Thomas Whitham Sixth Form

Further

Mathematics

UUnniitt FFPP11

Polynomial Equations

Mathematical Induction

Finite Series

Complex Numbers

Derivative by formula & by

using Logarithms

Matrices

1

Polynomial Equations

1. An nth

degree polynomial equation will have n roots

Example 072 2 xx is a quadratic equation. It is 2nd

degree

and it has two roots.

Example 01623 xx is a cubic equation. It is 3rd

degree and it

has three roots.

Example 0323 234 xxxx is a quartic equation. It is 4th

degree and has four roots.

2. Irrational roots will occur in pairs qp where 0q and is not a

perfect square.

Example The equation 0112 23 baxxx has one root

32 . Find the values of a and b, and hence solve the

equation completely.

If 32x is one root then 32x is a second root.

It follows that 3232 xx is a quadratic factor; this

simplifies to 142 xx

Hence cxxxbaxxx 214112 223

2x c 811 3c

x ca 42 14a

const cb 3b

The equation reduces to 0314112 23 xxx

Or 032142 xxx

By inspection

2

Full solution set 2

3 ,32 x

3. Imaginary roots will occur in conjugate pairs iba .

4. It follows from 1 and 3 that any polynomial equation of odd degree

will have at least one real root.

5. Recall the Factor Theorem (P2), restated her.

“If 0abP then bax is a factor of xP ; it follows that

a

bx is a solution of 0xP ”

6. If is a repeated root of 0xP , then is also a root of 0 xP .

NB the converse is not generally true.

Example Solve the equation 025412 23 xxx given that it

has two equal roots.

Let 25412 23 xxxxP

518125836 2 xxxxxp

0 xp for 2

1x and

18

5x

but 021 p

2

1x is a double root of the equation.

(you can readily check that 0185 p )

It follows that 231225412223 xxxxx

HHeennccee tthhee rroooottss ooff tthhee eeqquuaattiioonn aarree 21 ,, 2

1 ,, -- 23

By inspection

3

7. Graphical interpretations Real and distinct roots to 0xp

correspond to intersections of the graph of xpy with the x axis.

Repeated roots occur where the graph touches the x axis. Some

illustrative examples follow

Quadratics

Cubics If a cubic equation is to have more than one real root, the

corresponding cubic function must have two stationary points. This is a

necessary condition but is not in itself sufficient.

0 x

y

0 x

y

0 x

y

No real roots Real distinct roots Double roots

0 x

y

0 x

y

0 x

y

Three real and

distinct roots

Only one real root Adouble + one more

root

0 x

y

0 x

y

(a, 0)

inflexion ifpoint

stationary a

0

0

0

ap

ap

ap

Only one real root

(no stationary points)

4

Quartics Quartic graphs may have one, two or three stationary points. In

each of the following illustrations, the coefficient of 4x has been taken

to be positive so that, as x y

One SP

Two SPs

Three SPs

0 x

y

0 x

y

0 x

y

0 x

y

0 x

y

0 x

y

No real roots Two real distinct roots Four equal roots

Two real distinct roots

Triple +one root

Root (poi)

Double root

Two roots

Two distinct roots + one double root

Four roots

No real roots

5

8. Relationship between roots and coefficients of a polynomial

equation.

Quadratic 02 cbxax

Cubic 023 dcxbxax

Quartic

0234 edxcxbxax This time, using notation

a

e

a

d

a

c

a

b

Example Solve the equation 0367 23 xx given that one root

is double the other.

Let the roots be , 2 and

73

022 2

032

a

ca

b

Product

roots of sum

a

da

ca

b

product

pairs ofproduct

sum

6

0 032

Solve simultaneously 3 , 2

Check “” a

d 36263.2.

Therefore roots are 2,6,3 x

Example (i) Show that 22222

(ii) The roots of the equation 0133 xx are , , ; Find

the values of (a) 222 (b)

111

(i) 22222

2

222

222

222

(ii) 0

1

3

(a) Using the result (1) 320 2222

6222

(b) 31

3111

9. Equations with related roots Given the roots of a polynomial

equation the equation can be formed as follows.

Quadratic 02 xx

Cubic 023 xxx

7

Quartic 0234 xxxx

Etc..

Example The roots of the equation 05423 xxx are ,

and . Find the cubic equations whose roots are

(i) 2 , 2 , 2

(ii) 1 , 1 , 1

1

3

1

(i) For new equation 22222

164222222

408222

Equation 040162 23 xxx

(ii) For new equation

43111

1324

32

111

111111

71145

1

1

111

Equation 074 23 xxx

8

Mathematical Induction

Example Prove by induction that, where n is a positive integer

(i) 152 n is divisible by 24

(ii) 1216

1

1

2

nnnrn

r

(i) Let 152 nnf

nf

nf

n

nn

nnn

2

22

2212

5.24

155.24

1512415.25151

Hence if nf is divisible by 24, so is 1nf

When n = 1, 24125151 2 f , so the result is true for 1n

Hence it will be true successively for ,...4,3,2n

i.e. it is universally true.

(ii) Let the result be true for some value of kn say

i.e. kfkkkrk

r

1216

1

1

2

then

67216

1

161216

1

11216

1

2

2

21

1

2

kkk

kkkk

kkkkrk

r

132216

1 kfkkk

Hence if the result is true for kn , it is also true for 1 kn

9

With 1n LHS = 1

RHS = 13216

1

Hence result is true for 1n , and so will be successively true for

,...4,3,2n

Summation of finite series

(i) 12

1.....321

1

nnnrn

r

summing the AP

(ii) 1216

1.....321 2222

1

2

nnnnrn

r

Proof by induction, or as follows, using the identity

1331 233 nnnn

1131312

1232323

........

........

........

1232321

113131

13 31

233

233

233

233

233

nnnn

nnnn

nnnn

10

121

121

132

326

2136662

1333

3311

612

2

232

223

23223

22

nnnr

nnn

nnn

nnnr

nnnrnnn

nnnrnnn

nrrn

(iii) 2213 1 nnr

proof by induction or by the identity 14641 2344 nnnnn

Example Find the sum of n terms of the series ... 53 32 11

The rth

term 12 rrU r so we require

rrrrrrn

r

22

1

2212

141

31221

1121212

61

61

21

61

1

nnn

nnn

nnnnnrrn

r

(iv) nn

r

r nxxxxrx

....32.1 32

1

Here the coefficients are in AP and the terms in x are in GP. Let the sum

be nS

11

x

nx

x

xxS

nxxxxxSx

nxxnxxxS

nxxxxS

nn

n

nn

n

nn

n

n

n

11

1

........1

1.......2

.......32

1

2

132

132

32

Method to be learnt, not as a formula!

Note that if 1x then as n 0nx

21 x

xSn

i.e.

21 x

xS

In other words the series is convergent with a finite sum of 21 x

x

so

long as 1x

Method of differences This is based upon the following result or similar.

If 1 rfrfU r 01

fnfUn

r

r

It isn‟t really necessary to learn this as a formula

Example Find

n

r rr1 1

1

111

1

1

rrrr Cover up rule.

1

11

1

1

1 rrrr

n

r

12

11

11

1

11

......11

1141

31

31

21

21

n

n

n

n

n

nn

[As was stated, similar, 1 rfrfU r ]

COMPLEX NUMBERS

Real numbers consist of integers, rational numbers and irrational numbers; they

can be represented by points on a (real) number line.

A complex number is one which can be written in the form ibaz

where a and b are real and 1i

(i) 0,0 ba ; the number ibaz is said to be imaginary

(ii) 0,0 ba ; the number ibz is said to be purely imaginary

(iii) 0,0 ba ; the number az is said to be real

Given ibaz , a is known as the real part of z, and b is known as the

imaginary part of z. –both real!

Powers of i 12 i , ii 3 , 14 i , ii 5 , 16 i , etc…

Given ibaz , the imaginary number ibaz is said to be the

conjugate of z. Imaginary roots of polynomial equations occur in

conjugate pairs.

Example 0422 zz has roots 312

1642iz

i.e the roots are 31 i and 31 i

The zero complex number If ibaz and 0z then both 0a and

0b . i.e. both real parts and imaginary parts are zero.

13

Equal complex numbers If iqpiba then pa and qb . i.e.

real parts are equal and imaginary parts are equal.

Some algebraic operations on complex numbers

Addition qbipaiqpiba

Subtraction qbipaiqpiba

Multiplication by real number ikbkaibak

Product of two complex nos. aqbpibqapiqpiba

Example Show that for the complex number z and its conjugate z

(i) the sum is real

(ii) the difference is purely imaginary

(iii) the product is a real positive number.

Let ibaz ibaz

(i) azz 2 real

(ii) bizz 2 purely imaginary

(iii) 22 bazz real and positive

Example Find real numbers x and y such that 052143 iyix

yxiyxiyix 245352143

RP = 0, 053 yx

IP = 0, 024 yx

Division

idcidc

idciba

idc

ibaetc..

10,5

yx

Makes real denominator

14

Example 13

7

13

3

13

73

49

276

2323

232

23

22

2

ii

i

ii

ii

ii

i

i

Graphical representation of a complex number is by a point in the

Argand diagram, otherwise called the z plane.

P represents ibaz

Note the vector like use of the line

segment OP to denote the position

of P.

NB Not an i to be seen!

Addition

Parallelogram law

Subtraction 2121 zzzz

So use above law.

The modulus and argument of a complex number

Let P(x, y) represent iyxz

Modulus 22 yxrz

Argument zarg where is a directed

angle, in radians, and in the angle made by

OP with Ox measured from Ox where

P(a, b)

z

x

y

0

Axis for the

imaginary

part

(imaginary

axis)

Axis for the real part (real axis)

2z

x

y

0

1z

21 zz

P(x, y)

r

x

y

0

15

Take care with argument!

x

y1tan can lead to the wrong angle.

Always plot a point in the argand diagram to find the argument.

Example Find the modulus and argument of iz 3

213 z

3

1tan

6

6

5arg

z

NB If you are going to use your calculator when exact values are

required, work in degrees and convert to radians.

Note that

3

1tan 1 has a principle value of –30, not the angle which

we want… now refer to the diagram.

Modulus/ Argument form of a complex number

Using simple right angled triangle trig

cosrx and sinry

sincos irriyxz

sincos irz

P r

x

y

0

1

3

P(x, y)

r

x

y

0

16

Geometrically the difference of two complex numbers is very important

2121 PPzz

21arg zz angle between Ox and 12 PP

measured from Ox

Product and Quotient of two complex numbers

Given 1111 sincos irz and 2222 sincos irz it is quite

straight forward to show that

21212121 sincos irrzz

So that 2121 zzzz and 2121 argargarg zzzz

2121

2

1

2

1 sincos ir

r

z

z

So that 2

1

2

1

z

z

z

z and 21

2

1 argargarg zzz

z

Geometrically the quotient of two complex numbers is important

21

2

1 argargarg zzz

z

and is the angle

between 1OP and 2OP measured from

2OP

2z

r

1z

r

1P

x

y

0

2P

2z

r

1z

r

1P

x

y

0

2P

17

Example (a) Express iz 11 and 312 iz each in the form

sincos ir by plotting points 1P and 2P

representing 1z and 2z in an argand diagram. Show that

12arg zz is the acute angle where

2

13tan 1

(b) Write 2

1

z

z in the form iba and hence show that

13

13

12

5tan

2111 z

4

3arg 1

z

4

3sin

4

3cos21

iz

2312 z 3

arg 2

z

3sin

3cos22

iz

(a) 2

13

11

13gradtan 21

PP

2

13tan 1

(b)

ii

i

i

i

i

z

z

4

13

4

13

4

1313

31

31

31

1

2

1

2

1argz

z

3,12P

x

y

0

1,11 P

18

Since 12

5

34

3argargarg 21

2

1

zz

z

z then if r

z

z

2

1

4

13

12

5cos

r and

4

13

12

5sin

r

13

13

4

13

4

13

12

5tan

Extensions of products

Let kkkk rz sincos nk ,...3,2,1

Then

nnnn rrrrzzzz ..sin..cos...... 221221321321

Now if zzzz n ....21 say where sincos irz

ninrz nn sincos

Example If 31 iz express 5z in modulus argument form

231 z 325 z

32arg z

3

10arg 5

z ; but it is usual to express

argument in the range so we would take

32arg z

32

325 sincos32 z

32

3

(-1, 3)

19

Loci in the argand diagram Let P(x, y) represent iyxz in the agrand

diagram.

(i) Associated with modulus

Type azz 1 where 1z is a

constant complex number

represented by 1P ; and a is a real

constant

Locus of P is circle centre 1P , rad a.

Type 21 zzzz

Locus of P is perpendicular bisector

of 21PP

The above two are readily recognised, and their equations follow

Example 32 iz rewritten 32 iz gives a circle

with equation 91222 yx

However the equations of these, and other perhaps unfamiliar loci, can be

determined algebraically.

Example 32 ziz

Let iyxz

32 iyxiiyx

iyxyix 321

1P z

1z

P

a

1P

2P

P

20

35224330

3624412

341

321

22

2222

2222

2222

yxyx

yxxyyx

yxyx

yxyx

08335

3222 yxyx

This is the equation of a circle –the centre and radius can be found by

completing the square.

Example 64 zz

Let iyxz

64 iyxiyx 64 2222 yxyx

ellipsean 1

59

2

45925

459205

4202599

253

82012

12364

64

22

22

22

222

22

22

222222

2222

yx

yx

yxx

xxyx

xyx

xyx

yxyxyx

yxyx

(ii) Associated with argument

Type 1arg zz . The locus of P(x, y) will be a part line through P,

(representing 1z ) making an angle with 0x.

21

Example 4

1arg

iz

Write 4

1arg

iz and we have the locus as a part line through

(1, –1) making 4

with 0x.

Gradient = 14

tan

Equation. 111 xy

2 xy for 1x

Algebraic approach gives the equation of the line but will not reveal the

restriction. As above

4

1arg

iz 4

11arg

yix

4

tan1

1

x

y 1

1

1

x

y Hence 2 xy

So the diagram is essential!

Transformations of the Argand diagram

If iyxz is represented by P(x, y) in the z plane and zfw then

the function f represents a mapping of the point P to another point Q

where Q(u, v) represents ivuw . Consequently the locus of P

transforms to the locus of Q.

The procedure in finding the locus of Q is illustrated in the following:

4

1z

r

1P

x

y

0

P

22

Example The locus of P representing iyxz in the line 1x .

Show that the locus of Q representing w where z

w1

is a

circle which passes through the origin.

zw

1

iyxivu

1

iyx

iyx

iyxivu

1

2222 yx

yi

yx

xivu

22 yx

xu

,

22 yx

yv

Locus P is 1x 21

1

yu

,

21 y

yv

yuv i.e. u

vy

2

2

1

1

u

vu

22

2

vu

uu

uvu 22 022 vuu 4122

21 vu

Circle centre 0,21 radius =

21 and passing through 0.

Example The locus of P representing iyxz is the rectangular

hyperbola 422 yx . If 2zw find the locus of the

point Q representing w.

2zw xyiyxiyxivu 2222

,22 yxu xyv 2

but 422 yx 4u . Hence the locus of Q is the line 4u

23

CALCULUS: FURTHER DIFFERENTIATION

General definition of a derivative

x

xfxxf

xxf

0

lim

Proof –in essence a first principles procedure.

Let xfy

xf

x

xfxxf

x

y

x

xfxxf

x

y

xfxxfy

xxfyy

limlim

The formula might be otherwise expressed as

h

xfhxf

hxf

0

lim

Example Find using the formula the derivative of 12

1

x

1212

12212

0

lim

1212

2

0

lim

1212

12212

0

lim

12

1

12

1

0

lim

xhxh

hxx

h

xhxh

h

h

xhxh

hxx

h

h

xhx

hxf

24

212

2

1212

2

x

xx

Logarithmic differentiation

Example Obtain the derivative of xxsin

x

xxxx

x

xxxy

dx

dy

xxxx

dx

dy

y

xxy

xy

xy

x

x

x

sinlncos

sinlncos

1.sinlncos.

1

lnsinln

lnln

sin

sin

sin

MATRICES

3 x 3 matrices

(1) The determinant of

333

222

111

cba

cba

cba

A written Adet or A is given

by 33

22

1

33

22

1

33

22

1detba

bac

ca

cab

cb

cbaA

The 2x2 determinants 33

22

33

22

33

22,,

ba

ba

ca

ca

cb

cb are called the minors of

111 ,, cba respectively.

Implicit Product rule

25

In expanded form

231312123132213321det cbacbacbacbacbacbaA

And can be obtained schematically and certainly not by formula.

If 0det A , the matrix is said to be singular.

(2) Minors and cofactors of detA. The minor of an element of Adet is

found by deleting the row and column containing the element, and

forming a determinant of the remaining four elements. The cofactor

is then found by multiplying by +1 or –1 according to

111

111

111

e.g. From (2), the minor of 2a is 33

11

cb

cb and cofactor is

33

11

cb

cb

(3) The Adjugate matrix, written Aadj

Either (a) replace each element in A by its cofactor and transpose the

resulting matrix,

Or (b) transpose first and then replace elements by cofactors.

Example

150

132

011

A

1510

112

112

111

511

1022

adj

1

A

You could try it the other way round to check out the result!

(4) The inverse matrix, written 1

A is given by

AA

A adj.det

11

26

Example To find the inverse of

821

482

248

A

392)646416()168512(det A

adjA =

56124

286212

02856

56280

126228

412561

71

983

981

141

19631

983

141

71

1

0

56124

286212

02856

392

1A

You could check to see if IAA 1

(5) Solutions to simultaneous equations by matrices

Example Solve the equations

132

5

42

zyx

zyx

zyx

Write

1

5

4

132

111

121

z

y

x

1

5

4

132

111

1211

z

y

x

1

5

4

0

93

91

95

93

93

93

95

92

z

y

x

27

You will find

920x ,

31y ,

922z

Geometrically these equations represent three planes and this unique

solution corresponds to intersection of the planes in one point.

However other possibilities arise when solving

c

b

a

z

y

x

A

Here are some

(i) All three planes could be parallel

(ii) Two could be parallel and one not

(iii) The planes could intersect in three parallel lies to form an

enclosed triangular prism

In each of these cases there will be no values of x, y, z which satisfy the

equations simultaneously. The equations are said to be inconsistent and

0det A

(iv) The planes intersect in a line. In this case there is a line

solution and again 0det A . The equations are said to be not

independent.

Example Show that the equations 0352 zyx , 24 zyx

and 0457 zy are not independent.

First write as

0

2

0

570

411

352

z

y

x

and obtain 02556021010

570

415

352

det

A

Next

457

24

0352

zy

zyx

zyx

eliminate x 457 zy (this being the 3rd equation)

28

Hence the equations are not independent. The three planes intersect in a line, so

we can find a „line‟ solution.

In 457 zy let kz

7

45ky

Sub into 2nd

equation 247

45

k

kx

7

2310 kx

Line solution is

7

2310 kx ,

7

45ky , kz

You could check with the 1st equation if you wish!

Example Show that the equations 1 zyx , 4232 zyx

and 223 zyx are inconsistent .

First 0249463

123

232

111

det

A

Next

3rd and1st from eliminate

2first from eliminate

545

645

223

4232

1

x

x

zy

zy

zyx

zyx

zyx

These two equations are clearly inconsistent and consequently so are the

original three. Hence there are no solutions to these equations.

(6) Reduction to Echelon form

Example Solve the equations 7 zyx , 92 zyx ,

12 zyx using the reduction to echelon form.

First we write in augmented matrix form as follows:

29

1112

9211

7111

31 rr

1112

9211

8023

32 2rr

1112

11015

8023

21 2rr

1112

11015

14007

147 x 2x

115 yx 1y

12 zyx 4z

The matrix is said to be in echelon form when there are zeros above or

below the leading diagonal.

In the previous example, the matrix can be reduced to

28700

2120

7111

Try it and see!

30

Example Reduce the following equations to echelon form.

,132 zyx 32 zyx , zyx 554

Consider the following

(a) 1 (b) 7,1 (c) 7,1

554

321

1312

you will find reduces to

72200

53220

1312

(a) 1 There will be a unique solution to the equations.

(b) 7,1 Gives

7000

5120

1312

and the equations are inconsistent

(c) ) 7,1 Gives

0000

5120

1312

and the equations are consistent but not independent.

In this case there will be a line solution. You could find it in the form

kx 58 , ky , 53 kz .

(7) Translations in the x/y plane

We let the point (x, y) be represented by

1

y

x

31

Transformation matrices will be of the form

100

rqp

nml

M

1

'

'

1100

y

x

rqypx

nmylx

rqp

nml

z

y

x

M

so that rqypxnmylxyx ,,

In particular consider

(i)

100

10

01

b

a

M

11100

10

01

by

ax

y

x

b

a

byaxyx ,, which is a translation

b

a

The following 2x2 matrices can be found in the formula booklet

(ii) A rotation through anticlockwise as represented by

cossin

sincos becomes

100

0cossin

0sincos

(iii) A reflection in tanxy as represented by

2cos2sin

2sin2cos becomes

100

02cos2sin

02sin2cos

32

Invariance

For an invariant point, if any, put xx ' and yy ' so that

11

y

x

rqypx

nmylx

Example Find a 33 matrix corresponding to

(i) a rotation of anticlockwise about 0 (where is acute and 43tan )

followed by a translation

1

1

(ii) a translation

1

1 followed by a rotation of anticlockwise about 0

(where is acute and 43tan ).

For each transformation find the image of (x, y) and identify invariant

points, if any.

Can you describe either or both of these transformations in simpler

form?

(i) The rotation 2x2 matrix is

cossin

sincos where

53sin ,

54cos

The 3x3 matrix is

100

0

0

54

53

53

54

33

The translation matrix is

100

110

101

The matrix for the combined transformations will be

100

1

1

100

0

0

100

110

101

54

53

53

54

54

53

53

54

1

1

1

1100

1

1

54

53

53

54

54

53

53

54

yx

yx

y

x

1,1,54

53

53

54 yxyxyx

For invariance

1

2

05354351

05353451

54

53

53

54

y

x

yxyxyyxy

yxyxxyxx

(-2, -1) is an invariant point. Hence the transformation is an

anticlockwise rotation about (-2, -1)

(ii) The matrix is

100100

110

101

100

0

0

51

54

53

57

53

54

54

53

53

54

11100

51

54

53

57

53

54

51

54

53

57

53

54

yx

yx

y

x

51

54

53

57

53

54 ,, yxyxyx

34

For invariance

2

1

0131435

0737345

51

54

53

57

53

54

y

x

yxyxyyxy

yxyxxyxx

(-1, -2) is an invariant point. Hence the transformation is an

anticlockwise rotation about (-1, -2)

Example To find a single 3x3 matrix for reflection in the line

643 xy and the image of point (x, y)

This line is 234 xy

First translate

2

0 with matrix

100

210

001

Next reflect in xy34 with matrix

100

02cos2sin

02sin2cos

where 34tan and, therefore

2524

916

34

1

2sin

,

257

916

916

1

1cos

Finally translate

2

0 with matrix

100

210

001

The single matrix will be given by

y

x

(0,2)

0

35

100

100100

210

001

100

210

001

100

0

0

100

210

001

2536

257

2524

2548

2524

257

2514

257

2524

2548

2524

257

257

2524

2524

257

11100

2536

257

2524

2548

2524

257

2536

257

2524

2548

2524

257

yx

yx

y

x

2536

257

2524

2548

2524

257 ,, yxyxyx

Linear Transformations All a foregoing transformations are called linear

because straight lines map to straight lines.

An invariant line is one which all points map to some other point on the

same line.

A line of invariant points is one in which all points on the line map onto

themselves.

Example The transformation T of the x/y plane is equivalent to a

reflection in xy followed by a translation in which (1, 2) is mapped

onto (0, 1).

a) find the 3x3 matrix representing T

b) determine the fixed points of T

c) find the image of the line cxy

36

d) find the image of parabola xy 42

The reflection matrix is

100

001

010

The translation is given by

1

1

2

1

1

0

a) the matrix for T is

100

101

110

100

001

010

100

110

101

b)

1

1

1

1

x

y

y

x

c) Let the lien be described by parametric equations tx , cty

1

1

1

1

t

ct

y

x

d) Let the parabola be described by tytx 2,2

1

1

12

1

2t

t

y

x

For invariant points 11

1

yx

xy

yx

1 yx is the line of invariant points

cxyty

ctx

1

1

cxy is an invariant line for all c.

2

41

211

1

12

xy

ty

tx

another parabola

37

Notes

38

Notes

39

Notes