e z 5z + 6 = 0 6 Solve the following quadratic equations over C using the formula for the solution of a quadratic equation. a z 10z + 5 = 0 z 10z + 5 = 0 c z + z + 7 = 0 d z 7z + 6 = 0 e 3z 7z + 7 = 0 f z + z 6 = 0 7 Factorise the following quadratic expressions over C without using the completion of the square method. a z + 8z + 16 z 8z + 8 c z + 3z d z + z 3 e z z f 1z + 10z + 1 8 Factorise the following quadratic expressions over C, and then solve the given quadratic equations. a z 5z + 3 = 0 z z + = 0 c z + 5z + = 0 d z 6z + 5 = 0 e 3z z 1 = 0 9 Expand the following. a (z ( + 3i)) (z ( 3i)) (z ( + 3i)) c (z + 3i) (z 3 i) 10 Solve the following quadratic equations over C, using the formula for the solution of a quadratic equation. a iz 6z + 5i = 0 ( + i)z iz ( i) = 0 11 Solve 3iz (1 + i)z + 5i = 0. Topic 1 Numer systems: real and complex 37

Master 1 Using the smallest set from Q, I and C, the solutions of z 5z + 6 = 0 and 5z 11z + 5 = 0, respectively, elong to the sets: A C, C B C, Q C C, I D I, I E I, Q 13 The factors of z + 6z + 11 and z z + 3, respectively, are: A (z + 3 i) (z + 3 + i), z 1 i z 1 + i B (z + 3 i) (z + 3 + i), z 1 i z 1 + i C (z 3 i) (z 3 + i), z 1 i z 1 + i D (z + 3 i) (z + 3 + i), z + 1 i z + 1 + i E (z 3 i) (z 3 + i), z + 1 i z + 1 + i 1 i is a solution of x x + k = 0. What is the value of k? A 3 B 5 C 5 D 3 E + i 15 The solutions to the quadratic equation: x x + 3 = 0 are: A x = 3 or x = 1 B x = + i or x = i C x = 1 + i or x = 1 i D x = 1 + i or x = 1 i E x = 3i or x = i 16 Solve x + x + 7 = 0. 17 If 10x x + = A(Bx + Cx + D), find A, B, C and D. x 3 + x x(x i)(x + i) 18 Solve x + 13x + 36 = 0. 38 MATHS QUEST 11 SPECIALIST MATHEMATICS VCE Units 1 and

ONLINE ONLY 1.8 Review www.jacplus.com.au the Maths Quest review is availale in a customisale format for you to demonstrate your knowledge of this topic. the review contains: Multiple-choice questions providing you with the opportunity to practise answering questions using CAS technology short-answer questions providing you with the opportunity to demonstrate the skills you have developed to efficiently answer questions using the most appropriate methods studyon is an interactive and highly visual online tool that helps you to clearly identify strengths and weaknesses prior to your exams. You can then confidently target areas of greatest need, enaling you to achieve your est results. Extended-response questions providing you with the opportunity to practise exam-style questions. A summary of the key points covered in this topic is also availale as a digital document. REVIEW QUESTIONS Download the Review questions document from the links found in the Resources section of your ebookplus. Units 1 & Numer systems: real and complex Sit topic test Topic 1 NUMER SySTEMS: REAL ANd CoMPLEx 39