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.