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HSC Mathematics Extension 2

Complex Numbers

Week 1

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HSC Mathematics Extension 2 2

Complex Numbers

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Week 1 – Theory Arithmetic of Complex Numbers and solving quadratic equations

The need for complex numbers arises from the impossibility of solving some equations using only

the real set of numbers.

A simple example is the quadratic equation, 𝑥2 + 1 = 0.

Solving for 𝑥 gives, 𝑥 = ±√−1

However we know that it is impossible to take the square root of any negative number using only

the real number system.

Thus, this number cannot belong to the set of real numbers.

Now we define the set of imaginary numbers (or purely Imaginary numbers), which are of the form

𝑘𝑖 where 𝑖 = √−1 and where k is some constant.

Now since 𝑖 = √−1 then 𝑖2 = −1.

We can now define the set of complex numbers which are of the form 𝑎 + 𝑖𝑏 where 𝑎 & 𝑏 are real

numbers. i.e. Complex Numbers are a combination of both real and imaginary numbers.

From this definition we can see that 𝑎, 𝑏 can be 0.It follows that all numbers whether Imaginary or

Real belong to the set of complex numbers. Real numbers have their Imaginary part set as 0 (i.e of

the form 𝑎 + 0𝑖 where 𝑎 ∈ 𝑅, 𝑎 ≠ 0) and Purely Imaginary numbers have their Real part set as 0 (i.e

of the form 0 + 𝑏𝑖 where 𝑏 ∈ 𝑅, 𝑏 ≠ 0). The following Venn diagram illustrates this.


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Real and Imaginary parts of Complex Numbers

Within any complex number there are two separate parts that need to be considered. These are the

Imaginary part and the Real part. This concept can be illustrated using the following example.

Example:

Determine the Real and Imaginary Parts of 1 + 2𝑖.

Example:

We can see that the constant in front of the 𝑖 is 2. This is the imaginary part. Note that the imaginary

part is not 2𝑖 but is 2. The real part of this number is 1.

Using symbols this is represented as, 𝑅𝑒(1 + 2𝑖) = 1

𝐼𝑚(1 + 2𝑖) = 2

Hence, for any complex number 𝑎 + 𝑏𝑖 𝑅𝑒(𝑎 + 𝑏𝑖) = 𝑎

𝐼𝑚(𝑎 + 𝑏𝑖) = 𝑏

The Basic Operations involving Complex Numbers

Since all Imaginary Numbers and most Complex Numbers essentially involve a surd (𝑖 = √−1 ), the

operations involving Imaginary and Complex Numbers are very similar to those of surds.

The following examples illustrate the operations with Complex Numbers,

Let 𝑧1 = 𝑎 + 𝑏𝑖, 𝑧2 = 𝑐 + 𝑑𝑖.

𝑧1 + 𝑧2 = (𝑎 + 𝑏𝑖) + (𝑐 + 𝑑𝑖)

= (𝑎 + 𝑐) + 𝑖(𝑏 + 𝑑)

𝑧1 − 𝑧2 = (𝑎 + 𝑏𝑖) − (𝑐 + 𝑑𝑖)

= (𝑎 − 𝑐) + 𝑖(𝑏 − 𝑑)


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𝑧1𝑧2 = (𝑎 + 𝑏𝑖)(𝑐 + 𝑑𝑖)

= 𝑎𝑐 + 𝑎𝑑𝑖 + 𝑏𝑐𝑖 + 𝑖2𝑏𝑑

= (𝑎𝑐 − 𝑏𝑑) + 𝑖(𝑎𝑑 + 𝑏𝑐) (Note that 𝑖2 = −1)

𝑧12 = (𝑎 + 𝑏𝑖)2

= 𝑎2 + 𝑖2𝑏2 + 𝑖2𝑎𝑏

= 𝑎2 − 𝑏2 + 𝑖2𝑎𝑏

𝑧 1

𝑧2=

𝑎+𝑏𝑖

𝑐+𝑑𝑖

=(𝑎 + 𝑏𝑖)

(𝑐 + 𝑑𝑖)×

(𝑐 − 𝑑𝑖)

(𝑐 − 𝑑𝑖)

=(𝑎𝑐 + 𝑏𝑑) + 𝑖(𝑏𝑐 − 𝑎𝑑)

𝑐2 + 𝑑2

In the last example, the idea of rationalising the denominator wth surdic expressions, is extended to

making the denominator real. The same process would be carried out for 1

𝑧 . It is also improtant to

note that in the last example, 𝑐 − 𝑑𝑖 is the conjugate of 𝑐 + 𝑑𝑖 and is represented by placing a dash

above the letter representing the complex number.

i.e. If 𝑧 = 𝑎 + 𝑏𝑖,

Then, 𝑧̅ = 𝑎 − 𝑏𝑖

Also it is important to note that 𝑧𝑧̅ is real for any Complex Number 𝑧.

(Note that we use the difference of two

squares in the denominator)


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Equalities of Complex Numbers

Two Complex Numbers are equal if and only if their Real parts are equal and their Imaginary parts

are equal.

i.e. 𝑧1 = 𝑧2 if and only if, 𝑅𝑒(𝑧1) = 𝑅𝑒(𝑧2)

𝐼𝑚(𝑧1) = 𝐼𝑚(𝑧2)

The use of this in a question is illustrated by the example on the next page.

Example:

For what values of x is the complex number (𝑥 + 2) + 𝑖(𝑥 + 𝑦) equal to 2𝑦 + 𝑖 ?

Solution:

For 2 Complex Numbers to be equal, their Imaginary parts must be equal and their Real parts must

be equal.

Equating Real parts gives, 𝑥 + 2 = 2𝑦

Equating Imaginary parts gives, 𝑥 + 𝑦 = 1

Now, Solving Simultaneously gives, 𝑥 = 0, 𝑦 = 1.

Thus only for the pair values of 𝑥 = 0, 𝑦 = 1, (𝑥 + 2) + 𝑖(𝑥 + 𝑦) is equal to 2𝑦 + 𝑖.


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Square Roots of Complex Numbers

The method used for finding the Square Roots of Complex Numbers is straightforward and can be

applied to any Complex Number.

The technique used involves equating a Complex Number 𝑥 + 𝑖𝑦 to the square root of the Complex

Number and then solving the equations obtained by equating Real and Imaginary parts.

Example:

Find the square roots of −7 + 24𝑖 .

Solution:

Let 𝑥 + 𝑖𝑦 = √−7 + 24𝑖 (𝑥, 𝑦 ∈ 𝑅)

(𝑥 + 𝑖𝑦)2 = −7 + 24𝑖 (Squaring Both Sides)

𝑥2 − 𝑦2 + 𝑖2𝑥𝑦 = −7 + 24𝑖

𝑥2 − 𝑦2 = −7 𝑎𝑛𝑑 𝑥𝑦 = 12 (Equating Real and Imaginary parts)

𝑥2 − (12

𝑥)

2= −7 (Solving Simultaneously)

𝑥2 −144

𝑥2= −7

𝑥4 + 7𝑥2 − 144 = 0

(𝑥2 + 16)(𝑥2 − 9) = 0

𝑥2 ≠ −16 , 𝑥2 = 9 (𝑥 ∈ 𝑅)

∴ 𝑥 = ±3

∴ 𝑦 = ±4

Thus, the square roots of −7 + 24𝑖 are ±(3 + 4𝑖).

Note: The symbol ∈ means “Element of” and is used if a term belongs to a set. In this case the set is

the Real Numbers, designated by the letter 𝑅. Other symbols for other sets include, 𝐶 for Complex

Numbers, 𝑀for purely Imaginary Numbers, 𝑄 for rational numbers & 𝑍 for Integers.


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Solving Equations of the form 𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄 = 𝟎 where 𝒂, 𝒃, 𝒄 are both Real and Complex

The method for solving equations of the form 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0 is the same for both real and

complex coefficients.

Both methods involve the use of the quadratic formula, 𝑥 =−𝑏±√𝑏2−4𝑎𝑐

2𝑎

When solving quadratic equations with real coefficients, if the solutions are complex (i.e. of the form

𝑎 + 𝑏𝑖), it can be guaranteed that the solutions will be conjugates of each other. This theorem will

be further investigated in the Polynomials module and will not be further investigated here.

Example:

Solve the equation for 5𝑥2 − 8𝑥 + 5 = 0, over the Complex set of numbers.

Solution:

We solve this equation like we would solve any other Quadratic.

So, firstly 𝑥 =8±√(−8)2−4.5.5

2.5 (Note that the point in between numbers means multiplication)

∴ 𝑥 =8 ± √−36

10

∴ 𝑥 =8±6𝑖

10 (Note that √−36 = √36 × √−1 = 6𝑖)

∴ 𝑥 =4

5±

3

5𝑖

Note that the solutions of this equation are of the form 𝑧, 𝑧̅. i.e. the solutions are conjugates. As

stated earlier this is due to the coefficients of the quadratic being Real.


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Example:

a) Find the square roots of 16 − 30𝑖.

b) Hence solve the equation 𝑧2 − (3 − 𝑖)𝑧 − 2 + 6𝑖 = 0.

Solution:

a) Firstly we find the square roots of 16 − 30𝑖

Hence let 𝑥 + 𝑖𝑦 = √16 − 30𝑖 (𝑥, 𝑦 ∈ 𝑅)

𝑥2 − 𝑦2 + 𝑖2𝑥𝑦 = 16 − 30𝑖 (Squaring both sides)

𝑥2 − 𝑦2 = 16 , 𝑥𝑦 = −15 (Equating Real and Imaginary parts)

𝑥2 − (−15

𝑥)

2= 16 (Solving Simultaneously)

𝑥4 − 16𝑥2 − 225 = 0

(𝑥2 − 25)(𝑥2 + 9) = 0

𝑥2 ≠ −9 , 𝑥2 = 25 (𝑥 ∈ 𝑅)

∴ 𝑥 = ±5

∴ 𝑦 = ∓3

∴ √16 − 30𝑖 = ±(5 − 3𝑖)

b) We now solve the equation, but first we will deal with the discriminant.

𝑧2 − (3 − 𝑖)𝑧 − 2 + 6𝑖 = 0.

∆ = 𝑏2 − 4𝑎𝑐

= [−(3 − 𝑖)]2 − 4.1. (−2 + 6𝑖)

= 9 + 𝑖2 − 6𝑖 + 8 − 24𝑖

= 16 − 30𝑖

∴ √∆ = √16 − 30𝑖

∴ √∆ = ±(5 − 3𝑖) (From part (a))

∴ 𝑧 =(3−𝑖)±√∆

2.1 (Using the Quadratic Formula)

∴ 𝑧 =(3 − 𝑖) ± (5 − 3𝑖)

2

∴ 𝑧 = 4 − 2𝑖, −1 + 𝑖

Note that dealing with the discriminant separately is not necessary but tidies up your Mathematics.


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Representing Complex Numbers on an Argand diagram

In previous years the idea of a Number Line was introduced as a continuous line in which real

numbers could be placed according to their magnitude and sign.

We now must introduce the concept of a Number Plane to accommodate for the Imaginary &

Complex numbers that we have now established.

This plane is called an Argand diagram and has an horizontal axis which corresponds to the real part

of the number and a vertical axis corresponding to the Imaginary part.

For example the Complex Number 𝑧 = 2 + 3𝑖 would be represented by the point (2, 3).

The Modulus and Argument of a Complex Number

So far we have only represented Complex Numbers in the form 𝑎 + 𝑏𝑖 where 𝑎, 𝑏 ∈ 𝑅. However

Complex numbers can also be represented in Modulus – Argument form, or mod – arg form.

The Modulus of a Complex Number is the displacement the point has from the origin. It is

represented by |𝑧|, √𝑧𝑧̅, or 𝑚𝑜𝑑 𝑧. There is a basic formula used to find the Modulus of a Complex

Number,

|𝑧| = √(𝑅𝑒(𝑧))2

+ (𝐼𝑚(𝑧))2

, where 𝑧 is any Complex Number.


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This formula arrives from applying Pythagoras’ theorem to the right angled triangle formed by the

Complex Number, represented geometrically on the Argand diagram. For example, take the Complex

Number 1 + 2𝑖. Using the general formula,

|𝑧| = √12 + 22

|𝑧| = √5 Using the geometrical representation,

The Argument of a Complex Number is the angle which the line joining the Origin and the Complex

Number makes with respect to the positive real axis. Note that the argument can only be in the

interval −𝜋 ≤ arg 𝑧 ≤ 𝜋.

Using the same example, the Argument of the Complex Number 1 + 2𝑖 is represented by 𝜃.


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It can be seen that 𝜃 can be found by using simple trigonometry.

tan 𝜃 =2

1

∴ 𝜃 = tan−1 2 (Normally we would not go further than this step)

∴ 𝜃 ≅ 1.11 rads.

∴ arg 𝑧 ≅ 1.11 rads

In general, the Argument of some Complex Number 𝑎 + 𝑏𝑖 is found by,

arg 𝑧 = tan−1 (𝑏

𝑎)

It is important to note which quadrant the Complex Number is in since the range is −𝜋 ≤ arg 𝑧 ≤ 𝜋.

For example the Complex Number −1 − 𝑖 represented on the Argand diagram, has not an

Argument of 5𝜋

4, rather has an Argument of −

3𝜋

4 , since the range for arg 𝑧 is −𝜋 ≤ arg 𝑧 ≤ 𝜋.

You may find it easier to find the argument after making a quick sketch of the complex number on

an Argand diagram. This will remove any faults brought about by quadrant errors.

Representing Complex Numbers in modulus – argument form

We can represent Complex Numbers in a form which utilises the modulus and argument, alongside

the form

𝑎 + 𝑏𝑖.

If, |𝑧| = 𝑟 and arg 𝑧 = 𝜃,

Then the complex number 𝑧 in modulus – argument form is represented by, 𝑧 = 𝑟(cos 𝜃 + 𝑖 sin 𝜃)

OR 𝑧 = 𝑟. 𝑐𝑖𝑠 𝜃. Where 𝑐𝑖𝑠 𝜃 = cos 𝜃 + 𝑖 sin 𝜃.

(Note that this is not an exact value

and that it is better to leave your

answer in exact form)


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Example:

Express √3 − 𝑖 in modulus – argument form.

Solution:

Let 𝑧 = √3 − 𝑖.

Firstly we find the modulus.

𝑚𝑜𝑑 𝑧 = √(√3)2

+ (−1)2

= 2

Secondly we find the argument.

Now 𝑧 is in the 4th quadrant, from the diagram.

∴ 𝑎𝑟𝑔 𝑧 = − 𝑡𝑎𝑛−1 |𝑏

𝑎| (Fourth Quadrant)

= − 𝑡𝑎𝑛−1 |−1

√3|

= −𝜋

6

Now we express z in modulus - argument form.

∴ 𝑧 = 2 (cos (−𝜋

6) + 𝑖 sin (−

𝜋

6))

i.e. 𝑧 = 2 (cos𝜋

6− 𝑖 sin

𝜋

6) (Noting the sign of the trigonometric functions in the 4th

Quadrant)


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Week 1 – Homework 1. Determine the Real and Imaginary parts of 𝑧 where 𝑧 =

1+𝑖

1−𝑖.

2. Write 1 +3−2𝑖

6+7𝑖 in the form 𝑎 + 𝑏𝑖 where 𝑎, 𝑏 ∈ 𝑅.

3. Find the two square roots of 5 − 12𝑖.


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4. Using the answer from Question 3, solve 𝑧2 + 𝑧 − 1 + 3𝑖 = 0 leaving your answer in the form

𝑥 + 𝑖𝑦 where 𝑥 & 𝑦 are Real.

5. If 𝑧 = 2 − 3𝑖 and 𝜔 = 12 − 7𝑖,

Find in the form 𝑎 + 𝑏𝑖;

a) 𝑧 + 𝜔

b) 𝑧 − 𝜔

c) 𝑧𝜔

d) 1

𝑤

e) 𝑤

𝑧


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6. Evaluate the pair values of 𝑥, 𝑦 for which the expression

(𝑥2 − 4) + 𝑖(𝑥 + 𝑦) = (𝑥 + 𝑦) − 𝑖(𝑥2 + 2) maintains equality.

7. If 𝑧 = 1 is one of the roots of the equation 𝑖𝑧2 + (𝑖𝑥 − 1)𝑧 + (𝑖 − 𝑦) = 0, where 𝑥 & 𝑦 are Real,

find:

a) The values of 𝑥 & 𝑦.

b) The other zero of the equation.

8. Find 𝑥 & 𝑦, when (𝑥 + 𝑖𝑦)(2 + 3𝑖) = 5 + 6𝑖.


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9. Let 𝑧 = 𝑥 + 𝑖𝑦, where 𝑥, 𝑦 ∈ 𝑅, express 𝑧 +1

𝑧 in the form 𝑎 + 𝑖𝑏 where 𝑎, 𝑏 ∈ 𝑅.

10. Determine the values of 𝑥 for which (𝑥+2

𝑥+1) + 𝑖 (

𝑥2+𝑥−2

𝑥) is a positive Real number.

11. Evaluate (2 + 3𝑖)3 − (2 + 𝑖)3.


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12. Show that 𝑖𝑛 where 𝑛 ∈ 𝑍, has one of the four values ±1 , ±𝑖.

13. Prove that (4+3𝑖)√3+4𝑖

3+𝑖=

5

2(1 + 𝑖).

14. Find 𝑥 in the domain 0 < 𝑥 <𝜋

2, if

√2(cos 𝑥−𝑖 sin 𝑥)

2+𝑖=

1−3𝑖

5.


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15. Divide 𝑥3 − 2 − 2𝑖 by 𝑥 + 1 − 𝑖 and hence prove that the 3 cube roots of 2 + 2𝑖 are −1 + 𝑖

and 1

2[1 − 𝑖 ± (√3 + √3𝑖)].

16. On an Argand diagram, sketch the following points,

a) 1 + 𝑖

b) 1 − 𝑖

c) 𝑖(1 + 𝑖)

d) √3 + 𝑖

e) √2 𝑐𝑖𝑠 (−𝜋)

f) 𝑖2 (𝑐𝑖𝑠 𝜋

2)

g) −5𝑖

h) 6 𝑐𝑖𝑠 𝜋

6


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17. Express the following Complex Numbers in Modulus - Argument form,

a) 1 + 𝑖

b) 𝑖3

c) 1 + √3𝑖

d) (1 + √3𝑖)(1 + 𝑖)

e) 𝑖5(√3 + 𝑖)

f) (√3 − 𝑖)5

g) 1+𝑖

√3+𝑖

18. If 𝑧 = 𝑧̅, what can you say about the geometrical position of 𝑧?


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19. If 𝑧̅ = −𝑧, what can be said about the geometrical position of 𝑧?

20.

a) Find the roots of the Quadratic Equation 𝑧2 + 2𝑖 and express them in the form 𝑎 + 𝑏𝑖.

b) Hence, find the value of arg 𝑧1 + arg 𝑧2 and also find the relationship between the moduli of

the 2 roots.

End of Homework

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