Extension 2 Mathematics

Notes about these Notes

These notes are from the 2008 NSW HSC course so some parts may not be relevant to the new course

16 October 2007 Ext 2 Maths
Date: 16 October 2007 7:14 PM
Category: School

Mod-Arg Form.

Modulus = r
Arguement =

The Cartesian plane for Complex numbers follows a square grid. There is a polar grid for complex numbers

  • cisθ = cosθ + i sinθ
  • On the polar grid the first number is the radius then the turn (Modulus and Argument)

22 October 2007 Ext 2 Maths
Date: 22 October 2007 8:30 PM
Category: School

Converting Cartesian to Mod-Arg and back again.

z, z bar, cz, iz In mod-arg form

z = a complex number.
z bar (line on top of the z meaning conjugate) = z but mirrored in the Real axis.
cz (constant x z) = an extension of r, the point has a bigger radius but the same turn.
iz = z rotated 90˚ anticlockwise

Multiplication in mod-arg form

Let z
1 = rcisθ
Let z
2 = rcisØ

Find z
We add arguments and multiply moduli

n = rncis(nθ)

23 October 2007 Ext 2 Maths
Date: 23 October 2007 6:14 PM
Category: School

z = cis
then z
3 = cis3θ

But z3 = (cosθ + isinθ)3

*expand using binomial theorem*
Turn the expression you get into a complex number (Cartesian form) and substitute where the cos and sin were.


A complex number in mod-arg form is like a physics vector. You can add them like physics. To subtract one you flip it about the x axis then add.

30 October 2007 Ext 2 Maths
Date: 30 October 2007 6:11 PM
Category: School

Triangle Inequality

1 + z2| ≤ |z1| + |z2|

This proves the shortest way to anything is a straight line.

More work on vectors.

In a square where x is a side and y is the diagonal:
x = _
√2 y

y = √2 x


|z| = the modulus/radius of the complex
arg(z) = the argument/angle of the complex number

13 November 2007 Ext 2 Maths
Date: 13 November 2007 7:18 PM
Category: School

This is very long.

An ellipse: The locus of a point where the sum of the distances to two fixed points is constant.

Sketch locus of
|z-2-2i| + |z-5-2i| = 5

Therefore the two foci are at (2+2i) and (5 + 2i) and the distance to the points is added up to be 5.


1|/|z2| = |z1/z2|

1z2| = |z1| |z2|

|a+ib| = √a



Sketch the region on the Argand Diagram where |z| ≥ 1 & 0 ≤ argz ≤

De Moivre's Theorem

n = (rcosθ + risinθ)n
= cis(nθ) for r=1
n can be real

The theorem holds for n<0

Roots of unity

z = cosø + isinø
= cos (ø + 2k
π) + isin (ø + 2kπ)
z(1/n) = cos ((ø+2kπ)/(n)) + isin ((ø+2kπ)/(n))

Now you can find all the n roots of 1 including complex ones.

Roots of the cube root of 1 are 1, ω, ω

All roots of ±1 lie on a circle with radius 1 and they form an n sided polygon.
When finding the roots of +1 it will always have a root lying on the real axis and the roots of -1 will never be on the real axis.

5 February 2008 Ext 2 Maths
Date: 5 February 2008 3:06 PM
Category: School

y = b + a sin (nx + ø)

b moves the graph vertically
a changes the amplitude of the graph
The period of the graph is:

(this squishes or lengthens the graph)
ø is the phase shift of the graph by:
if - then moves y axis left
is +then moves y axis right.

Graphs of the form y=f(x)+c

c moves the x axis up or down.

When adding two functions you simply add all the y values together to get a new graph.

8 February 2008 Ext 2 Math
Date: 8 February 2008 3:09 PM
Category: School

changes amplitude

anything bellow the x axis is reflected up in the x axis

CLONES and reflects the 1 and 4 quadrants into the 3 and 2 quadrants. i.e. the 2 quadrant will look the same as the 1st but reflected in the y axis. Makes an even function. The 2 and 3 quadrents are overwritten or ignored.

y= -f(x)
reflection of entire function in x-axis

reflection of entire function in y-axis

11 February 2008 Ext 2 Maths
Date: 11 February 2008 5:58 PM
Category: School


It looks like sin x but the crest and trough always hits the line y=x

zeros of f(x) will give an asymptote.
If f(x) approaches infintiy
function approaches 0
Guess what happens when f(x) approaches zero.


13 February 2008 Ext 2 Maths
Date: 13 February 2008 6:01 PM
Category: School

y = g(x)/f(x)

Horizontal asymptotes when degree of g(x) is equal to the degree of f(x) (deg[g(x)] = deg[f(x)])
Oblique asymptotes if deg[g(x)] > deg[f(x)] Try a division to see why

y = sqroot(f(x))

Exists only for f(x) ≥ 0

2 = f(x)

The graph of y = root(f(x)) but also reflected in the x axis

y = [f(x)]

The zeros of f(x) are stationary points of [f(x)]
You can show this by diffrentiating

Implicit differentiation

Differentiate xy = 1
(Use the chain rule ie. vu’ + uv’ rule)
u’ = 1 v’ = y’
xy’ + y = 0
therefore y’ = -y/x

Implicit differentiation is like normal differentiation but for non-functions.

y = e
Where f(x) = 0 then e
f(x) = 1

21 February 2008 Ext 2 Maths
Date: 21 February 2008 6:50 PM
Category: School


1. If a polynomial is of degree n it will have at most n real roots and exactly n complex roots (remember complex roots can also be real) Complex roots occur in conjugate pairs if all coefficients are real (unconfirmed).
2. If the degree of P(x) is odd then P(x) has at least one real root. This can be factored into one linear factor and some quadratic factors.
3. If a polynomial P(x) degree is even then it can be factored into a series of quadratics.
4. If a polynomial P(x) has a root α of multiplicity n then (x - α)
n is a factor and (x - α)n-1 is a factor of the derivative. You can prove this through the derivative of P(x) = (x - α)n . Q(x)

22 February 2008 Ext 2 Maths
Date: 22 February 2008 7:13 PM
Category: School

You will be asked this question:

Polynomial P(x)=...
It has roots α, β, γ
Find the equation that has the opposite OR reciprocal OR squares of these roots.

To do this: say m is the roots of the new equation. If they ask you to find the equation whose roots are the squares of the original roots then
therefore √m=x

now substitute √m into the original polynomial equal to zero. Solve. Present as a nice polynomial.

27 February 2008 Ext 2 Maths
Date: 27 February 2008 2:53 PM
Category: School

Use De Moivre’s Theorem to express cos4ø in terms of cosø
Use the result to solve
4 = 8x2 + 1 = 0
Hence show that cosπ/8 + cos3π/8 + cos 5π/8 + cos7π/8 = 0

De Moivre’s Theorem is that one that says (rcisø)
n = rncisnø

So to do this question:

(cosø + isinø)
4 = *expand with binomial thorem*
You will get a bunch of crap that you can separate into real part and imaginary part. Completely ignore the imaginary part and just use the real part to equal cos4ø
Change the sin
2ø into 1-cos2ø and rearrange to make the entire thing look like 8x4 = 8x2 + 1 = 0 but where x = cosø
so cos4ø = 0
4ø = π/2, 3π2 etc.
ø = π/8, 3π/8, 5π/5, 7π/8

so the roots of the equation 8x
4 = 8x2 + 1 = 0 are cosπ/8, cos3π/8, cos5π/5, cos7π/8

so all roots added together = -b/a = 0

Partial Fractions

Express (5x+1)/(x-1)(x+2) in the form A/(x-1) + B/(x+2)

If you get that second bit of the question and try to turn it into the first bit of the question you get A(x+2) + B(x-1) on top. That should equal 5x+1 right? Now simultaneous equations.

There is the possibility that A and B could be polynomials so sometimes you will have to write them as Ax + C for example.

If the degree of the top polynomial is larger than the bottom polynomial then do a polynomial division before splitting into fractions.

12 March 2008 Ext 2 Maths
Date: 12 March 2008 3:30 PM
Category: School


Definition 1
The locus of a point P such that the sum of the distances from two other fixed points is always the same. The fixed points are called foci (singular: focus)

Definition 2
The ratio of distance from P to a focus to distance from P to the directrix is always the same. And less than 1.

Say the foci are named F and F’ and R and R’ is a point on the directrix such that PR is perpendicular to the directrix then
PF = PF’ = eccentricity (e)

e is a measure of the deviation from a circle. (e = 0)

The equation of an ellipse is
2/a2 + y2/b2 = 1

Chords are lines that touch two points on the ellipse.
Focal chords are chords through one of the foci.
A diameter chord runs through the center of the ellipse.
The semi-major axis is the axis that runs through the long side of the ellipse, usually the x-axis
The semi-minor axis is the axis that runs through the short side.
The coordinates of the foci are ae and -ae
The ellipse cuts the x-axis at a and -a and the y-axis at b and -b
The equation of the directrices is a/e and -a/e

2/a2 = 1 - e2

13 March 2008 Ext 2 Maths
Date: 13 March 2008 12:16 PM
Category: School

The equation of a tangent to a ellipse is given by
xx1 + yy1 = 1
2 b2

Use implicit differentiation to do this.

2x2 + a2y2 = a2b2
2 - b2 = a2e2

The equation of a normal to a point on an ellipse is:

a2x - b2y = a2 - b2
1 y1

Parametric Definition

Reflective property of the ellipse

If you stood on the focus in an elliptical room the a person could hear everything you say very clearly if they are standing on the other focus.

30 April 2008 Ext 2 Maths
Date: 30 April 2008 6:05 PM
Category: School

The Chord of Contact for an ellipse from an external point:

xx1 + yy1 = 1
2 b2

The equation for a normal tangent is the same.

The equation for the tangent and chord of contact to a hyperbola is the same but with a minus sign instead of a plus.

Locus: The difference of the distances to two fixed points is equal

The fixed points are the foci.

The ratio of the distance of a point P to the focus to the distance to the directrix is > 1
This ratio is the eccentricity (e)

Foci: (±ae, 0)
Directrices: x = ±a/e
Asymptotes: y = (±b/a)x
2/a2 = e2 - 1

1 May 2008 Ext 2 Maths
Date: 1 May 2008 8:22 PM
Category: School

A rectangular hyperbola is one where the asymptotes are perpendicular. y = 1/x is one of them.

conjugate axis is the one with the b’s on it while the transverse axis has the a’s on it. Diameter goes through the origin

A conjugate hyperbola is one that occurs on the other sides of the asymptotes and just means the x2/a2 is swapped with the y2/b2

Parametric definition of a hyperbola: x = asecø y = btanø

16 May 2008 Ext 2 Maths
Date: 16 May 2008 1:38 PM
Category: School


You can split the fraction into multiple fractions. Try to manipulate the fraction to make the bottom resemble a standard integral or the top the derivative of the bottom.

28 May 2008 Ext 2 Maths
Date: 28 May 2008 3:11 PM
Category: School


∫u dv = uv - ∫v du

It’s a very useful formula

29 May 2008 Ext 2 Maths
Date: 29 May 2008 3:13 PM
Category: School

Recurrence Integrals

e.g. ∫cos
nx dx

Split into cosx . cos
Use integration by parts

Why not use the t results when integrating trigonometric functions! Hours of fun!

5 June 2008 Ext 2 Maths
Date: 5 June 2008 3:10 PM
Category: School


I have a circle. It is of radius a around the origin.
This isn’t going to work.

12 June 2008 Ext 2 Maths
Date: 12 June 2008 8:23 PM
Category: School

Volumes by circular cross-sections

Want to spin the curve y = 2√x around y=4? Here’s how.
Taking vertical slices:

radius of slices: 4 - x
area: π(4-x)
volume: π(4-x)
2∆y (∆y being a tiny height of each slice)

When integrating, make sure you either change the x into a y or dy into dx somehow.

13 June 2008 Ext 2 Maths
Date: 13 June 2008 11:53 AM
Category: School

The curve of y=cosx is rotated around the line y =1 from x = 0 to π/2
Find the volume generated.

Answer is 3π
2/4 - 2π

y = √(x-1) rotated around x=0 from x=0 to 5
You have to find the inner/outer radius/area/volume then take one away from the other.

Volume of the washer is π x
22∆y = π x12∆y
V = ∫x
22 - x12 dy (from 0 to 1/4 (I cant write it properly))
but y = x-x
use x
1 and x2 as the roots to this with the whole alpha beta = c/a
Answer is π/6

25 June 2008 Ext 2 Maths
Date: 25 June 2008 11:59 AM
Category: School

Theorem of Pappus.

If a symmetrical area is rotated about a line outside the area then the volume generated is given by
Area times distance traversed by the axis of symmetry.

General formula:
2π∫xy dx
Don’t be fooled. x is actually the radius of each shell. So if you are rotating around something other than x=0 then x will not be x but the radius (eg maybe 1-x)

30 July 2008 Ext 2 Maths
Date: 30 July 2008 12:09 PM
Category: School

Banked tracks.
N is sideways, mg is straight down and sideways is provided by mv
Vertical component of N counterbalances mg therefore is the track is banked at ø then Nsinø = mv
2/r and Ncosø = mg

31 July 2008 Ext 2 Maths
Date: 31 July 2008 12:11 PM
Category: School

Resolving forces on banked tracks could involve multiplying by sin or cos or squaring everything.
Then add the two equations you have together and solve.

6 August 2008 Ext 2 Maths
Date: 6 August 2008 12:59 PM
Category: School

Resisted motion!

When shot out of a canon upwards: forces acting on a particle are -mg and -mkv

a = - g - kv

Change the a to a v(dv/dx) and integrate to get displacement. Remember to find the constant of integration.

7 August 2008 Ext 2 Maths
Date: 7 August 2008 1:01 PM
Category: School

Motion downwards (resisted):

A particle is dropped from a height.
Forces acting with: g
Forces acting against: -kv

a = g - kv

You can find terminal velocity (which is g/k) by integrating the above equation if a = dv/dt
Once you have t (with the constant figured out) you could say as t approaches infinity v approaches ...
Or isolate the v and make t = infinity.