1. Complex Numbers 12 Learning Outcomes
In this chapter you have learned how to:
Plot complex numbers on the Argand diagram
Find the modulus of a complex number and calculate the complex conjugate
Add, subtract, multiply and divide complex numbers
Use complex numbers to perform transformations
Solve equations with complex roots and use the Conjugate Root Theorem
Write complex numbers in polar form
Prove de Moivre’s Theorem
Use de Moivre’s Theorem to evaluate (a + bi)n, a, b ∈ R, n ∈ Z
Use de Moivre’s Theorem to find the roots of complex numbers

2. 12 Complex Numbers Definition
A Complex Number z is a number of the form 𝑧=𝑎+𝑏𝑖, 𝑎, 𝑏∈𝑅, 𝑖 2 =−1.
a is called the real part of z, Re z . b is called the imaginary part of z, Im z .
Examples: 2+3𝑖, 5−2𝑖, 𝑖, 2 −3𝑖
Note:
The complex numbers 𝑧 1 =𝑎+𝑏𝑖 and 𝑧 2 =𝑐+𝑑𝑖 are equal if 𝑎=𝑐 and 𝑏=𝑑.
Simplify 𝒊 𝟒 .
𝑖 4 =( 𝑖 2 ) 2 =(−1 ) 2 =1
𝑖=𝑖
𝑖 2 =−1
𝑖 3 =−𝑖
𝑖 4 =1
𝑖 5 =𝑖
𝑖 6 =−1
𝑖 7 =−𝑖
𝑖 8 =1
𝑖 9 =𝑖
Solve the equation 𝒛 𝟐 +𝟑𝟔=𝟎.
𝑧 2 +36=0
𝑧 2 =−36
𝑧=± −36
𝑧=± −1
𝑧=±6𝑖
Note the repeating pattern.

3. 12 Complex Numbers The Argand Diagram
A complex number can be represented on a diagram called the Argand diagram.
Here are the complex numbers 𝑧 1 =3+2𝑖, 𝑧 2 =4−3𝑖
and 𝑧 3 =−2−5𝑖 represented on an Argand diagram. -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 Re Im
𝑧 1 =3+2𝑖
𝑧 2 =4−3𝑖
𝑧 3 =−2−5𝑖
Modulus of a Complex Number
Definition
The modulus of a complex number, 𝑎+𝑏𝑖, is its distance from the origin.
The modulus of 𝑎+𝑏𝑖 is written |𝑎+𝑏𝑖|.
|𝒂+𝒃𝒊|= 𝒂 𝟐 + 𝒃 𝟐
If 𝒛 𝟏 =𝟕+𝟐𝟒𝒊 find | 𝒛 𝟏 |.
𝑧 1 = 7+24𝑖 = = = 625 =25

4. Addition, Subtraction, Multiplication and Division of Complex Numbers12
Complex Numbers
Addition, Subtraction, Multiplication and Division of Complex Numbers
Addition
𝐈𝐟 𝒛 𝟏 =𝟐−𝟑𝒊 𝐚𝐧𝐝 𝒛 𝟐 =𝟏𝟏+𝟓𝒊 evaluate 𝒛 𝟏 + 𝒛 𝟐 .
𝑧 1 + 𝑧 2 = 2−3𝑖 +(11+5𝑖)
=2−3𝑖+11+5𝑖
Add the real parts to the real parts and the imaginary parts to the imaginary parts.
𝑧 1 + 𝑧 2 =13+2𝑖
Subtraction
𝐈𝐟 𝒛 𝟏 =𝟏𝟖+𝟏𝟔𝒊 𝐚𝐧𝐝 𝒛 𝟐 =𝟏𝟒−𝟐𝒊 evaluate 𝒛 𝟏 − 𝒛 𝟐 .
𝑧 1 − 𝑧 2 = 18+16𝑖 −(14−2𝑖)
=18+16𝑖−14+2𝑖
Add the real parts to the real parts and the imaginary parts to the imaginary parts.
𝑧 1 − 𝑧 2 =4+18𝑖

5. 12 Complex Numbers Multiplication
𝐈𝐟 𝒛 𝟏 =𝟐−𝟕𝒊 𝐚𝐧𝐝 𝒛 𝟐 =𝟑+𝟕𝒊 evaluate 𝒛 𝟏 𝒛 𝟐 .
𝑧 1 𝑧 2 =(2−7𝑖)(3+7𝑖)
=2(3+7𝑖)−7𝑖(3+7𝑖)
=6+14𝑖−21𝑖−49 𝑖 2
=6+14𝑖−21𝑖−49(−1)
=6+14𝑖−21𝑖+49
=55−7𝑖
Conjugate of a Complex Number
Definition
If 𝑧=𝑎+𝑏𝑖, then the 𝐜𝐨𝐧𝐣𝐮𝐠𝐚𝐭𝐞 of 𝑧 written as 𝑧 =𝑎−𝑏𝑖.
Rule: Change the sign of the imaginary part. Example: If 𝑧=−2−12𝑖, then 𝑧 =−2+12𝑖.

6. 12 Complex Numbers 𝟐+𝟏𝟏𝒊 𝟐+𝒊 . Division Calculate
𝟐+𝟏𝟏𝒊 𝟐+𝒊 .
Note: When dividing by a complex number multiply the numerator and the denominator by the
conjugate of the denominator.
Step 1
Write down the conjugate of the denominator:
2+𝑖 =2−𝑖
Step 2
Multiply the denominator 2+𝑖 by 2−𝑖.
2+𝑖 2−𝑖 =2 2−𝑖 +𝑖(2−𝑖)
=4−2𝑖+2𝑖− 𝑖 2 =5
Step 3
Multiply the numerator by 2−𝑖.
2+11𝑖 2−𝑖 =2 2−𝑖 +11𝑖(2−𝑖)
=4−2𝑖+22𝑖−11 𝑖 2
=4+20𝑖+11
Step 4
2+11𝑖 2+𝑖 = 15+20𝑖 5
=15+20𝑖
=3+4𝑖

7. 12 Complex Numbers Transformations
The addition of complex numbers is the equivalent of a translation on the complex plane.
If 𝒛 𝟏 =𝟐+𝟒𝒊 and 𝝎=𝟏+𝒊 plot 𝒛 𝟏 + 𝝎 on an Argand diagram.
Describe the transformation that maps the addition of 𝒛 𝟏 to 𝒛 𝟏 +𝝎.
The multiplication of a complex number by 𝒊 is the equivalent of rotating the complex number anti-clockwise through 𝟗𝟎 ° about the origin.
𝐈𝐟 𝒛 𝟏 =𝟐+𝒊 𝐚𝐧𝐝 𝝎=𝒊(𝟐+𝒊) describe the transformation that maps 𝒛 𝟏 to 𝝎. -8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8 Re Im
𝑧 1 +𝜔=3+5𝑖
𝑧 1 =2+4𝑖 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 Re Im
𝜔=1+𝑖
𝜔=−1+2𝑖
𝑧 1 =2+𝑖
𝑧 1 →𝑧 1 + 𝜔 is a translation.
𝑧 1 → 𝜔 is a+90° rotation.

8. Quadratic Equations with Complex Roots12
Complex Numbers
Quadratic Equations with Complex Roots
𝑎 𝑥 2 +𝑏𝑥+𝑐=0, 𝑎, 𝑏, 𝑐∈𝑅, 𝑎≠0 is solved by using the formula
𝑥= −𝑏± 𝑏 2 − 4𝑎𝑐 2𝑎 .
In this formula, 𝑏 2 −4𝑎𝑐 is the discriminant.
If 𝑏 2 −4𝑎𝑐<0, then the solutions (roots) will be complex.
Solve the equation 𝒛 𝟐 −𝟔𝒛+𝟏𝟑=𝟎. -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 Re Im
𝑧= 6± (−6 ) 2 −4(1)(13) 2(1)
= 6± 36−52 2
𝑧=3+2𝑖
= 6± −16 2
= 6±4𝑖 2
𝑧=3−2𝑖
∴𝑧=3±2𝑖
The Conjugate Root Theorem
If the complex number 𝑧=𝑎+𝑏𝑖, 𝑎, 𝑏∈𝑅, is a root of a polymomial with real coefficients, then 𝑧 =𝑎−𝑏𝑖 (conjugate of z) is also a root.

9. Polynomials with Complex Roots12
Complex Numbers
Polynomials with Complex Roots
If 𝟏+𝟐𝒊 is a root of 𝟑 𝒛 𝟑 −𝟓 𝒛 𝟐 +𝟏𝟑𝒛+𝟓, find the other roots.
1+2𝑖 is a root ∴1−2𝑖 is a root. (Conjugate Root Theorem)
Step 1 Form a quadratic polynomial with these two roots.
𝑧 2 − sum of roots 𝑧+product of roots=0
Sum of roots= 1+2𝑖 + 1−2𝑖 =2
Product of roots= 1+2𝑖 1−2𝑖 =5
So the quadratic polynomial is 𝑧 2 −2𝑧+5=0
Step 2 Divide this function into the cubic function.
Step 3 3𝑧+1 is a linear factor.
Therefore, the solution to the equation 3𝑧+1 = 0 gives the real root.
3𝑧+1=0 ⟹𝑧=
− 1 3
The three roots are 𝑧=1+2𝑖, 𝑧=1−2𝑖, 𝑧=− 1 3

10. √ 12 Complex Numbers Polar Form of a Complex Number (− 3 ) 2 +(1 ) 2Re Im
Definition
𝑟=|𝑥+𝑦𝑖|
𝑧=𝑥+𝑦𝑖
The polar form of a complex number z
is 𝑟 𝑐𝑜𝑠𝜃+𝑖 𝑠𝑖𝑛𝜃 , where r is the
modulus of the complex number
and 𝜃 is its argument
anticlockwise angle .
tan 𝜃 = 𝑦 𝑥 ⟹𝜃= tan −1 𝑦 𝑥 𝑟 𝑦 𝜃 𝑥
Write the complex number − 𝟑 +𝒊 in polar form. Re Im
𝜃= tan −1 𝑦 𝑥
𝑟=|𝑥+𝑦𝑖|
𝑧=− 3 +𝑖
𝛼= tan −
𝑟=|− 3 +𝑖| 1 𝜃 𝛼 𝑟= √
(− 3 ) 2 +(1 ) 2
𝛼= 𝜋 6 3
(0,0)
𝑟= 3+1
𝜃=𝜋− 𝜋 6 = 5𝜋 6
𝑟=2
Polar Form: 𝑟 cos𝜃+𝑖 sin𝜃 =2(cos 5𝜋 6 +𝑖 sin 5𝜋 6 )

11. 12 Complex Numbers De Moivre’s Theorem Theorem
If 𝑧=𝑟( cos 𝜃+𝑖 sin 𝜃) then 𝑧 𝑛 = 𝑟 𝑛 ( cos 𝑛𝜃+𝑖 sin 𝑛𝜃), for 𝑛∈𝑍.
i.e: [𝑟( cos 𝜃+𝑖 sin 𝜃) ] 𝑛 = 𝑟 𝑛 ( cos 𝑛𝜃+𝑖 sin 𝑛𝜃), for 𝑛∈𝑍.
The formal proof of De Moivre’s Theorem may be asked in the exam.
Use de Moivre’s Theorem to write (𝟏+𝒊 ) 𝟏𝟎 in the form 𝒙+𝒚𝒊, 𝒙,𝒚∈𝑹.
Step 1 Write 1+𝑖 in polar form, 𝑟 cos𝜃+𝑖 sin𝜃 . Re Im
𝜃= tan −1 𝑦 𝑥
𝜃= 𝜋 4
𝜃= tan −1 1
1+𝑖
Polar Form: 𝑟 cos𝜃+𝑖 sin𝜃 = 2 (cos 𝜋 4 +𝑖 sin 𝜋 4 ) 1
[𝑟( cos 𝜃+𝑖 sin 𝜃) ] 𝑛 = 𝑟 𝑛 ( cos 𝑛𝜃+𝑖 sin 𝑛𝜃) 𝜃
(0,0) 1
Step 2 Apply de Moivre’s Theorem.
= ( 2 ) 10 [cos ( 𝜋 4 )+𝑖 sin ( 𝜋 4 ) ] 10
= ( 2 ) 10 [cos 10( 𝜋 4 )+𝑖 sin 10( 𝜋 4 )]
= cos ( 5𝜋 2 )+𝑖 sin ( 5𝜋 2 )
𝑟=|𝑥+𝑦𝑖|
𝑟= (1 ) 2 +(1 ) 2
𝑟= 2
= 32[0+1𝑖]
= 0+32𝑖

12. √ 12 Complex Numbers (2 ) 2 +(2 3 ) 2
If 𝝎=−𝟐−𝟐 𝟑 𝒊, solve the equation 𝒛 𝟒 −𝝎=𝟎.
𝑧=(−2−2 3 𝑖 ) 1 4
𝑧 4 =𝜔
𝑧 4 =−2−2 3 𝑖
Step 1 Express −2−2 3 𝑖 in general polar form:
𝑟[( cos 𝜃+2𝑛𝜋 + 𝑖 sin (𝜃+2𝑛𝜋)]
𝜔=−2−2 3 𝑖 𝜃 𝛼
𝛼= tan − ⟹𝛼= 𝜋 3 𝑟= √
(2 ) 2 +(2 3 ) 2
𝜃=−𝜋+ 𝜋 3 =− 2𝜋 3
𝑟= 16 =4
General polar form:
𝜔=4 cos − 2𝜋 3 +2𝑛𝜋 + 𝑖 sin − 2𝜋 3 +2𝑛𝜋
Step 2 Apply de Moivre’s Theorem:
𝑧=4 cos − 2𝜋 3 +2𝑛𝜋 + 𝑖 sin − 2𝜋 3 +2𝑛𝜋
1 4
=4 [ cos − 𝜋 6 + 𝑛𝜋 2 + 𝑖 sin (− 𝜋 6 + 𝑛𝜋 2 )]
1 4

13. 12 Complex Numbers 𝑧=4 cos − 𝜋 6 + 𝑛𝜋 2 + 𝑖 sin − 𝜋 6 + 𝑛𝜋 2 . 1 4
Step 3 Find the four solutions of
For 𝒏=𝟎
For 𝒏=𝟐
𝑧= 2 cos − 𝜋 6 + 𝑖 sin − 𝜋 6
𝑧= 2 cos − 𝜋 6 +𝜋 + 𝑖sin − 𝜋 6 +𝜋
= − 1 2 𝑖
= cos 5𝜋 6 + 𝑖sin 5𝜋 6
𝒛 𝟏 = 𝟔 𝟐 − 𝟐 𝟐 𝒊
= 2 − 𝑖
𝒛 𝟑 =− 𝟔 𝟐 + 𝟐 𝟐 𝒊
For 𝒏=𝟏
For 𝒏=𝟑
𝑧= 2 cos − 𝜋 6 + 3𝜋 2 + 𝑖sin − 𝜋 6 + 3𝜋 2
𝑧= cos − 𝜋 6 + 𝜋 2 + 𝑖sin − 𝜋 6 + 𝜋 2
= 2 cos 𝜋 3 + 𝑖sin 𝜋 3
= cos 4𝜋 3 + 𝑖sin 4𝜋 3
= 𝑖
𝒛 𝟐 = 𝟐 𝟐 − 𝟔 𝟐 𝒊
= 2 − 1 2 − 𝑖
𝒛 𝟒 =− 𝟐 𝟐 − 𝟔 𝟐 𝒊