Complex Numbers: four rules

December 3, 2010 Leave a comment

Consider complex numbers in the form a+bi

Adding complex numbers

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

Examples

(2+i)+(4+i)=(2+4)+(1+1)i=6+2i

(3+5i)+(-2-3i)=(3-2)+(5-3)i=1+2i

(4-i)+(-4+7i)=(4-4)+(-1+7)i=6i

Subtracting complex numbers

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

Examples

(2+i)-(4+i)=(2-4)+(1-1)i=-2

(3+5i)-(-2-3i)=(3+2)+(5+3)i=5+8i

(4-i)-(-4+7i)=(4+4)+(-1-7)i=8-8i

Multiplying complex numbers

 When multiplying complex numbers we need to remember that i\times i = -1

Examples

(2+i)(4+i)=8+2i+4i+i^2=7+6i

(3+5i)(-2-3i)=-6-9i-10i-15i^2=9-19i

(4-i)(-4+7i)=-16+28i+4i-7i^2=-9+32i

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

\displaystyle z=\frac{2+8i}{1+i}

\displaystyle \implies z=\frac{2+8i}{1+i}\times \frac{1-i}{1-i}

\displaystyle \implies z=\frac{10+6i}{2}=5+3i


\displaystyle w=\frac{3-i}{4+i}

\displaystyle \implies w=\frac{3-i}{4+i}\times \frac{4-i}{4-i}

\displaystyle \implies w=\frac{11-7i}{17}

\displaystyle \implies w=\frac{11}{17}- \frac{7}{17}i

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Categories: Complex numbers

Sketching derivative graphs (cubics)

December 1, 2010 Leave a comment

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:

  1. A cubic differentiates to a quadratic (parabola)
  2. A the turning points have a gradient of zero and will move to the x axis on the derivative sketch
  3. A point with a positive gradient will be above the x axis in the sketch
  4. 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 x=0\mbox{ and }x=\frac{4}{3}, because this is where the turning points occur.

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

For x<0 the gradient is positive.

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

For 0<x<\frac{4}{3} 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

Categories: Differentiation

Simultaneous Equations: Substitution Method

November 30, 2010 Leave a comment

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

Process:

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

Example

Solve the system of equations

y=4x-3\rightarrow 1

3y+x=7\rightarrow 2

Substitute eqn 1 into eqn 2

\implies 3(4x-3)+x=7

\displaystyle \implies 12x-9+x=7\implies 13x=16\implies \frac{16}{13}

Substituting into eqn 1 gives

\displaystyle y=4\left(\frac{16}{13}\right)-3=\frac{25}{13}


Solve the system of equations

y-3=x^2+4x\rightarrow 1

2x=y-6\rightarrow 2

Rearrange eqn 2 to make y the subject

\implies y=2x+6\rightarrow 3

Substitute eqn 3 into eqn 1

\implies (2x+6)-3=x^2+4x\implies x^2+2x-3=0

Factorising gives

(x+3)(x-1)=0\implies x=-3\mbox{ or }x=1

Substituting these values into eqn 3 gives

y=0\mbox{ and }y=8

So the solutions are (-3,0)\mbox{ and }(1,8)


Solve the system of equations

3x+2y=13\rightarrow 1

2x-y=-3\rightarrow 2

Rearranging eqn 2 to make y the subject gives

y = 2x+3\rightarrow 3

Substituting eqn 3 into eqn 1 gives

3x+2(2x+3)=13\implies 7x+6=13\implies x=1

Substituting the value of x into eqn 3 gives

y=5

Categories: Equations

Simple trig questions

November 30, 2010 Leave a comment

Find the value of x

Categories: Home

Simultaneous equations: Elimination method

November 30, 2010 Leave a comment

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

\displaystyle ax+by=c\rightarrow 1

\displaystyle dx+ey=f\rightarrow 2

With these equations we can do one of three things

  1. Multiply by a scaler.
  2. Add the equations together
  3. Subtract one equation from another

Multiplying by a scaler

eqn 1 multiplied by 6 gives

\displaystyle 6ax+6by=6c\rightarrow 3

Adding two equations

eqn 1 + eqn 2 gives

\displaystyle (a+d)x+(b+e)y=c+f\rightarrow 4

Subtract one equation from another

eqn 1 – eqn 2 gives

\displaystyle (a-d)x+(b-e)y=c-f\rightarrow 5

Type 1 Simultaneous equations

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

To solve these we:

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

Examples

Solve

\displaystyle 3x+4y=18\rightarrow 1

\displaystyle 7x-4y=2\rightarrow 2

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

\displaystyle 10x=20

\displaystyle \implies x=2

Substitute x=2 into eqn 1

gives \displaystyle 6+4y=18

\implies 4y=12

\implies y=3

Check in eqn 2

14-12=2 true


Solve

\displaystyle 2x-5y=13\rightarrow 1

\displaystyle 3x+5y=7\rightarrow 2

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

\displaystyle 5x=20

\displaystyle \implies x=4

Substitute x=4 into eqn 2 (it is easier to deal with positives)

gives \displaystyle 12+5y=7

\implies 5y=-5

\implies y=-1

Check in eqn 1

8--5=13 true

Type 2 Simultaneous equations

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

To solve these we:

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

Examples

Solve

\displaystyle 4x+2y=20\rightarrow 1

\displaystyle 7x+2y=23\rightarrow 2

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

So eqn 2 – eqn 1 gives

\displaystyle 3x=3

\implies x=1

Substitute x=1 into eqn 1

gives \displaystyle 4+2y=20

\implies 2y=16

\implies y=8

Check in eqn 2

\displaystyle 7+16=23 true


Solve

\displaystyle 2x-4y=6\rightarrow 1

\displaystyle 5x-4y=3\rightarrow 2

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

So eqn 2 – eqn 1 gives

\displaystyle 3x=-3\rightarrow 1

\implies x=-1

Substitute x=-1 into eqn 1

gives \displaystyle -2-4y=6

\implies 4y=-8

\implies y=-2

Check in eqn 2

\displaystyle -5--8=3 true

Type 3 Simultaneous equations

With this type of problem the y terms are different.

To solve these we:

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

Examples

Solve

\displaystyle 3x+4y=-5\rightarrow 1

\displaystyle 5x+2y=1\rightarrow 2

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

\displaystyle 6x+8y=-10\rightarrow 3

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

\displaystyle 20x+8y=4\rightarrow 4

Equations 3 and 4 now make a type 2 problem

Subtracting eqn 3 from eqn 4 gives

\displaystyle 14x=14

\displaystyle \implies x=1

Substitute x=1 into eqn 2

\displaystyle \implies 5+2y=1

\displaystyle \implies 2y=-4

\displaystyle \implies y=-2

Check in eqn 1

\displaystyle 5-4=1 true


Solve

\displaystyle 4x-2y=7\rightarrow 1

\displaystyle x+3y=10\rightarrow 2

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

\displaystyle 12x-6y=21\rightarrow 3

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

\displaystyle 2x+6y=20\rightarrow 4

Equations 3 and 4 now make a type 1 problem

Adding eqn 3 and eqn 4 gives

\displaystyle 14x=41\rightarrow 4

\displaystyle x=\frac{41}{14}

Substitute x=\frac{41}{14} into eqn 2

\displaystyle \implies \frac{41}{14}+3y=10

\displaystyle \implies 3y=\frac{99}{14}\implies y=\frac{33}{14}

Check in eqn 1

\displaystyle \frac{164}{14}-\frac{66}{14}=7 true

Categories: Equations

Circle theorems 2: Questions

November 24, 2010 Leave a comment

Set 1

Set 2

Categories: Geometry

Circle Theorems 2

November 23, 2010 Leave a comment

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 of x
Since PA is a tangent \angle PAO = 90^o
\implies x = 180^o-90^o-65^o=25^o \because 180^o \mbox{in a }\triangle

Given that PA and PB are tangents find the value of x
Since PA and PB are tangents they must both be of equal length.  Therefore \triangle ABP is isosceles.
\implies \angle ABP =\frac{180^o-49^o}{2}=65.5^o \because base angles in an isosceles triangle are equal.
\implies x=90^o-65.5^o=24.5^o \because the angle between a tangent and a radius is a right-angle.

Find the value of x
Since \triangle ABC \mbox{ is isosceles}\implies \angle ACB = 50^o
\implies x=50^o \because \mbox{Angle subtended by the cord}

Questions

Categories: Geometry