Pure FP1 Revision Notes, Lecture notes of Complex analysis

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Pure FP

Revision Notes

March 2013

• 1 Complex Numbers..................................................................................................
  • Definitions and arithmetical operations
  • Complex conjugate
    • Properties
  • Complex number plane, or Argand diagram
  • Complex numbers and vectors
  • Multiplication by i
  • Modulus of a complex number
  • Argument of a complex number
  • Equality of complex numbers
  • Square roots
  • Roots of equations
• 2 Numerical solutions of equations
  • Accuracy of solution
  • Interval bisection....................................................................................................................
  • Linear interpolation................................................................................................................
  • Newton-Raphson
• 3 Coordinate systems
  • Parabolas
    • Parametric form
    • Focus and directrix
    • Gradient..........................................................................................................................................
    • Tangents and normals
  • Rectangular hyperbolas........................................................................................................
    • Parametric form
    • Tangents and normals
• 4 Matrices - Order of a matrix ............................................................................................................................ - Identity matrix ................................................................................................................................
  • Determinant and inverse
  • Linear Transformations
    • Basis vectors...................................................................................................................................
    • Finding the geometric effect of a matrix transformation ...............................................................
    • Finding the matrix of a given transformation.................................................................................
    • Rotation matrix...............................................................................................................................
    • Determinant and area factor ...........................................................................................................
• 5 Series
• 6 Proof by induction
  • Summation
  • Recurrence relations
  • Divisibility problems
  • Powers of matrices...............................................................................................................
• 7 Appendix
  • Complex roots of a real polynomial equation
  • Formal definition of a linear transformation
  • Derivative of xn , for any integer
• 8 Index

1 Complex Numbers

Definitions and arithmetical operations

i = √− 1 , so √−16 = 4𝑖𝑖, √− 11 = √ 11 i , etc. These are called imaginary numbers

Complex numbers are written as z = a + bi , where a and b ∈ ℝ. a is the real part and b is the imaginary part.

+, –, × are defined in the ‘sensible’ way; division is more complicated.

( a + bi ) + ( c + di ) = ( a + c ) + ( b + d ) i ( a + bi ) – ( c + di ) = ( a – c ) + ( b – d ) i ( a + bi ) × ( c + di ) = ac + bdi^2 + adi + bci = ( ac – bd ) + ( ad +bc ) i since i^2 = –

So (3 + 4 i ) – (7 – 3 i ) = –4 + 7 i and (4 + 3 i ) (2 – 5 i ) = 23 – 14 i

Division – this is just rationalising the denominator.

3+4𝑖𝑖 5+2𝑖𝑖 =^

3+4𝑖𝑖 5+2𝑖𝑖 ×^

5 − 2 𝑖𝑖 5 − 2 𝑖𝑖 multiply top and bottom by the complex conjugate

=

23+14𝑖𝑖 25+ =

23 29

14 29 𝑖𝑖

Complex conjugate

z = a + bi The complex conjugate of z is z * = 𝑧𝑧 = a – bi

Properties

If z = a + bi and w = c + di , then

(i) {( a + bi ) + ( c + di )}* = {( a + c ) + ( b + d ) i }*

= { ( a + c ) – ( b + d ) i }

= ( a – bi ) + ( c – di )

⇔ ( z + w )* = z * + w *

Multiplication by i

i (3 + 4 i ) = –4 + 3 i – on an Argand diagram this would have the effect of a positive quarter turn about the origin.

In general;

i ( a + bi ) = – b + ai

Modulus of a complex number

This is just like polar co-ordinates.

The modulus of z is  z  and

is the length of the complex number

 z  = √𝑎𝑎 2 + 𝑏𝑏 2.

z z * = ( a + bi )( a – bi ) = a^2 + b^2

⇒ z z * =  z ^2.

Argument of a complex number

The argument of z is arg z = the angle made by the complex number with the positive x -axis.

By convention, – π < arg z ≤ π.

N.B. Always draw a diagram when finding arg z****.

Example: Find the modulus and argument of z = –6 + 5 i.

Solution: First sketch a diagram (it is easy to get the argument wrong if you don’t).

 z  = (^) √ 62 + 5^2 = (^) √ 61

and tan α =

5

6 ⇒^ α^ = 0⋅^694738276

⇒ arg z = θ = π – α = 2.45 to 3 S. F.

Re

Im

z

iz x^ ( a ,^ b )

x (– b , a )

Re

Im  z 

x (–6, 5)

θ 6

5 α

Re

Im

 z 

x z = a + bi

θ

Equality of complex numbers

a + bi = c + di ⇒ a – c = ( d – b ) i

⇒ ( a – c )^2 = ( d – b )^2 i^2 = – ( d – b )^2 squaring both sides

But ( a – c )^2 ≥ 0 and – ( d – b )^2 ≤ 0

⇒ ( a – c )^2 = – ( d – b )^2 = 0

⇒ a = c and b = d

Thus a + bi = c + di

⇒ real parts are equal ( a = c ), and imaginary parts are equal ( b = d ).

Square roots

Example: Find the square roots of 5 + 12 i , in the form a + bi , a , b ∈ ℝ.

Solution: Let (^) √5 + 12𝑖𝑖 = a + bi

⇒ 5 + 12 i = ( a + bi )^2 = a^2 – b^2 + 2 abi

Equating real parts ⇒ a^2 – b^2 = 5, I

equating imaginary parts ⇒ 2 ab = 12 ⇒ a = 6 𝑏𝑏

Substitute in I ⇒ � 6 𝑏𝑏 �

2 − 𝑏𝑏 2 = 5

⇒ 36 – b^4 = 5 b^2 ⇒ b^4 + 5 b^2 – 36 = 0

⇒ ( b^2 – 4)( b^2 + 9) = 0 ⇒ 𝑏𝑏 2 = 4

⇒ b = ± 2, and a = ± 3

⇒ √5 + 12𝑖𝑖 = 3 + 2 i or –3 – 2 i.

Roots of equations

( a ) Any polynomial equation with complex coefficients has a complex solution.

The is The Fundamental Theorem of Algebra , and is too difficult to prove at this stage.

Corollary: Any complex polynomial can be factorised into linear factors over the complex numbers.

2 Numerical solutions of equations

Accuracy of solution

When asked to show that a solution is accurate to n D.P ., you must look at the value of f ( x ) ‘half’ below and ‘half’ above, and conclude that

there is a change of sign in the interval , and the function is continuous , therefore there is a solution in the interval correct to n D. P.

Example: Show that α = 2⋅0946 is a root of the equation f ( x ) = x^3 – 2 x – 5 = 0, accurate to 4 D.P.

Solution:

f (2.09455) = –0⋅0000165…, and f (2.09465) = +0⋅ 00997

There is a change of sign and f is continuous

⇒ there is a root in [2 ⋅ 09455, 2 ⋅ 09465] ⇒ root is α = 2⋅ 0946 to 4 D.P.

Interval bisection

(i) Find an interval [ a , b ] which contains the root of an equation f ( x ) = 0.

(ii) x = 𝑎𝑎+𝑏𝑏 2 is the mid-point of the interval [ a ,^ b ]

Find 𝑓𝑓 � 𝑎𝑎+𝑏𝑏 2 �^ to decide whether the root lies in^ �𝑎𝑎,^

𝑎𝑎+𝑏𝑏 2 �^ or^ �^

𝑎𝑎+𝑏𝑏 2 ,^ 𝑏𝑏�^.

(iii) Continue finding the mid-point of each subsequent interval to narrow the interval which contains the root.

Example: (i) Show that there is a root of the equation f ( x ) = x^3 – 2 x – 7 = 0 in the interval [2, 3]. (ii) Find an interval of width 0.25 which contains the root.

Solution: (i) f (2) = 8 – 4 – 7 = –3, and f (3) = 27 – 6 – 7 = 14

There is a change of sign and f is continuous ⇒ there is a root in [2, 3].

(ii) Mid-point of [2, 3] is x = 2⋅5, and f (2⋅5) = 15⋅625 – 5 – 7 = 3⋅ 625

⇒ root in [2, 2⋅5]

Mid-point of [2, 2⋅5] is x = 2⋅25, and f (2⋅25) = 11⋅390625 – 4⋅5 – 7 = –0⋅ 109375

⇒ root in [2⋅25, 2⋅5], which is an interval of width 0⋅ 25

Linear interpolation

To solve an equation f ( x ) using linear interpolation.

First, find an interval which contains a root,

second, assume that the curve is a straight line and use similar triangles to find where it crosses the x -axis, third, repeat the process as often as necessary.

Example: (i) Show that there is a root, α , of the equation

f ( x ) = x^3 – 2 x – 9 = 0 in the interval [2, 3].

(ii) Use linear interpolation once to find an approximate value of α.

Give your answer to 3 D. P.

Solution: (i) f (2) = 8 – 4 – 9 = –5, and f (3) = 27 – 6 – 9 = 12

There is a change of sign and f is continuous ⇒ there is a root in [2, 3].

(ii) From (i), curve passes through (2, –5) and (3, 12), and we assume that the curve is a straight line between these two points.

Let the line cross the x -axis at ( α, 0)

Using similar triangles

3−𝛼𝛼 𝛼𝛼−2 =^

12 5

⇒ 15 – 5α = 12 α – 24

⇒ α = 39 17 = 2^

5 17

⇒ α = 2⋅294 to 3 D.P.

(2, –5)

(3, 12)

2 3 5

α^ 3 –^ α

12

α – 2

3 Coordinate systems

Parabolas

y^2 = 4 ax is the equation of a parabola which passes through the origin and has the x -axis as an axis of symmetry.

Parametric form

x = at^2 , y = 2 at satisfy the equation for all values of t. t is a parameter, and these equations are the parametric equations of the parabola y^2 = 4 ax.

Focus and directrix

The point S ( a , 0) is the focus , and

the line x = – a is the directrix.

Any point P of the curve is equidistant from the focus and the directrix, PM = PS.

Proof: PM^2 = at^2 – (– a ) = at^2 + a

PS^2 = ( at^2 – a )^2 + (2 at )^2 = a^2 t^4 – 2 a^2 t^2 + a^2 + 4 a^2 t^2

= a^2 t^4 + 2 a^2 t^2 + a^2 = ( at^2 + a )^2 = PM^2

⇒ PM = PS.

Gradient

For the parabola y^2 = 4 ax , with general point P , ( at^2 , 2 at ), we can find the gradient in two ways:

1. y^2 = 4 ax ⇒ 2 y 𝑑𝑑𝑑𝑑 𝑑𝑑𝑥𝑥 = 4 a^ ⇒^

𝑑𝑑𝑑𝑑 𝑑𝑑𝑥𝑥 =^

2𝑎𝑎 𝑑𝑑

2. At P , x = at^2 , y = 2 at ⇒ 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 = 2 a ,^

𝑑𝑑𝑥𝑥 𝑑𝑑𝑑𝑑 = 2 at

⇒

𝑑𝑑𝑑𝑑 𝑑𝑑𝑥𝑥 =

𝑑𝑑𝑑𝑑 �𝑑𝑑𝑑𝑑 𝑑𝑑𝑥𝑥 (^) �𝑑𝑑𝑑𝑑 =^

2𝑎𝑎 2𝑎𝑎𝑑𝑑 =^

1 𝑑𝑑

y

Directrix x = −a

Focus S, (a, 0)

M

+ x

+ (^) P ( at (^2) , 2 at )

Tangents and normals

Example: Find the equations of the tangents to y^2 = 8 x at the points where x = 18, and show that the tangents meet on the x -axis.

Solution: x = 18 ⇒ y^2 = 8 × 18 ⇒ y = ± 12

2 y

𝑑𝑑𝑑𝑑 𝑑𝑑𝑥𝑥 = 8^ ⇒^

𝑑𝑑𝑑𝑑 𝑑𝑑𝑥𝑥 =^ ±^

1 3 since^ y^ =^ ±^12

⇒ tangents are y – 12 = 1 3 ( x^ – 18)^ ⇒^ x^ – 3 y^ + 18 = 0^ at (18, 12)

and y + 12 = − 1 3 ( x^ – 18)^ ⇒^ x^ + 3 y^ + 18 = 0.^ at (18, –12)

To find the intersection, add the equations to give

2 x + 36 = 0 ⇒ x = –18 ⇒ y = 0

⇒ tangents meet at (–18, 0) on the x -axis.

Example: Find the equation of the normal to the parabola given by x = 3 t^2 , y = 6 t.

Solution: x = 3 t^2 , y = 6 t ⇒ 𝑑𝑑𝑥𝑥 𝑑𝑑𝑑𝑑 = 6 t ,^

𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 = 6,

𝑑𝑑𝑑𝑑 𝑑𝑑𝑥𝑥 =

𝑑𝑑𝑑𝑑 �𝑑𝑑𝑑𝑑 𝑑𝑑𝑥𝑥 �𝑑𝑑𝑑𝑑

6 6 𝑑𝑑 =^

1 𝑑𝑑

⇒ gradient of the normal is – t

⇒ equation of the normal is y – 6 t = – t ( x – 3 t^2 ).

Notice that this ‘general equation’ gives the equation of the normal for any particular value of t :– when t = –3 the normal is y + 18 = 3( x – 27) ⇔ y = 3 x – 99.

Rectangular hyperbolas

A rectangular hyperbola is a hyperbola in which the asymptotes meet at 90o.

xy = c^2 is the equation of a rectangular hyperbola in which the x -axis and y -axis are perpendicular asymptotes.

Parametric form

x = ct , y = 𝑐𝑐 𝑑𝑑 are parametric equations of the hyperbola xy = c^2_._

x

y

4 Matrices

You must be able to add, subtract and multiply matrices.

Order of a matrix ............................................................................................................................

An r × c matrix has r rows and c columns;

the fi R st number is the number of R ows

the se C ond number is the number of C olumns.

Identity matrix ................................................................................................................................

The identity matrix is I = �

Note that MI = IM = M for any matrix M.

Determinant and inverse

Let M = �𝑎𝑎^ 𝑏𝑏 𝑐𝑐 𝑑𝑑

� then the determinant of M is

Det M = | M | = ad – bc.

To find the inverse of M = �𝑎𝑎^ 𝑏𝑏 𝑐𝑐 𝑑𝑑

Note that M –1 M = M M –1^ = I

(i) Find the determinant, ad – bc. If ad – bc = 0, there is no inverse.

(ii) Interchange a and d (the leading diagonal) Change sign of b and c , (the other diagonal) Divide all elements by the determinant, ad – bc.

⇒ 𝑴𝑴−1^ =

1 𝑎𝑎𝑎𝑎−𝑏𝑏𝑏𝑏 (^) � 𝑑𝑑^ −𝑏𝑏 −𝑐𝑐 𝑎𝑎

Check:

M –1 M =

1 𝑎𝑎𝑎𝑎−𝑏𝑏𝑏𝑏 (^) � 𝑑𝑑^ −𝑏𝑏 −𝑐𝑐 𝑎𝑎

1 𝑎𝑎𝑎𝑎−𝑏𝑏𝑏𝑏 (^) �𝑑𝑑𝑎𝑎^ − 𝑏𝑏𝑐𝑐^0 0 −𝑏𝑏𝑐𝑐 + 𝑎𝑎𝑑𝑑

Similarly we could show that M M –1^ = I.

Example: M = �

� and MN = �−^1 2 1

�. Find N.

Solution: Notice that M –1^ ( MN ) = ( M –1 M ) N = IN = N multiplying on the left by M –

But MNM –1^ ≠ IN we can not multiply on the right by M –

First find M –

Det M = 4 × 3 – 2 × 5 = 2

⇒ 𝑴𝑴−1^ =

1 (^2) � 3 −^2 − 5 4

Using M –1^ ( MN ) = IN = N

⇒ N =

1 (^2) � 3 −^2 − 5 4

� �−^1

1 (^2) �−^7 13 − 6

� = �−3.5^2

Linear Transformations

A matrix can represent a transformation, but the point must be written as a column vector before multiplying by the matrix.

Example: The image of (2, 3) under T = �

� is given by �

⇒ the image of (2, 3) is (23, 8).

Note that the image of (0, 0) is always (0, 0)

⇔ the origin never moves under a matrix (linear) transformation

Basis vectors...................................................................................................................................

The vectors i = �^1 0

� and j = �^0 1

� are called basis vectors, and are particularly important in describing the geometrical effect of a matrix, and in finding the matrix for a particular geometric transformation.

�𝑎𝑎^ 𝑏𝑏

� �^1

� and �𝑎𝑎^ 𝑏𝑏 𝑐𝑐 𝑑𝑑

� �^0

i = �

𝑐𝑐�, the^ first^ column, and^ j^ =^ �

�, the second column

This is a more important result than it seems!

j = �^0 1

�. This will be the second column of the matrix �

⇒ Matrix of the reflection is � 0 −^1 − 1 0

Rotation matrix...............................................................................................................................

From the diagram we can see that

i = �^1 0

� → �cos^ 𝜃𝜃 sin 𝜃𝜃

j = �^0 1

� → �−^ sin^ 𝜃𝜃 cos 𝜃𝜃

These will be the first and second columns of the matrix

⇒ matrix is 𝑅𝑅𝜃𝜃 = �cos^ 𝜃𝜃^ −^ sin^ 𝜃𝜃 sin 𝜃𝜃 cos 𝜃𝜃

Determinant and area factor ...........................................................................................................

For the matrix 𝐴𝐴 = �

�𝑎𝑎^ 𝑏𝑏

and �𝑎𝑎^ 𝑏𝑏 𝑐𝑐 𝑑𝑑

� �^0

⇒ the unit square is mapped on to the

parallelogram as shown in the diagram.

The area of the unit square = 1.

The area of the parallelogram = ( a + b )( c + d ) – 2 × ( bc + 1 2 ac^ +^

1 2 bd )

= ac + ad + bc + bd – 2 bc – ac – bd

= ad – bc = det A.

All squares of the grid are mapped onto congruent parallelograms

⇒ area factor of the transformation is det A = ad – bc.

A (1, 0)

A ′ (cos θ , sin θ)

B (0, 1) B ′ (–sin θ, cos θ)

x

y

θ

θ

cos θ

cos θ sin θ

sin θ

×

× × ×

b

d

c

a

( b , d )

( a , c )

b

d

c

a

x

y

1

1

5 Series

You need to know the following sums

� 𝑟𝑟

𝑛𝑛

𝑟𝑟=

= 1 + 2 + 3 + ⋯ + 𝑛𝑛 =

1 2

𝑛𝑛(𝑛𝑛 + 1)

� 𝑟𝑟 2

𝑛𝑛

𝑟𝑟=

= 1^2 + 2^2 + 3^2 + ⋯ + 𝑛𝑛 2 =

1 6 𝑛𝑛(𝑛𝑛 + 1)(2𝑛𝑛 + 1)

� 𝑟𝑟 3

𝑛𝑛

𝑟𝑟=

= 1^3 + 2^3 + 3^3 + ⋯ + 𝑛𝑛 3 =

1 4 𝑛𝑛 2 (𝑛𝑛 + 1)^2

= � 1 2 𝑛𝑛(𝑛𝑛^ + 1)�^

2 = a fluke, but it helps to remember it

Example: Find.

Solution:

= 14 𝑛𝑛 2 (𝑛𝑛 + 1)^2 − 3 × 12 𝑛𝑛(𝑛𝑛 + 1)

1 4 𝑛𝑛(𝑛𝑛^ + 1){𝑛𝑛(𝑛𝑛^ + 1)^ −^ 6}

= 1 4 𝑛𝑛(𝑛𝑛^ + 1)(𝑛𝑛^ + 3)(𝑛𝑛 −^ 2)

Example: Find Sn = 22 + 4^2 + 6^2 + … + (2 n )^2.

Solution: Sn = 22 + 4^2 + 6 2 + … + (2 n )^2 = 2^2 (1^2 + 2^2 + 3^2 + … + n^2 )

= 4 × 1 6 𝑛𝑛(𝑛𝑛^ + 1)(2𝑛𝑛^ + 1)^ =^

2 3 𝑛𝑛(𝑛𝑛^ + 1)(2𝑛𝑛^ + 1)^.

Example: Find

Solution: notice that the top limit is 4 not 5

1 6 (𝑛𝑛^ + 2)(𝑛𝑛^ + 2 + 1)(2(𝑛𝑛^ + 2) + 1)^ −^

1 6 × 4 × 5 × 9

= 1 6 (𝑛𝑛^ + 2)(𝑛𝑛^ + 3)(2𝑛𝑛^ + 5)^ −^30.

� 𝑟𝑟(𝑟𝑟^2 − 3 )

𝑛𝑛

𝑟𝑟= 1

�� 𝑟𝑟

𝑛𝑛

𝑟𝑟= 1

�

2

� 𝑟𝑟(𝑟𝑟^2 − 3 )

𝑛𝑛

𝑟𝑟= 1

= � 𝑟𝑟^3

𝑛𝑛

𝑟𝑟= 1

− 3 � 𝑟𝑟

𝑛𝑛

𝑟𝑟= 1

� 𝑟𝑟^2

𝑛𝑛+ 2

𝑟𝑟= 5

� 𝑟𝑟^2

𝑛𝑛+ 2

𝑟𝑟= 5

= � 𝑟𝑟^2

𝑛𝑛+ 2

𝑟𝑟= 1

− � 𝑟𝑟^2

4

𝑟𝑟= 1