## Complex Numbers: four rules

Consider complex numbers in the form

### Adding complex numbers

When adding complex numbers we add the real part and then we add the imaginary part

Examples

### Subtracting complex numbers

When subtracting complex numbers we subtract the real part and then we subtract the imaginary part

Examples

### Multiplying complex numbers

When multiplying complex numbers we need to remember that

Examples

### Dividing complex numbers

Dividing complex numbers requires us to get rid of the imaginary part from the denominator.

We can do this by multiplying the denominator and numerator by the complex conjugate of the denominator.

Examples

## Sketching derivative graphs (cubics)

The important thing to remember here is that we are doing a sketch. It does not have to be accurate, but id does have to conatian all the key points.

Hints:

- A cubic differentiates to a quadratic (parabola)
- A the turning points have a gradient of zero and will move to the x axis on the derivative sketch
- A point with a positive gradient will be above the x axis in the sketch
- A point with a pnegative gradient will be below the x axis in the sketch

Examples

Describe and sketch the derivative graph for the function below

The critical points are , because this is where the turning points occur.

These points will be on the x axis in the sketch.

For the gradient is positive.

These points will be above the x axis in the sketch.

For the gradient is negative .

These points will be below the x axis in the sketch.

The graph is a cubic so the sketch will be a parabola

### Practice

## Simultaneous Equations: Substitution Method

This is exactly the same method we use to find where two lines intersect.

Process:

- Rearrange one of the equations to make one of the variables a subject
- Substitute for this variable into the other equation
- Solve the resulting equation
- Substitute these solutions into one of the original equations to find the values of the other variable

Example

Solve the system of equations

Substitute eqn 1 into eqn 2

Substituting into eqn 1 gives

Solve the system of equations

Rearrange eqn 2 to make y the subject

Substitute eqn 3 into eqn 1

Factorising gives

Substituting these values into eqn 3 gives

So the solutions are

Solve the system of equations

Rearranging eqn 2 to make y the subject gives

Substituting eqn 3 into eqn 1 gives

Substituting the value of x into eqn 3 gives

## Simultaneous equations: Elimination method

Simultaneous equations are sytems of equations where there are a number of variables.

Our task is to find the values of the variables that satisfy all the equations.

For this method we will solve equations of the type

With these equations we can do one of three things

- Multiply by a scaler.
- Add the equations together
- Subtract one equation from another

Multiplying by a scaler

eqn 1 multiplied by 6 gives

Adding two equations

eqn 1 + eqn 2 gives

Subtract one equation from another

eqn 1 – eqn 2 gives

### Type 1 Simultaneous equations

With this type of problem the y terms are the same size but different signs.

To solve these we:

- Add the equations together (this will cancel out the y terms)
- Solve the resulting equation to find the value of x
- Substitute the x value into one of the original equations
- Solve the resulting equation to find the value of y
- Check by substituting x and y into the other equation

Examples

Solve

This is a type 1 problem because we have two 4y’s, but one is positive and the other is negative.

So eqn 1 + eqn 2 gives

Substitute into eqn 1

gives

Check in eqn 2

true

Solve

This is a type 1 problem because we have two 5y’s, but one is positive and the other is negative.

So eqn 1 + eqn 2 gives

Substitute into eqn 2 (it is easier to deal with positives)

gives

Check in eqn 1

true

### Type 2 Simultaneous equations

With this type of problem the y terms are the same size and sign.

To solve these we:

- Subtract one equation from the other (this will cancel out the y terms)
- Solve the resulting equation to find the value of x
- Substitute the x value into one of the original equations
- Solve the resulting equation to find the value of y
- Check by substituting x and y into the other equation

Examples

Solve

This is a type 2 problem because we have two +2y’s.

So eqn 2 – eqn 1 gives

Substitute into eqn 1

gives

Check in eqn 2

true

Solve

This is a type 2 problem because we have two -4y’s.

So eqn 2 – eqn 1 gives

Substitute into eqn 1

gives

Check in eqn 2

true

### Type 3 Simultaneous equations

With this type of problem the y terms are different.

To solve these we:

- Multiply each equation by the other equations positive y coefficient
- Solve the resulting type 1 or type 2 problem

Examples

Solve

Multiply eqn 1 by 2 (that is the number of y’s in eqn 2)

Multiply eqn 2 by 4 (that is the number of y’s in eqn 1)

Equations 3 and 4 now make a type 2 problem

Subtracting eqn 3 from eqn 4 gives

Substitute into eqn 2

Check in eqn 1

true

Solve

Multiply eqn 1 by 3 (that is the number of y’s in eqn 2)

Multiply eqn 2 by 2 (that is the number of y’s in eqn 1)

Equations 3 and 4 now make a type 1 problem

Adding eqn 3 and eqn 4 gives

Substitute into eqn 2

Check in eqn 1

true

## Circle Theorems 2

Tangents to circles

#### Investigation

### Theorem 1

The length of the tangents from a point to a circle are of equal length

### Theorem 2

The Angle between a tangent and a radius is a right-angle

### Theorem 3

The angel between a tangent and a cord is equal to the angle subtended by the cord

## Examples

Given that PA is a tangent find the value ofSince PA is a tangent

Given that PA and PB are tangents find the value ofSince PA and PB are tangents they must both be of equal length. Therefore is isosceles.base angles in an isosceles triangle are equal.the angle between a tangent and a radius is a right-angle.

Find the value ofSince