FP1 – Revision Notes
Complex Numbers
1  i

i 2  1
An imaginary number is in terms of i whereas a complex number has a real part and an imaginary
part.
Eg: Imaginary number 2i, 4i,14i
Complex Number 2  3i, 5  9i
If z  a  bi then z*  a  bi is the complex conjugate
Addition and Subtraction: Add and subtract real and imaginary parts.
Multiplication: Multiply like quadratic expressions
Division: Like rationalising the denominator – multiply top and bottom by the complex conjugate of
the denominator
Complex roots of polynomial equations occur in conjugate pairs – ie if a  bi is a solution, then so is
a  bi
If a  bi is a solution to the equation f  x   0 then  x  (a  bi)  is a factor as is  x  (a  bi) 
An Argand Diagram represents complex numbers on a 2D plane. The x-axis is also called the real axis
whereas the y-axis is the imaginary axis. The number 3  2i would then be represented by the
coordinate (3, 2) . The number 5  3i would be represented by the coordinate (-5, 3).
The modulus of a complex number z  a  bi is given by z  a2  b2
The argument of a complex number is the angle between the positive x-axis and the line
representing the number z on the Argand Diagram. Answers are usually required in the range
    
Im
Im
r
θ
Re
Re
r
θ
The modulus argument- form of a complex number z  a  bi is z  r  cos  i sin   where r is the
modulus of z and  is the argument.
For two complex numbers z1 , z2 then z1 z2  z1 z2
To find the square root of a complex number a  bi set a  bi  x  yi . Squaring both sides gives
a  bi   x  yi  . Multiply out the brackets and equate the real and imaginary parts.
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Cubic equations: ax3  bx2  cx  d  0
This has either one real root or 2 complex roots (which will be a conjugate pair) OR 3 real
roots
Quartic equations: ax4  bx3  cx2  dx  e  0
This will have 4 real roots OR 2 real roots and 2 complex roots (as a conjugate pair) OR 4
complex roots (in 2 conjugate pairs)
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FP1 – Revision Notes
Numerical Methods
(1)
Interval Bisection

 ab
Take the mid point 
 of an interval  a, b  as the first approximation
 2 

 ab
Evaluate f 

 2 

ab
 ab
 ab
If f  a   0 and f 
. If f  a   0 and f 
 0 replace a with

  0 , replace
2
 2 
 2 
ab
ab
 ab
. If f 
.
 0 replace whichever is positive from a or b with

2
2
 2 
Repeat the process until you reach the required degree of accuracy
b with

(2)
Linear Interpolation


Calculate f  a  and f  b 
1st approximation
f b
a
f a
b
x1
root
Then
(3)
x1  a b  x1

. Solve this for x1 to get the next approximation.
f a
f b

DO NOT TAKE THE SIGNS OF f  a  and f  b  INTO ACCOUNT. YOU ONLY WANT THE

DISTANCE.
Repeat this process until you get a solution to the required accuracy.
Newton-Raphson process
f  xn 

xn 1  xn 

This method involves finding the tangent to the curve at a given point. The
intersection of the tangent with the x-axis gives the next root.
It does not always work!

f '  xn 
(given in formula book)
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FP1 – Revision Notes
Coordinate Systems
To find the Cartesian equation of a curve given parametrically, eliminate the parameter by solving
simultaneously.
Directrix
X
P(x,y)
(-a,0)
S(a,0)
focus
y2  4ax
x=-a
For a parabola PX=SP
The parametric equations for a parabola are x  at 2 , y  2at . This means that a general point on this

curve has coordinates at 2 , 2at

A rectangular hyperbola has parametric equations
c
x  ct , y  . This means that the general point on this
t
 c
curve has coordinates  ct , 
 t
xy  c2
Matrices
An n  m matrix has n rows and m columns
To add or subtract matrices add or subtract the corresponding elements in the two matrices. You
can only add or subtract matrices that have the same dimensions.
To multiply matrices by a scalar, multiply each element by the scalar
To multiply matrices together, multiply the rows of the first with the columns of the second
In terms of dimensions  n  m    m  p    n  p 
Not all matrices can be multiplied together and the order of multiplication is important.
A square matrix has the same number of rows as columns
Transformations:
A linear transformation is one where the transformation vector involves only linear expressions in x
and y.
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FP1 – Revision Notes
To work out a transformation matrix consider the effect of the transformation on the points (1, 0)
and (0, 1). The image of these points become the first a second columns of the transformation
matrix respectively.
To work out what a transformation matrix does, apply it to (1,0) and (0,1) and consider how they
have been transformed.
When multiplying transformation matrices together the first transformation matrix goes on the
right, the second to the left of the first, the third to the left of the second etc.
So if transformation A is followed by transformation B and then transformation C the order of
multiplication is CBA
You can find the inverse of square matrices.
a b
If A  
 then det  A   ad  bc
c d
If det  A   0 the matrix is said to be singular and has no inverse.
A 1 
1  d b 


ad  bc  c a 
1 0
AA -1  A -1 A  I  
 This is the identity matrix (it works a little like multiplying by 1)
0 1
The determinant of a matrix gives the scale factor of the areas following a transformation by that
matrix
Area of image=Area of object  det M
Series
n
U
r 1
r
 U 1  U 2  U 3  ..........  U n
n
n
k 1
r k
r 1
r 1
U r  U r  U r
You can use the standard formulae to sum more complex series
n
1  n
r 1
n
 k  kn
r 1
n
n
 r  2  n  1
r 1
n
r
2

n
 n  12n  1
6
3

n2
2
 n  1
4
r 1
n
r
r 1
  2r
n
Then
r 1
3

n
n
n
n
r 1
r 1
r 1
r 1
 3r 2  6r  1  2 r 3  3 r 2  6 r   1
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FP1 – Revision Notes
Mathematical Induction




Prove the statement is true for n  1
Assume the statement is true for n  k
Show the statement is true for n  k  1 if it is true for n  k
Write out the conclusion that shows that it is true for all positive integers n
Make sure that when you carry out the third step you use the fact that the statement is true for
nk
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