# A-level Mathematics/AQA/MFP1
## Roots and coefficients of a quadratic equation

Given the equation ax$^2$+bx+c=0 has roots $\alpha$ and $\beta$:
$\mathrm {sum\ of\ roots} =\alpha +\beta ={{-b} \over {a}}$
$\mathrm {product\ of\ roots} =\alpha \beta ={{c} \over {a}}$

### Useful results

$\alpha ^{2}+\beta ^{2}=(\alpha +\beta )^{2}-2\alpha \beta$ $\alpha ^{3}+\beta ^{3}=(\alpha +\beta )^{3}-3\alpha \beta (\alpha +\beta )$ ${1 \over \alpha }+{1 \over \beta }={\alpha +\beta \over \ \alpha \beta }$
This last one comes from that when you add two fractions. Generally the formula is:
${a \over b}+{c \over d}={ad+cb \over bd}$

## Complex Numbers

A complex number is different 'type' of number to what we'd normally think of as a number. There are infinitely many complex numbers, like there are infinitely 'normal', or real numbers. To clarify, a real number is any number on the 'number line' : integers (...-2, -1, 0, 1, 2...), rational numbers (can be written as fractions, such as 1.2, 5.75), irrational numbers (can't be written as fractions, such as π, √2) - they're all real numbers. Complex numbers 'go beyond' these ordinary numbers.
Consider the discriminant of a quadratic - from the formula: $b^{2}-4ac$. Remember, the discriminant of a quadratic determines the nature of its roots (solutions). A discriminant ≥ 0 implies that there are real root(s), and a discriminant < 0 implies that there are no real roots. Why is this?
Recall the quadratic formula: $x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}$
From this, we can see if $b^{2}-4ac<0$, then $x$ involves the square root of a negative number. Taking the quadratic $x^{2}+1=0$ as an example, we can see no solution for $x$. We need $x^{2}=-1$, so what can $x$ be? It can't be 1, because 1 squared is 1. It can't be -1, either, as -1 squared is also 1. What can we do?

### The Imaginary Unit, i

To get round this, we 'invent' a number, called $i$. $i$ does not 'exist' as a real number - so you cannot have $i$ apples in your hand. Thus $i$ is dubbed an imaginary number, which is what the i stands for. The key property of $i$ is that $i^{2}=-1$. So in our above example, the solution was $x=i$. Right?
Well, not quite. We can use multiples of $i$ : 2i, 3i, 4i, 0.4i, -0.4i, -1.4i... . We can have positive or negative, integer, rational or irrational multiples of $i$. Let's keep it simple and think of, $-i$. What's $(-i)^{2}$?
The answer is -1. Why? In short, a minus times a minus makes a plus, and $i$ times $i$ makes -1 - so the result is just -1. If this is a little confusing, think of the problem as finding $(ab)^{2}$, and then substituting $a=-1$ and $b=i$. In short, $i^{2}=(-i)^{2}=-1$. Putting this to use in the problem above gives us two (non-real) solutions: $x=i$ and $x=-i$.
In the same way that we could say the square root of 1 is 1 or -1, the square root of -1 is i or -i.

### Square Roots of negative numbers

We've already looked at ${\sqrt {-1}}$. But this isn't too useful by itself alone. What, for example, is ${\sqrt {-4}}$ equal to? We can treat this problem in the same way we simplify surds.
${\sqrt {ab}}={\sqrt {a}}{\sqrt {b}}$ If ab = -2, we could let a=-1, and b =2. ${\sqrt {-4}}={\sqrt {-1}}{\sqrt {4}}$
* powers of i
* a + bi
* addition, subtraction
* multiplication
* 'division'
* conjugates, z, z* notation
* conjugates as roots of a quadratic.
* considering real and imaginary parts
* lead into complex numbers from imaginary.
(image and caption of mandlebrot set, mention use of complex numbers)

### The Power Tower of i

Part of the definition of i is that $i^{2}=-1$, and based only on this fact alone we can find i to the power of any integer.
Let's go up a power. What's $i^{3}$? Well, we can write $i^{3}$ as $i^{2}*i$. We already know that $i^{2}=-1$, so this simplifies to $-1*i=-i$. Easy!
Let's try the next power - $i^{4}$. We treat this in the same way, writing $i^{4}=i^{3}*i$. And it's fairly easy to substitute $i^{3}=-1$ to arrive at the conclusion that $i^{4}=1$. This makes sense, when you think about it: what do you get if you square both sides of $i^{2}=-1$?
Now we can climb back down the power tower. $i^{1}$ is just i - there's nothing complex there. If you wanted to, you could let $i^{1}=x$, and the solve for x in $i^{2}=i*x$.
We can also use this to find $i^{0}$ - let's denote this by y. Then:
$i^{1}=i*y$ $i=i*y$

### Complex Numbers, more formally

Complex numbers are numbers of the form $a+bi$, where a and b are both real numbers, and i is the imaginary unit. We call a the re
So if you were to have the quadratic formula ${\frac {2\pm {\sqrt {-2^{2}-4\cdot 1\cdot 16}}}{2}}$. You would get ${\frac {2\pm {\sqrt {-64}}}{2}}$. Knowing that ${\sqrt {-1}}=i$ It would simplify down to ${\frac {2\pm 8i}{2}}$ and then to $1\pm 4i$ where the 1 is the real part and the 8i is the imaginary part.

### Complex Arithmetic
#### Addition and Subtraction

Addition and subtraction of complex numbers is fairly intuitive: you sum the real parts and imaginary parts separately. (Or condense into on expression and group like terms.)
Example (adding) $(3+4i)+(4-5i)=(3+4)+(4-5)i=7-i$
Example (subtracting) $(3+4i)-(4-5i)=(3-4)+(4+5)i=-1+9i$

#### Multiplication

The easiest way to multiply complex numbers is to form an expression with the numbers in brackets and then to expand the brackets and simplify. For example:
$\displaystyle (3+2i)\times (4-i)$
Expanding the brackets gives;
$\displaystyle 12-3i+8i-2i^{2}$
Then collect like terms and replace $i^{2}$ with $-1$;
$\displaystyle 12+5i-2\times -1$
$\displaystyle 12+5i+2$
Therefore the final result is;
$\displaystyle 14+5i$

#### Division

Division is not covered in FP1, but is covered in other Further Pure modules.

### Complex Conjugates

Each complex number has a complex conjugate which is formed by taking the original complex number and changing the sign of the imaginary part. For example, the complex conjugate of $3+2i$ would be $3-2i$. These numbers are often referred to as a conjugate pair.
The conjugate of z is written as z*.
Conjugate pairs have the property that when one is subtracted from the other, the result is purely imaginary. Also if they are added or multiplied together the result is a real number. This means that for a quadratic equation with real coefficients but imaginary roots, the roots will be a pair of complex conjugates.
Example:
$\displaystyle x^{2}+2x+14=0$
This equation cannot be factorised, therefore we use the quadratic formula ($\scriptstyle x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}$) to solve it;
$x={\frac {-2\pm {\sqrt {2^{2}-4\times 1\times 14}}}{2}}$
You will find that the portion of the equation under the root sign is negative, this means that the roots of the quadratic are not real.
$x={\frac {-2\pm {\sqrt {-52}}}{2}}$
Write the root in terms of i and simplify the surd.
$x={\frac {-2\pm 2i{\sqrt {13}}}{2}}$
Now you can cancel the fraction to give the answers.
$\displaystyle x=-1\pm i{\sqrt {13}}$

### Solving Equations with Complex Numbers
## Inequalities

* Multiplying by the square of the denominator to ensure a positive multiplication
* Critical Values
* Truth tables and graphs of inequalities
* Factorising

## Trigonometry

* Special angles for sin, cos, tan
* Radians
* Finding general solutions as opposed to specific solutions
* Use of sin/cos === tan, cos^2+sin^2 === 1

## Matrices and Transformations

* Definition of matrices
* Order of a matrix
* Matrix addition, subtraction
* Matrix multiplication
** When is it defined
* Using a matrix to represent a transformation
** Reflection, Reflection, Enlargement
* Composite Transformations
* Formulae for AC rotation through theta, reflection in y = x * tan theta

## Linear Laws

* Log Laws
* Use of log laws to reduce to a linear relationship for
** y=a^x + b
** y=x^a + b

## Series

Simply put, a series is the result of adding the terms of a sequence together, for example the series produced by summing the terms of the sequence 2n+1 would be;
$\displaystyle 3+5+7+9...$
This is an infinite sequence as it has an infinite number of terms. However if we limit it to the first 5 terms of the sequence we create a finite sequence. This is much easier to work with and find the sum, thus;
$\displaystyle 3+5+7+9+11=35$
To write this simply we use sigma notation.

### Sigma Notation

The capital Greek letter sigma, $\Sigma$ is used to mean 'sum of'. It is written with the range of the variable above and below it (maximum value above and minimum below) and the letter representing the variable to be changed below it. Therefore the second example shown above would be written;
$\sum _{n=1}^{5}(2n+1)$
The first example shown above can also be written using sigma notation, with the sign for infinity )$\infty$) above the sigma.

### Summing with an initial counter value not 1

If the number under the sigma is not 1 then you should split expression, using two sigma signs, for example:
$\sum _{n=5}^{10}(2n+1)$
becomes;
$\sum _{n=1}^{10}(2n+1)-\sum _{n=1}^{4}(2n+1)$
Note that in the second part of the expression, the maximum value of the variable is one less than the minimum in the original expression.

### Forumlae for r, r^2, r^3

Special formulae are used to find the values of series with variables above the sigma.

#### Formula for r

The expression to add the first n natural numbers can be written using sigma notation thus;
$\sum _{r=1}^{n}r$
This can be written as an expression using only n;
${\frac {n(n+1)}{2}}$

#### Formula for $r^{2}$

The expression to add the first n square numbers can be written using sigma notation;
$\sum _{r=1}^{n}r^{2}$
It too can be written as an expression in terms of n
${\frac {n(n+1)(2n+1)}{6}}$

#### Formula for $r^{3}$

The expression to add the first n cube numbers can be written using sigma notation;
$\sum _{r=1}^{n}r^{3}$
It too can be written as an expression in terms of n
${\frac {n^{2}(n+1)^{2}}{4}}$

### Manipulating Sigma Notation

Using combinations of the formulae above we can find values for more complicated expressions. For example;
$\sum _{r=1}^{10}(r^{3}+5r-7)$
Firstly, you can split the terms of the expression, you can also take the coefficients of the terms before the sigma signs.
$\sum _{r=1}^{10}r^{3}+5\sum _{r=1}^{10}r-\sum _{r=1}^{10}7$
Using the above formulae you can replace the first 2 terms with their equivalent expressions. For the final term, you see that it is the sum of ten sevens, or $\scriptstyle 7\times 10$
${\frac {10^{2}(10+1)^{2}}{4}}+5{\frac {10(10+1)}{2}}-7\times 10$
Now this expression can be solved like any other;
$\displaystyle 3025+275-70$
$\displaystyle 3230$

## Calculus

* First-principals approach to differentiation
* Difference quotient
* Improper Integration
** Why improper
** Does the integral exist?

## Rational Functions

* Definition, examples
* Asymptote - definition
* How to find vertical and horizontal asymptotes.
* Sketching functions by considering
** Asymptotes
** x- and y-intercepts
** Behaviour as approaching an asymptote

## Further Functions

* Using the discriminant of a quadratic to find maxima and minima
** NB Differentiation by Quotient rule is explicitly excluded

## Hyperbolae, Parabolae and Ellipses

* Definition, examples of, generalised formulae of
* Transformations of
* Asymptotes of hyperbolae
* Relation of an ellipse to a circle
** Foci.

## See also

* AQA Mathematics Specification 2009 (pp. 48-50)