chapter 1

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25 JUNE – 3 JULY 2018 ℝ

snash’s

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be able to:

ℚ′

ℤ Upon completion of this chapter, readers should

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1. Define and state the rules of indices, surds

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ℂ

2. Perform the algebraic operations of indices,

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and logarithms.

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3. Solve equations involving indices, surds, logarithms and complex numbers. ℝ

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surds, logarithms and complex numbers.

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TYPES OF NUMBER

INTERVAL OF REAL NUMBER

COMPLEX NUMBER

INDEX RULES / OPERATIONS

OPERATIONS

SURD SOLVING EQUATION

SOLVING EQUATION

snash’s

LOG

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TYPES OF NUMBERS REAL; â„?

A number that has a decimal representation.

Examples: 1, 0.5, 2.11111 , 2/3 Number that can’t be Number that can be represented as fraction represented of two as a fraction of RATIONAL; â„š IRRATIONAL; ℚ’ đ?‘Ž integers. , and a,b ∈ đ?‘?, đ?‘? ≠0 . đ?‘?

0 & +ve integers Examples: 2.3154‌ , e,Examples: 3 -1, 35 , -ve 0.2,integers, 1. 23 Examples: ‌, -1, 0, 1, 2, ‌ INTEGER; ℤ

WHOLE;W NATURAL; ℕ snash’s

Zero and Natural numbers

Examples: 0, 1, 2, ‌ Counting numbers Example: 1,2,..

Example 1: List the numbers in the set 4 {−3, , 0, 0.12, 2, đ?œ‹, 2.1515 ‌ , đ?‘’, 10} that are 3

a) b) c) d) e) f)

snash’s

Natural numbers Whole numbers Integers Rational numbers Irrational numbers Real numbers

Example 2: Express the following decimal numbers as fractions. c) 1.651

snash’s

d) 0.725

REAL NUMBER LINE & INTERVALS

INTERVALS

OPEN

snash’s

CLOSE

HALF-OPEN/ HALF-CLOSE

INFINITE

INTERVAL

a x b

[ -2, 1]

a< x <b

(a,b) ( -2 , 1 )

a  x <b

-2  x <1 a<x b

-2< x  1

x>a @ a< x< x<a @ -< x<a x > -2 @ - 2 < x < 

REAL NUMBER LINE

[a,b]

-2  x  1

-2 < x <1

snash’s

SYMBOL

[a,b) [ -2 , 1 ) (a,b] ( -2 , 1 ] ( a , ) ( - , a ) ( -2 ,  )

TYPE

Close interval

a

b Open interval

a

a a

b b b

Half open interval Half open interval Infinite interval

Example 3: Rewrite each of the following inequalities by using interval notation and illustrate them on the real number line.

snashâ&#x20AC;&#x2122;s

a) â&#x2C6;&#x2019;3 < đ?&#x2018;Ľ < 5

b) â&#x2C6;&#x2019;1 < đ?&#x2018;Ľ â&#x2030;¤ 0

c) đ?&#x2018;Ľ â&#x2030;Ľ â&#x2C6;&#x2019;2

d) đ?&#x2018;Ľ < 0

Exercise: Example 4: Write each of the following intervals as inequalities. Use đ?&#x2018;Ľ as the variable.

snashâ&#x20AC;&#x2122;s

a) (â&#x2C6;&#x2019;â&#x2C6;&#x17E;, 5]

b) [â&#x2C6;&#x2019;2, 3]

c) [3, â&#x2C6;&#x17E;)

d) (â&#x2C6;&#x2019;1, 7)

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ℂ

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INDICES If a is any real number, a ď&#x192;&#x17D; R (a > 0) and n is a positive integer, then đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; = đ?&#x2018;&#x17D; Ă&#x2014; đ?&#x2018;&#x17D; Ă&#x2014; đ?&#x2018;&#x17D; â&#x20AC;Ś Ă&#x2014; đ?&#x2018;&#x17D; đ?&#x2018;&#x203A; đ?&#x2018;Ąđ?&#x2018;&#x2013;đ?&#x2018;&#x161;đ?&#x2018;&#x2019;đ?&#x2018;

The integer n is called the index or exponent and a is the base. ( Read an as â&#x20AC;&#x2DC;a to the nth powerâ&#x20AC;&#x2122;) snashâ&#x20AC;&#x2122;s

n  Z   ve integer 

n0

a

n



n Z (-ve integer)

14 snash’s

For n=0

a 1 , a  0 0

0

0 is not defined.

For n positive integer

a n  a  a  a 1st

snash’s

2nd

nth

 n factors of a

For n is negative integer ; i.e: n = –m

a

m

a b  

1  m a

;

m

m

b   a

a0 ; a  0, b  0

x For n is in the form of fraction ;i.e: n  y x y

y

1 y

y

a  a  a  a snash’s

x

 a y

x

, a  0.

, a0

RULES FOR INDICES 1) 2)

aman  amn am

 amn

n

a

3)

 

4)

 ab 

n

m

a

m

5) snash’s

EXAMPLE 5c)

a

EXAMPLE 5e)

 ambm

EXAMPLE 6c)

mn

m

a b  

EXAMPLE 5b)



m

a

m

b

, bEXAMPLE  0 6e)

Example 7: Simplify: 1 3

â&#x2C6;&#x2019;

a) 27

1 3

b) 4 2 5

c) đ?&#x2018;&#x17D; â&#x2C6;&#x2122; đ?&#x2018;&#x17D; á đ?&#x2018;&#x17D;

snashâ&#x20AC;&#x2122;s

1 2

d)

3

1 2 5

đ?&#x2018;&#x17D; á đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; â&#x2C6;&#x2122; (đ?&#x2018;&#x17D;â&#x2C6;&#x2019;1 )

1 2

Example 10: đ?&#x2018;Ľ

đ?&#x2018;Ś

đ?&#x2018;§

If 2 = 6 = 12 , prove that đ?&#x2018;§ =

snashâ&#x20AC;&#x2122;s

đ?&#x2018;Ľđ?&#x2018;Ś đ?&#x2018;Ľ+đ?&#x2018;Ś

.

Exercise: Example 8: Simplify: b) 122đ?&#x2018;Ľ+3 â&#x2C6;&#x2122; 6đ?&#x2018;Ľâ&#x2C6;&#x2019;5 â&#x2C6;&#x2122; 82đ?&#x2018;Ľâ&#x2C6;&#x2019;1

c)

â&#x2C6;&#x2019;2 3

3 4

đ?&#x2018;&#x17D; đ?&#x2018;?2 Ă&#x2014; đ?&#x2018;&#x17D; đ?&#x2018;?

2 1 3 2

1 6

á đ?&#x2018;&#x17D; đ?&#x2018;?

10 3

Example 11: Show that 2đ?&#x2018;&#x203A; + 2đ?&#x2018;&#x203A;+1 + 2đ?&#x2018;&#x203A;+2 is divisible by 7 if đ?&#x2018;&#x203A; is a positive integer. snashâ&#x20AC;&#x2122;s

ℝ

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ℂ

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ℕℚ′ℝ ℝℂ ℚ′

ℚ′ ℕ ℂ

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snash’s

SURDS ď&#x201A;§ Surd is a number that contain one or more root sign;

(or radical sign)

ď&#x201A;§ It cannot be simplified into a fraction of two integers,

đ?&#x2018;&#x17D; . đ?&#x2018;?

ď&#x201A;§ Example, 2 is a surd but 4 = 2. ď&#x201A;§ Surd is an irrational number. snashâ&#x20AC;&#x2122;s

4 is not a surd since

SURDS n

n

a



a a

nth root of a

1 n

Surd is expressed in simplest form. For examples:  8 = 4(2) = 2 2 

12 25

=

4 3 25

=

2 3 5

Examples of surd: a) 3 b) d) 3+1 d) snash’s

4 5 3+ 5

c) e)

2− 7 3 9

PROPERTIES OF SURD 1)

n

a a m

m

a 3) m  b

m

m n

a b

2) m a m b 

4)

n m

ab

a  mn a

5) a c  b c   a  b  snash’s

m

c

Notes: 𝑎 𝑎=𝑎

4 ≠ ±2 but

4 = 2 and − 4 = −2

𝑎±𝑏 ≠ 𝑎± 𝑏 𝑎2 ± 𝑏2 ≠ 𝑎 ± 𝑏 snash’s

Example 12: Identify which of the followings are surd. a) d)

snashâ&#x20AC;&#x2122;s

6 48

b) e)

3

8 3

c) 64

25

Example 13: Simplify: b) 3 10 2 5 d) 3 40 + 20 5

snash’s

18 − 2

Example 14: Simplify:

c)

snashâ&#x20AC;&#x2122;s

3

27đ?&#x2018;Ľ 6 đ?&#x2018;Ś 3 3đ?&#x2018;Ľ 2 đ?&#x2018;Ś 4

Exercise: Example 13: a)

5 3 3

Example 14:

b)

snashâ&#x20AC;&#x2122;s

đ?&#x2018;Ľđ?&#x2018;Ś 3 81đ?&#x2018;Ľ 3 đ?&#x2018;Ś

125

CONJUGATE OF SURD

snash’s

a b

a b

a b

a b

a b

a b

RATIONALISING DENOMINATOR ď&#x192;&#x2DC; process of eliminating surd in denominator so that the denominator is a rational number. ď&#x192;&#x2DC; Simplest form ď&#x192; denominator free from surd

ď&#x192;&#x2DC; Multiplication between conjugates will result in a rational number (no surd expression exists): đ?&#x2018;&#x17D;+ đ?&#x2018;? đ?&#x2018;&#x17D;â&#x2C6;&#x2019; đ?&#x2018;? =đ?&#x2018;&#x17D;â&#x2C6;&#x2019;đ?&#x2018;? đ?&#x2018;&#x17D;+đ?&#x2018;? đ?&#x2018;&#x17D; â&#x2C6;&#x2019; đ?&#x2018;? = đ?&#x2018;&#x17D; â&#x2C6;&#x2019; đ?&#x2018;?2 đ?&#x2018;&#x17D; + đ?&#x2018;? đ?&#x2018;&#x17D; â&#x2C6;&#x2019; đ?&#x2018;? = đ?&#x2018;&#x17D;2 â&#x2C6;&#x2019; đ?&#x2018;? snashâ&#x20AC;&#x2122;s

RATIONALISING DENOMINATOR  Problem arise when surd exists in denominator

Solution:

snash’s



Multiply the numerator and denominator by suitable factor



Multiply the numerator and denominator with the conjugate of denominator

Example 15: Simplify the expressions: a)

snashâ&#x20AC;&#x2122;s

5 2đ?&#x2018;&#x17D;

c)

2+ 2 5+ 2

+

2â&#x2C6;&#x2019; 10 3

Example 16: b)

Given

đ?&#x2018;?+ đ?&#x2018;&#x17E; 1â&#x2C6;&#x2019; đ?&#x2018;?

=

4 đ?&#x2018;?+ đ?&#x2018;&#x17E; 1â&#x2C6;&#x2019;đ?&#x2018;?

value of đ?&#x2018;? + đ?&#x2018;&#x17E;.

snashâ&#x20AC;&#x2122;s

and đ?&#x2018;? > 1, find the

Exercise Example 15: Simplify the expression: b)

snash’s

2 2 2−2 3

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snash’s

LOGARITHM ď&#x201A;§ A number which is in an index form can be written in a logarithmic form. ď&#x201A;§ x is the logarithm of b to the base a is written as

đ?&#x2018;Ľ = log đ?&#x2018;&#x17D; đ?&#x2018;? and it is equivalent to đ?&#x2018;&#x17D; đ?&#x2018;Ľ = đ?&#x2018;?. đ?&#x2018;&#x17D;đ?&#x2018;Ľ = đ?&#x2018;? â&#x2020;&#x201D;

đ?&#x2018;Ľ = log đ?&#x2018;&#x17D; đ?&#x2018;?

where đ?&#x2018;&#x17D;, đ?&#x2018;? > 0 and đ?&#x2018;&#x17D; â&#x2030; 1. snashâ&#x20AC;&#x2122;s

LOGARITHM 𝑏>0 log 𝑎 𝑏 = 𝑥 𝑎 > 0, 𝑎 ≠ 1

snash’s

LOGARITHM a ď&#x20AC;˝b x

Index Form 32 = 9 91 = 9 5đ?&#x2018;Ľđ?&#x2018;Ś = đ?&#x2018;§ â&#x2C6;&#x2019; 1 đ?&#x2018;Ľ 2 =4

snashâ&#x20AC;&#x2122;s

loga b ď&#x20AC;˝ x Logarithmic Form log 3 9 = 2 log 9 9 = 1 log 5 (đ?&#x2018;§ â&#x2C6;&#x2019; 1) = đ?&#x2018;Ľđ?&#x2018;Ś log 2 4 = đ?&#x2018;Ľ

LOGARITHM ď&#x192;&#x2DC; Logarithmic with base 10 is called common logarithm and it can be written without the base 10:

log10 đ?&#x2018;Ľ = log đ?&#x2018;Ľ ď&#x192;&#x2DC; Logarithmic with base e is called natural logarithm and it can be written as đ?&#x2018;&#x2122;đ?&#x2018;&#x203A;: log đ?&#x2018;&#x2019; đ?&#x2018;Ľ = ln đ?&#x2018;Ľ snashâ&#x20AC;&#x2122;s

LOGARITHMIC RULES Let đ?&#x2018;&#x17D;, đ?&#x2018;Ľ, đ?&#x2018;Ś > 0, đ?&#x2018;&#x17D; â&#x2030; 1 and đ?&#x2018;&#x203A; â&#x2C6;&#x2C6; đ?&#x2018;&#x2026;

1/

loga xy ď&#x20AC;˝ loga x ď&#x20AC;Ť loga y

x 2 / loga ď&#x20AC;˝ loga x ď&#x20AC;­ loga y y 3 / loga xn ď&#x20AC;˝ nloga x

snashâ&#x20AC;&#x2122;s

4/

loga a ď&#x20AC;˝ 1

5/

loga 1 ď&#x20AC;˝ 0

6/

aloga x ď&#x20AC;˝ x

Example 19: Evaluate without using calculator: b)

snashâ&#x20AC;&#x2122;s

9 log 3 27

d) đ?&#x2018;&#x2019;

ln 4â&#x2C6;&#x2019;ln 24 +2đ?&#x2018;&#x2122;đ?&#x2018;&#x203A;3

Example 20: Expand using the rules of logarithm. a) đ?&#x2018;&#x2122;đ?&#x2018;&#x153;đ?&#x2018;&#x201D;

snashâ&#x20AC;&#x2122;s

đ?&#x2018;Ľ2đ?&#x2018;Ś 10

c) đ?&#x2018;&#x2122;đ?&#x2018;&#x203A;

đ?&#x2018;Ľ3đ?&#x2018;Ś đ?&#x2018;§

Example 21: Rewrite the following expressions in a single logarithmic expression. b) đ?&#x2018;&#x2122;đ?&#x2018;&#x153;đ?&#x2018;&#x201D;2 đ?&#x2018;Ľ + đ?&#x2018;Ś + đ?&#x2018;&#x2122;đ?&#x2018;&#x153;đ?&#x2018;&#x201D;2 đ?&#x2018;Ś â&#x2C6;&#x2019;

snashâ&#x20AC;&#x2122;s

1 đ?&#x2018;&#x2122;đ?&#x2018;&#x153;đ?&#x2018;&#x201D;2 3

đ?&#x2018;Ľ

CHANGE OF LOG BASE Let 𝑎, 𝑏, 𝑐 > 0, then 𝑙𝑜𝑔𝑐 𝑏 𝑙𝑜𝑔𝑎 𝑏 = 𝑙𝑜𝑔𝑐 𝑎

𝑙𝑜𝑔𝑎 𝑏 = snash’s

1 𝑙𝑜𝑔𝑏 𝑎

Example 22: Evaluate đ?&#x2018;&#x2122;đ?&#x2018;&#x153;đ?&#x2018;&#x201D;1 70 by converting into ln. Leave the 2

answer in four decimal places. đ?&#x2018;&#x2122;đ?&#x2018;&#x153;đ?&#x2018;&#x201D;1 70 2

=

đ?&#x2018;&#x2122;đ?&#x2018;&#x203A; 70 1

đ?&#x2018;&#x2122;đ?&#x2018;&#x203A;2

= â&#x2C6;&#x2019;6.1293

snashâ&#x20AC;&#x2122;s

loga  x  y   loga x  loga y x loga x loga  y loga y n n loga x   loga x  loga x loga x  loga y  loga y loga x  loga y   loga x  loga y 

loga  xy    loga x   loga y  loga  2 x   2loga x

snash’s

Exercise Example 19: Evaluate without using calculator: 1 c) log 5 125 â&#x2C6;&#x2019; log 2 + log 100 16

Example 20: Expand using the rules of logarithm. đ?&#x2018;Śâ&#x2C6;&#x2019;đ?&#x2018;Ľ b) đ?&#x2018;&#x2122;đ?&#x2018;&#x153;đ?&#x2018;&#x201D;3 đ?&#x2018;Ľđ?&#x2018;Ś

Example 21: Rewrite the following expressions in a single log expression. c) đ?&#x2018;&#x2122;đ?&#x2018;&#x153;đ?&#x2018;&#x201D;5 2đ?&#x2018;Ľ â&#x2C6;&#x2019; đ?&#x2018;&#x2122;đ?&#x2018;&#x153;đ?&#x2018;&#x201D;5 3 â&#x2C6;&#x2019; 5đ?&#x2018;&#x2122;đ?&#x2018;&#x153;đ?&#x2018;&#x201D;5 đ?&#x2018;Ľ snashâ&#x20AC;&#x2122;s

ℝ

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ℤ

ℚ

ℕ ℚ′

ℝ

ℚ′

ℂ

ℕ

ℂ

ℚ

ℚ

ℤ

ℚ′

ℚ ℝ

ℕℚ′ℝ ℝℂ ℚ′

ℚ′ ℕ ℂ

ℂ

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ℚ ℤ

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ℝ

snash’s

INDEX EQUATION COMMON BASE

 convert into common base  simplify  compare index EXAMPLE 9b:

COMMON BASE WITH ± BETWEEN TERMS

 convert into common base  substitution EXAMPLE 9c:

REMEMBER! an > 0 snash’s

DIFFERENT BASES

cannot be converted into common base take log(or ln) both sides  apply log property EXAMPLE 25a:

SURD EQUATION 1 SURD

2 SURDS

3 SURDS

1 surd at one side

1 surd each side

1 surd at one side, 2 surds at the other side

  

Square both sides and simplify Square both sides again (if there is a remaining surd) Solve the equation EXAMPLE 17b ?

check for validity  LHS  RHS snash’s

LOG EQUATION 1/COMMON LOG EXPRESSIONS

 simplify  use antilog or  change the base EXAMPLE 23a:

MORE THAN 1 LOG WITH ± BETWEEN TERMS

 change the base  apply property  substitution/ comparison EXAMPLE 23b:

check for validity  a, b > 0 for loga b snash’s

Example 26: Solve the simultaneous equations below. b)

snashâ&#x20AC;&#x2122;s

â&#x2C6;&#x2019;2 đ?&#x2018;Ľ â&#x2C6;&#x2019; â&#x2C6;&#x2019;2 đ?&#x2018;&#x2122;đ?&#x2018;&#x153;đ?&#x2018;&#x201D;đ?&#x2018;Ľ 125 đ?&#x2018;Ľ â&#x2C6;&#x2019;

đ?&#x2018;Ś+2

đ?&#x2018;Ś đ?&#x2018;&#x2122;đ?&#x2018;&#x153;đ?&#x2018;&#x201D;125 đ?&#x2018;Ľ

=0 =6

Example 18: Find the value(s) of x:

Exercise 3đ?&#x2018;Ľ + 1 â&#x2C6;&#x2019; 2đ?&#x2018;Ľ â&#x2C6;&#x2019; 1 = 1

Example 24: Solve the equation: (3đ?&#x2018;Ľ )2 â&#x2C6;&#x2019;3đ?&#x2018;Ľ+1 + 2 = 0

Example 25: Solve the equation and leave the answer correct to 3 s.f. b) đ?&#x2018;&#x2122;đ?&#x2018;&#x153;đ?&#x2018;&#x201D;4 3đ?&#x2018;Ľ = 2 + đ?&#x2018;&#x2122;đ?&#x2018;&#x153;đ?&#x2018;&#x201D;4 đ?&#x2018;Ľ + 1 Example 26: Solve the simultaneous equations below. a) 2đ?&#x2018;Ľ + 3đ?&#x2018;Ś = 41 2đ?&#x2018;Ľ+2 + 3đ?&#x2018;Ś+2 = 209 snashâ&#x20AC;&#x2122;s

EXPLORE INDEX You can explore this website for more information: 1. https://revisionmaths.com/gcse-maths-revision/algebra/indices 2. http://mathematics.laerd.com/maths/indices-1.php 3. https://www.youtube.com/watch?v=sbwSKpJkR2s

SURD You can learn more about simplifying surd, the simple fact and history about surd at: http://www.mathslearn.co.uk/core1surds.html LOGARITHM Browse the website to study about the proof of logarithmic rules: https://www.onlinemathlearning.com/logarithms-properties.html SOLVING EQUATIONS You can check your workings and answers for Examples 9, 17, 18, 23, 24 and 25 via the online equation calculator: https://www.symbolab.com/solver/equation-calculator snashâ&#x20AC;&#x2122;s

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ℤ

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ℝ

ℚ′

ℂ

ℕ

ℂ

ℚ

ℚ

ℤ

ℚ′

ℚ ℝ

ℕℚ′ℝ ℝℂ ℚ′

ℚ′ ℕ ℂ

ℂ

ℝ

ℕ

ℚ ℤ

ℚ

ℝ

snash’s

IMAGINARY NUMBER i  1

i  1 2

i 3  i 2 i  i

 

i  i 4

2

2

  1  1

i 5  i 41  i 4 i  i

snash’s

2

Example 28: Write the following imaginary numbers in the form of đ?&#x2018;&#x2013;. a) â&#x2C6;&#x2019;6 b) â&#x2C6;&#x2019; â&#x2C6;&#x2019;3 c) â&#x2C6;&#x2019;4.656 d) â&#x2C6;&#x2019;16

snashâ&#x20AC;&#x2122;s

COMPLEX NUMBER complex number = real part + (imaginary part)(imaginary number) z = a + bi a = Re (z) b = Im (z) a + bi

â&#x2C6;&#x2019;đ?&#x;?

real part imaginary part snashâ&#x20AC;&#x2122;s

COMPLEX NUMBER Set of complex number is denoted by C or Complex Real Part Number26

snashâ&#x20AC;&#x2122;s

Imaginary Part

Note

đ?&#x2018;&#x17D; and đ?&#x2018;? are real numbers đ?&#x2018;&#x17D; and đ?&#x2018;? are real numbers đ?&#x2018;&#x17D;=đ?&#x2018;? =0 Purely Real Purely Imaginary

2 â&#x2C6;&#x2019; 4đ?&#x2018;&#x2013;

2

-4

1 1 + đ?&#x2018;&#x2013; 4 3 0 + 0đ?&#x2018;&#x2013; 5 â&#x2C6;&#x2019;6i

1 4 0 5 0

1 3 0 0 â&#x2C6;&#x2019;6

Equality of a Complex Number:

đ?&#x2018;&#x17D; + đ?&#x2018;?đ?&#x2018;&#x2013; = đ?&#x2018;? + đ?&#x2018;&#x2018;đ?&#x2018;&#x2013; if and only if đ?&#x2018;&#x17D; = đ?&#x2018;? and đ?&#x2018;? = đ?&#x2018;&#x2018;.

Zeros of a Complex Number:

đ?&#x2018;§ = đ?&#x2018;&#x17D; + đ?&#x2018;?đ?&#x2018;&#x2013; = 0

snashâ&#x20AC;&#x2122;s

if and only if đ?&#x2018;&#x17D; = 0 and đ?&#x2018;? = 0.

CONJUGATE OF COMPLEX NUMBER Suppose z = a + bi, then the conjugate of z is given by:

z  a  bi • Conjugate of a complex number is a complex

number where i is being replaced by –i • i,e change the sign of imaginary part

snash’s

snashâ&#x20AC;&#x2122;s

Complex Number 2 + 4đ?&#x2018;&#x2013;

Conjugate 2 â&#x2C6;&#x2019; 4đ?&#x2018;&#x2013;

1 1 â&#x2C6;&#x2019; đ?&#x2018;&#x2013; 4 3

1 1 + đ?&#x2018;&#x2013; 4 3

â&#x2C6;&#x2019;1 + 9đ?&#x2018;&#x2013;

â&#x2C6;&#x2019;1 â&#x2C6;&#x2019; 9đ?&#x2018;&#x2013;

5

5

â&#x2C6;&#x2019;10đ?&#x2018;&#x2013;

10đ?&#x2018;&#x2013;

CONJUGATE OF COMPLEX NUMBER Suppose z = a + bi, then the conjugate of z is given by:

z  a  bi Properties:

z  z  2a z z a b 2

snash’s

2

Example: If ๐ ง = 2 + 3๐ , find:

snashโ s

a)

โ z

b)

๐ งาง

c)

z + ๐ งาง

d)

๐ ง๐ งาง

OPERATIONS OF COMPLEX NUMBERS

snash’s

ADDITION +

SUBTRACTION -

MULTIPLICATION ×

DIVISION ÷

Let z1  a  ib

and

z2  c  id

z1  z2

  a  ib    c  id   a  c   i b  d 

example 29 : example 30 :  3  2i   1  7i3  5i   4  3i

  3  1  i  2  7 3  4   i  5   3    4  9i

 1  8i operation

snash’s

z1z2

  a  ib  c  id   ac   i  bc  ad  i2  bd   ac  bd  i  bc  ad

example 32 :

 3  2i1  7i  3  2i  21i  14i2  3  23i  14  1  11  23i snash’s

operation

z1 z2

a  ib     c  id a  ib   c  id       c  id  c  id 

rationalizing denominator Multiply with conjugate

example 36 : 5  2i  7  4i  35  14i  20i  8i2 =   7  4i  7  4i  72  42

 

35  8  6i =  49  16 operation

snash’s

Example 31: Find

a)

Exercise

1 + 3đ?&#x2018;&#x2013; + 5 â&#x2C6;&#x2019; 2đ?&#x2018;&#x2013;

b)

2 3

1 2

3+đ?&#x2018;&#x2013; â&#x2C6;&#x2019; â&#x2C6;&#x2019; + đ?&#x2018;&#x2013;

Example 33: Multiply each of the following and write the answers in standard form. b) 3 â&#x2C6;&#x2019; 4đ?&#x2018;&#x2013; â&#x2C6;&#x2019;4 + 3đ?&#x2018;&#x2013; c) 5 â&#x2C6;&#x2019; 2đ?&#x2018;&#x2013; 7đ?&#x2018;&#x2013; Example 35: Find the product of the following complex number and its conjugate. a) 2 â&#x2C6;&#x2019; 3đ?&#x2018;&#x2013; 2 + 3đ?&#x2018;&#x2013; b) â&#x2C6;&#x2019;9 + 2đ?&#x2018;&#x2013; â&#x2C6;&#x2019;9 â&#x2C6;&#x2019; 2đ?&#x2018;&#x2013; Example 37: Express the following in the form of đ?&#x2018;&#x17D; + đ?&#x2018;?đ?&#x2018;&#x2013;: b) snashâ&#x20AC;&#x2122;s

4đ?&#x2018;&#x2013; 1+2đ?&#x2018;&#x2013;

d)

7â&#x2C6;&#x2019;4đ?&#x2018;&#x2013; 2đ?&#x2018;&#x2013;

ARGAND DIAGRAM • Complex numbers can be shown on the complex number plane known as an Argand diagram • Each number is represented by a point.

• The real part is plotted on the horizontal axis and the imaginary part on the vertical axis.

• Sometimes the number itself is represented as a line from the origin to the point. snash’s

𝐼𝑚(𝑧)

𝑏

(𝑎, 𝑏) 𝑧 = 𝑎 + 𝑏𝑖

𝑎

−𝑏

snash’s

(𝑎, −𝑏) 𝑧 = 𝑎 − 𝑏𝑖

𝑅𝑒(𝑧)

Example 38: Plot the following complex numbers on an Argand diagram. a) b) e)

snashâ&#x20AC;&#x2122;s

2 + 3đ?&#x2018;&#x2014; 2 â&#x2C6;&#x2019; 3đ?&#x2018;&#x2014; â&#x2C6;&#x2019;5 + 0đ?&#x2018;&#x2014;

b  b2  4ac x 2a

snash’s

EQUATION OF COMPLEX NUMBER

Solving Quadratic equation when discriminant is negative

Equality of Complex numbers Find the value(s) of unknown

z1  z2 a  bi  c  di  ac bd

Example 40: Solve the equation đ?&#x2018;Ľ 2 â&#x2C6;&#x2019; 4đ?&#x2018;Ľ + 13 = 0 in the complex

number system.

snashâ&#x20AC;&#x2122;s

Example: Solve for p where 𝑝 ∈ 𝐶 3 − 4𝑖 + 𝑝 2 + 𝑖 = 4 + 3𝑖 let p  a 3bi 4i  p  2  i   4  3i p  2  i   4  3i  3  4i 2 7i i2  i4  3i 3  4i   a  bi1 p 2i  2  i 3  4i   2a  2bi  ai  b   4  3i 2  14i  i  7  3  2a  bp i  4  2b  a   4  3i 4 1 3  2a  b  49 13  4  2b  a  3 p  i 2a  b  1 5 51 a  2b  7 snash’s

 2

Example 42: Find the square roots of 3 â&#x2C6;&#x2019; 4đ?&#x2018;&#x2013;.

snashâ&#x20AC;&#x2122;s

Exercise Example 38: Plot the following complex numbers on an Argand diagram. c) â&#x2C6;&#x2019;3 + 2đ?&#x2018;&#x2014; d) â&#x2C6;&#x2019;3 â&#x2C6;&#x2019; 2đ?&#x2018;&#x2014; f) 0 + 5đ?&#x2018;&#x2014; Example 41: Solve the equation đ?&#x2018;Ľ 2 â&#x2C6;&#x2019; 3đ?&#x2018;Ľ + 9 = 0 in the complex number system.

Example 42: â&#x20AC;Ś .Use the results to solve the quadratic equation (2 â&#x2C6;&#x2019; đ?&#x2018;&#x2013;)đ?&#x2018;Ľ 2 +(4 + 3đ?&#x2018;&#x2013;)đ?&#x2018;Ľ + (â&#x2C6;&#x2019;1 + 3đ?&#x2018;&#x2013;) = 0. snashâ&#x20AC;&#x2122;s