Argand Plane and Modulus Argument Form

Let's start with some terminology

If z = x +iy

Re(z) = x (real component of z)

Im(z) = y (imaginary component of z)

The Argand plane (also known as the complex plane) is used to plot complex numbers. Complex numbers 'z' are plotted on the Argand plane with Re(z) on the x axis and Im (z) on the y axis.

Eg.

A represents 2+i, B represents 1, C represents -1 +3i, D represents -1-2i.

Now let's look at absolute values again.

The absolute value of a number, also called the modulus of a number is the distance of the number from the origin.

If the number is a real number, then the distance from the origin is just the number without the sign.

Eg. 1

|4|= 4,

Eg. 2

|-3|=3

For a complex number, the modulus can be found using the distance formula.

Eg. 1

|3+4i| = √(3² +4²) = 5

 

Eg. 2

|5-12i|= √(5² + 12²) = 13

By definition, if z = x +iy, |x +iy|= √(x² +y²) , this is known as the modulus of z or sometimes shortened to mod z.

It is not enough to define a complex number by its modulus because there are infinite complex numbers that have the same modulus.

So we also define complex numbers by the angle that is created with the x-axis when that point is joined to the origin (known as the argument, often written as arg).

We can find the argument by using the fact that m(gradient) = tan(θ)

Eg. 1

If z = 1 + i√3

First we should determine which quadrant z is in. In this case z is in the first quadrant.

tanθ = √3/1

θ = π/3 + 2πn or 4π/3 +2πn (where n is an integer)

As we can see, there are infinite solutions for θ.

We know that 1 + i√3 is in the first quadrant and therefore 4π/3 + 2πn will not be solutions for θ, but we still have infinite possible solutions, so we limit the range of the answers to –π ≤ θ ≤ π.

So the final answer is arg z = π/3, this is known as the principle arguments, and the other solutions for θ are known as arguments.

Eg. 2

If z = –1 + i 

 

z is in the second quadrant

tanθ = 1/-1

θ = 3π/4 or -π/4 (for –π ≤ θ ≤ π)

Since z is in the second quadrant, arg z ≠ 7π/4 +2πn

Therefore arg z = 3π/4

Eg. 3

If z = -3 – 3i

z is in the third quadrant

tanθ = -3/-3

θ = π/4 or -3π/4 (for –π ≤ θ ≤ π)

Since z is in the third quadrant, arg z = -3π/4.

Now that we know how to work out the modulus and argument of a complex number, we can write it in modulus argument form.

The point z represents that complex number z = x + iy

Let θ = arg z and |z|= r

Therefore x = r cosθ and y = r sinθ

Therefore x +iy = r (cosθ + isinθ), sometimes written as r cisθ, (cis = cos + isin)

This is known as the modulus argument form. The reason why we write complex numbers in this form is that there are various properties that allow use to easily perform operations in complex numbers (will be shown in the next post).

The modulus and argument are sometimes known as polar coordinates. They are written in the form [r, θ] where r is the modulus and θ is the argument. Please note that square brackets are used instead of parentheses for polar coordinates.

Square Roots of Complex Numbers

Complex numbers can also be square rooted.

Example 1

Find the square root of 3 +4i

(a+ib)² = 3 +4i

a² + 2iab - b² = 3 +4i

By equating real and imaginary parts

a² - b² = 3

2ab = 4

By solving simultaneously

a = 2 or -2

When a = 2, b =1

When a = -2, b = -1

So 2 + i and -2 - i are the square roots of 3 +4i

Example 2

Find the square root of i

(a +ib)² = i

a² +2iab -b² = i

By equating real and imaginary parts

a² -b² = 0 (1)

2ab =1

a = 1/2b (2)

Substitute (2) into (1)

∴ (1/2b)² - b² = 0

1/4b² - b² = 0

1 - 4b⁴ = 0

(1+2b²)(1-2b²) = 0

(1+2b²)(1-√2b)(1+√2b)=0

b = 1/√2 or -1/√2

When b = 1/√2, a = 1/√2

When b = -1/√2, a = -1/√2

Square roots of i are ±(1/√2 + i/√2)

With this, we are able to solve any quadratic equation even if the coefficients are complex.

Example

x² -(1+2i)x - (1 - i) =0

Δ = b² - 4ac

= (1+2i)² -4(-1+2i)

= 1+4i-4 +4 -4i

= 1 (in this case Δ is a real number, there are cases where Δ is complex and the above method would have to be used to find the roots to Δ)

x = (-b ± √Δ)/2a

= [(1+2i) ± 1]/2

= i or 1+i

Introduction to Complex Numbers

Suppose we wanted to take the square √ of -1. We know that there aren't any real numbers that would give a suitable answer so mathematicians introduced a number "i" known as the imaginary unit.

i² = -1

√-1 = i

√-4 = 2i

√ -3 = i√3

These numbers are known as imaginary numbers. They are written in the form ib (where b is a real number. b is placed in front of i if it is a rational number, otherwise it is placed on the back)

A complex number is a number that has both a real and a imaginary component. They are written in the form a+ ib (where a and b are real numbers)

All numbers are complex numbers. Real numbers are simply complex numbers with a zero imaginary component. Eg 1 = 1+ 0i. Imaginary numbers are complex numbers with a zero real component. Eg. 4i = 0 +4i

Equal complex numbers

Two complex numbers are equal only if their real components are equal and their imaginary components are also equal.

Eg. (a + ib) = (c + id) iff a = c and b = d.

Fyi. Iff is a mathematical term which means if and only if.

An example question can be found here.


Quadratic with negative discriminant

With complex numbers we can solve any quadratic equation even if the discriminant is less than 0.

Eg.  x² + x + 1 =0

x = (-1 ± √-3)/2

x = -1/2 + i√3/2 or 1/2 - i√3/2

Conjugate of a complex number

If z = a + ib the conjugate of z, written as

= a – ib

z= (a+ ib) (a –ib) = a² – i²b²

=a² + b²

The conjugate of a complex number is used to change the denominator into a real number if it is a complex number.

Eg. 2/1-2i = 2 (1+2i)/(1-2i)(1+2i) = (2 + 4i)/5

Sum, difference, product and division of complex numbers.

You do not have to remember these as they can be easily worked out with actual numbers.

Sum:

(a+ib) + (c+id) = (a+c) + i(b+d)

Difference:

(a+ib) – (c+id) = (a-c) + i(b-d)

Product:

(a+ib)(c+id) = ac + iad + ibc + i²bd

=ac – bd + i(ad+bc)

Division:

(a+ib)/(c+id) = (a+ib)(c-id)/(c+id)(c-id) (multiply by conjugate)

= (ac –iad +ibc –i²bd)/(c² +d²)

= [ac + bd + i(bc – ad)]/(c² +d²)

Example questions

Let z = 5 –i, find:

i)           z² in the form of a+ib

ii)          z + 2 in the form of a+ib

iii)         i/z in the form of a+ib

Solution

i)      z² = (5-i)²

= 25 – 10i + i²

= 24 – 10i

ii)      = 5 +i

z + 2 = (5-i) + 2(5+i)

= 5 – i + 10 +2i

= 15 + i

iii)     i/z = i/(5-i)

= i(5+i)/(5-i)(5+i)

= (-1 + 5i)/26

= -1/26 + 5/26 i

Radians

Radian is a different unit system for measuring sizes of angles. It is the standard unit for angle size. Radians is the ratio of the arc length to the radius.

Eg. A semi circle with radius r would have an arc length of πr.

Angle in radians = arc length/radius = πr/r

= π

This tells us that π is equivalent to 180 degrees.

An angle of 1 radian means that the length of the radius is equal to the arc subtended on a circle.

The conversion is as follows:

360 degrees = 2π radians

1 degree = π/180 radians.

To convert degrees to radians we just multiply by π/180

To convert radians to degrees we just multiply by 180/π

The use of radians is extremely important as it allows certain trigonometric limits to work. This leads to certain identities in calculus to be true only if the units for angles used is radians.

When writing angles in radians the units “radians” is usually omitted.

Parabola as a Locus

The parabola is the main locus that is studied in Extension 1 Mathematics

You may be familiar with parabolas in the form of y = ax² + bx + c or y = (x-h)² +k

But in this topic we will be writing parabolas in a new form.

A parabola with the vertex at the origin will be written in the form of

x² = 4ay

The reason why we write parabolas in this new form is that we can easily find the vertex and focal length. This allows us to easily determine the point of the focus and the directrix. Just to recap a parabola is a locus of points that is equidistant from a fixed point called the focus and a line called the directrix.

Some definitions

Focal length: the length from the focus to the vertex, the length from the vertex to the directrix. It is equal to a.

Focus: the point a units from the vertex within the concave side of the parabola.

Directrix: the line a units from the vertex on the opposite side of the concave side.

Chord: a line that touches the curve twice.

Focal chord: a chord that goes through the focus.

Latus rectum: a focal chord that is perpendicular to the directrix. It has a length of 4a.

Diagram of parabola x² = 4ay

Four standard parabolas:

x² = 4ay

x² = -4ay

y² = 4ax

y² = -4ax

Vertex at point (h,k)

If the vertex of the parabola was at point (h,k) then we would replace x with (x-h) and y with (y-k) just like in transformation of graphs.

Example parabola question

1. Write down the equation of the parabola with vertex at (-1,2) and directrix at x=2

x=2 is vertical and the vertex is on the left side of the directrix.

This implies that the parabola will be in the form of (y-k)² = -4a(x-h)

(h,k) => (-1,2)

Therefore (y-2)² = -4a(x+1)

a = length from focus to vertex or length from directrix to vertex = 3units

so the equation of the parabola is (y-2)² = -12(x+1)

2. Find the focal length, focus, directrix and end points of the latus rectum of the parabola y= -3 -4x -x²

y = -(x² + 4x + 3)

   = -(x +4x + 4 -1)

   = -[(x+2)² - 1]

y - 1 = -(x+2)²

4a = 1

a = 1/4

Focal length is 1/4 units

Vertex at (-2 , 1)

The parabola is concave down

so the focus will be at (-2 , 3/4)

directrix is y = 5/4

Latus rectum is 4(1/4) = 1 units in length

Therefore the endpoints must be half unit on the left and right of the focus.

Endpoints are (-5/2 , 3/4) and (-3/2 , 3/4)

Locus

A locus (pronounced lock-us) is a set of points that satisfy a certain criteria. This may be something like a circle which is a locus of points that are equidistant from a certain point, or it could be a region that satisfies an inequality.

In the Extension 1 course it usually means using the distance formula on a variable point.

Common types of loci

Circle

A circle is a locus of points such that its distance from a point (centre) is constant.

Eg. The point P(x,y) moves such that its distance from A(1,2) is 2 units. Find the expression of the locus of P and describe it geometrically.

Distance of PA = √((x-1)²+(y-2)²) = 2

(x-1)² + (y-2)² = 4

The locus of P is a circle with a radius of 2 units and the centre at (1,2)

Straight line

A straight line is a locus of points that are equidistant from two points.

Eg. The point P(x,y) moves such that its distance from A(-1,2) and B(3,-1) are equal. Find the expression of the locus.

PA = √((x+1)² + (y-2)²)

PB = √((x-3)² + (y+1)²)

PA = PB

∴ (x+1)² + (y-2)² = (x-3)² + (y+1)²

x² + 2x + 1 + y² – 4y + 4 = x² -6x + 9 + y² +2y +1

8x – 6y – 5 = 0

Parabola

A parabola is a locus of points that is equidistant from a fixed point called the focus and a line called the directrix.

Eg. The point P(x,y) moves such that its distance from the point S(0,1) and the line y = -1 is equal. Find the locus of point P.

PS = √(x² + (y-1)²)

Distance from P to y=-1 is y + 1

∴ x² + (y-1)²= (y+1)²

x² + y² -2y + 1 = y² +2y +1

x² = 4y

y= x²/4

Ellipse

The ellipse is a locus of points that are the sum of the distances from two points (called foci) is constant.

i.e. PA + PB = c

Hyperbola

The hyperbola is a locus of points where the difference of the distances from two points (called foci) is constant.

i.e. PA – PB = c

The ellipse and hyperbola will not be studied in detail in Extension 1 Mathematics. It will be studied in Extension 2 Mathematics in a topic called Conics.

Further Coordinate Geometry

One of the best ways to learn how a concept works is to understand its proof.

Below are the proofs for coordinate geometry concepts co-ordinate geometry concepts.

Perpendicular Distance Formula

The line ax + by + c = 0 has gradient –a/b

ax₀ + by₁ + c = 0 (Point C lies on the line)

∴ x₀ = -(by₁ + c)/a

Distance of PC = (ax₁ + by₁ + c)/a

Sin θ = d/PC

d = PC sinθ

if tanθ = -a/b

sinθ = -a/√(a² + b²)

∴ d = [(ax₁ + by₁ + c)/a] x [-a/√(a² + b²)]

= -(ax₁ + by₁ + c)/√(a² + b²)

Since distance is positive, we take the absolute value of this so

|ax₁ + by₁ + c|/√(a² + b²)

Internal division of an interval

In ∆PAC and ∆BPD

∠PAC = ∠BPD (corresponding angles on parallel lines)

∠ APC = ∠ PBD (corresponding angles on parallel lines)

∴ ∆PAC ||| ∆BPD (equiangular)

(x-x₁)/(x₂-x) = k/l

l(x-x₁) = k(x₂-x)

kx +lx = lx₁ +kx₂

x = (lx₁ + kx₂)/k+l

Similarly

(y-y₁)/(y₂-y) = k/l

l(y-y₁) = k(y₂-y)

ky +ly = ly₁ +ky₂

y = (ly₁ + ky₂)/k+l

Angle between two intervals

Let the angle of inclination of l₁ be α₁

∴ tan α₁ = m₁

Let the angle of inclination of l₂ be α₂

∴ tanα₂ = m₂

Angle between the intervals = α₂ – α₁

tan (α₂ – α₁) = (tan α₂ – tan α₁)/1+tanα₁tanα₂

= (m₂ – m₁)/1+m₁m₂

Functions and relations

A function is a graph that has one y value for every x value. Hence a common test for whether a graph is a function is the vertical line test.

In function notation, instead of using y, we use f(x) which is more convenient for substituting values.

Eg. if f(x) = x²

f(2) = 2² = 4

f(x+h) = (x+h)²

Common graphs

Straight line graph

Parabola

Cubic graph

Hyperbola

Exponential

Logarithm

Square Root Graph

Semi Circle

Circle

Absolute Value Graph

Graphs can be translated, flipped and transformed.

The transformation of graphs is counter intuitive.

To move a graph along h units along the x axis we substitute f(x-h)

Eg. (x-2)² is an x² graph shifted along the x axis by 2 units on the positive side.

Similarly if we want to move the graph k units up the y axis, we subtract k from f(x)

Eg. f(x) – 2 = x² is an x² graph shifted up two units.

This is commonly written as x² + 2 but in this case, counter intuition no longer holds.

If we want to flip a graph along the x axis, we substitute –y

If we want to flip a graph along the y axis, we substitute –x

Even and odd functions

A function is:

Even if for every x value, f(x) = f(-x) that is that if the graph is flipped along the y axis, it would still be the same graph.

To test for even functions, we substitute –x into the original function and we check to see whether the resulting function is equal to the original.

Odd if for every x value, f(x) = -f(x) that is that the graph has rotational symmetry about the origin.

To test for odd functions, we substitute –x into the original and we check whether the resulting function is equal to –f(x).

When we substitute –x into f(x) and find it is no equal to f(x) or –f(x), then the function is neither even nor odd.

Absolute Values

The absolute value of a number is the value of the number ignoring whether the number is positive or negative.
Eg.
|9| = 9
|-3| = 3
|-x| = x

Absolute value expressions can be shown as a graph.
A y = |x| graph is basically a y=x graph where all the negative points are placed on the positive side.
y=x

y=|x|

This y=|x| can be described as being a y=x graph when x>0 and a y=-x graph when x<0

You may remember that a parabolic expression in the form of

Will be a y=x^2 with the axis on (k,h) and will be thinner or wider depending on the value of a.
This is the same case with absolute value graphs.
Eg, y = 2|x-2|+1

y = -|x+1|+2

Composite Absolute value graphs
Graphs such as y=|x+1|+|x-2| or y = |x-2|-|x+3| will require more effort to graph.
I will explain how to sketch these graphs as well as provide some examples below.

Addition of absolute value graphs.
Eg1. y=|x+1|+|x-2|

If you were to sketch both the graphs of y=|x+1| and y =|x-2| you may notice that

1. When x< -1   |x+1| is in the form of -(x+1) and |x-2| is in the form of -(x-2)
2. When -1<x<2    |x+1| is in the form of x+1 and |x-2| is in the form of -(x-2)
3. When x>2    |x+1| is in the form of x+1 and |x-2| is in the form of x-2

If you were to add up the 2 graphs from each of the 3 areas you would find

1. When x<-1  -(x+1) +  -(x-2) = -2x + 1
2. When -1<x<2  x+1 -(x-2) = 3
3. When x>2  x+1 + x-2 = 2x-1

Now that you know what the graph would look like in each section you just graph the 3 sections

Eg2. y = x+|x-2|

After sketching the two graphs you would notice

1. When x<2 y=x is in the form of y=x, y=|x-2| is in the form of -(x-2)
2. When x>2 y=x is in the form of y=x, y=|x-2| is in the form of (x-2)

After adding the resolving each section

1. When x<2  y = x - (x-2) = 2
2. When x>2  y= x + x-2 = 2x-2

After sketching each of the two parts it should look like this:

Subtraction of absolute value graphs

The concept for subtraction is the same except instead of adding we are subtracting.

Eg1. y = |x-2|-|x+3|

1. When x<-3  |x-2| is in the form of -(x-2)  |x+3| is in the form of -(x+3)
2. When -3<x<2  |x-2| is in the form of -(x-2)   |x+3| is in the form of (x+3)
3. When x>2  |x-2| is in the form of (x-2)   |x+3| is in the form of (x+3)

Therefore 

1. When x<-3   -(x-2) - -(x+3)= 5
2. When -3<x<2   -(x-2) - (x+3)= -2x -1
3. When x>2   (x-2) - (x+3) = -5

The graph would look like this

Absolute Value Equations and Inequalities

Eg1. |2x+7| =3

To solve an equation like this we have to break up the absolute value. |a| = |-a|
Therefore we can say that if |2x+7| = 3  2x+7 = ±3
Then we can just solve for 2x+7 =3 and 2x+7 = -3
So x = -2, -5
Before we say that the answer is x=-2 or -5 we must substitute those two values back into the original equation before there are special cases where the solutions may be invalid.
In this question both solutions are valid.

Eg2. |2x+3| = |x+6|
2x + 3 = x+6  or 2x+3 = -x-6
x=-3 or 3
Both solutions are valid.

Eg3. |x+2| =2x+1
x+2 = 2x+1  or x+2 = -2x-1
x = 1 or -1

When x = 1 LHS = 3 RHS = 3
When x = -1 LHS = 1 RHS = -1
Therefore x=1
This question shows that there are cases where the solutions may be invalid

Eg4. |2x+5|<7
Absolute value inequalities can be solved using test point.
Consider |2x+5|=7
2x+5 = 7 or 2x+5 = -7
x = 1 or -6

Those 2 points will be the points when LHS = RHS in this case we want LHS<RHS so we test a point for each of the 3 regions:

1. x<-6  False
2. -6<x<1  True
3. x>1  False

Therefore the inequality is true for when -6<x<1

This concludes my post.

Absolute values is actually quite a simple topic if you know the fundamentals.

Hope you have learnt something