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Chapter 2: COMPLEX NUMBERS
2.1 INTRODUCTION
In our earliest mathematical education we deal with REAL Numbers. At this stage we could solve
elementary equations of the forms:
3𝑥 = 9 where 𝑥 = 3
2
− 4 = 0 where
However, some of the equations had no real solutions. For example,
2
- 4 = 0 where 𝑥 = ±√−4 (no real solutions due to no perfect square can be negative in real
number system)
In order to solve this kind of equations, complex numbers form is needed.
Complex numbers is in 𝑎 + 𝑏𝑖 form where 𝒂 & 𝒃 are real numbers.
𝑖 was created which has a property that: 𝑖 = √− 1 or 𝑖
2
Thus, solutions for 𝑥
2
- 4 = 0 is = ±√−4 , i.e. 𝑥 = ±√4 × √−
A complex number of the form 𝑏𝑖 where 𝑏 ≠ 0 is called purely imaginary.
Example 1:
Write in terms of 𝑖:
(a) √− 16 (b) √− 8
(a) √
16 ×
(b) √
8 ×
4 × 2 ×
Example 2 :
Write as a product of linear factors:
(a) 𝑥
2
2
(a) 𝑥
2
2
2
2
2
(b) 4 𝑥
2
2
2
Example 3:
Solve for 𝑥: 𝑥
2
2
2
2.2 OPERATIONS WITH COMPLEX NUMBERS
If we write 𝑧 = 𝑎 + 𝑏𝑖 where 𝑎 & 𝑏 are real then:
𝒂 is the real part of z , and we write 𝑎 = 𝑅𝑒
𝒃 is the imaginary part of z , and we write 𝑏 = 𝐼𝑚
Addition:
Subtraction:
Multiplication:
2
Division:
𝑎+𝑏𝑖
𝑐+𝑑𝑖
𝑎+𝑏𝑖
𝑐+𝑑𝑖
𝑐−𝑑𝑖
𝑐−𝑑𝑖
𝑎𝑐−𝑎𝑑𝑖+𝑏𝑐𝑖−𝑏𝑑𝑖
2
𝑐
2
+𝑑
2
(multiply with conjugates of the denominator)
(𝑎𝑐+𝑏𝑑)+(𝑏𝑐−𝑎𝑑)𝑖
𝑐
2
+𝑑
2
Complex conjugates
Complex numbers 𝑎 + 𝑏𝑖 and 𝑎 − 𝑏𝑖 are called complex conjugates.
If 𝑧 = 𝑎 + 𝑏𝑖 we write its conjugates, 𝑧̅ = 𝑎 − 𝑏𝑖.
If a real quadratic equation has ∆ < 0 and 𝑎 + 𝑏𝑖 is a complex root then 𝑎 − 𝑏𝑖 is also a root.
Example 4 :
Find 𝑎 and 𝑏 if √ 2 + 𝑖 is a root of 𝑥
2
- 𝑎𝑥 + 𝑏 = 0 , where 𝑎 and 𝑏 are real.
Since 𝑎 and 𝑏 are real, the quadratic has real coefficients, ∴ √ 2 + 𝑖 is also a root
∴ sum of roots = √ 2 + 𝑖 + √ 2 − 𝑖 = 2 √ 2
Product of roots= (√ 2 + 𝑖) × (√ 2 − 𝑖) = 2 + 1 = 3
Thus 𝑎 = − 2 √ 2 and 𝑏 = 3
Example 5 :
If 𝑧 = 5 − 2 𝑖 and 𝑤 = 2 + 𝑖, find:
(a) 2 𝑧 − 3 𝑤 (b) 𝑤
2
(c) 𝑧𝑤 (d) 𝑖𝑧𝑤
(a) 2 𝑧 − 3 𝑤 = 2
(b) 𝑤
2
2
(c) 𝑧𝑤 =
2
(d) 𝑖𝑧𝑤 = 𝑖( 12 + 𝑖)
2
2.3 ARGAND DIAGRAMS
The plane of complex numbers (complex plane, or Argand plane), has a horizontal real axis and a
vertical imaginary axis.
Notice that:
Coordinates of A is (4, 6) and it represents 4 + 6𝑖
Coordinates of B is (8, −2) and it represents 8 − 2𝑖
Coordinates of C is (0, −6) and it represents −6𝑖
Coordinates of D is
and it represents −4 − 2𝑖
Example 8:
If 𝑎 = 3 + 𝑖 and 𝑏 = 1 − 4 𝑖, find algebraically and vectorially:
(a) 𝑎 + 𝑏 (b) 𝑎 − 𝑏
(a) 𝑎 + 𝑏
(b) 𝑎 − 𝑏
(a)
(b)
2.4 POLAR FORM
Suppose the complex number 𝑧 = 𝑥 + 𝑖𝑦 is represented by vector OP as shown below. 𝑥 + 𝑖𝑦 is
the Cartesian form of z.
Suppose also that OP makes an angle of with the positive real axis.
The angle is called the argument of z , or simply arg z , and
Any complex number z has Cartesian form 𝑧 = 𝑥 + 𝑖𝑦 or Polar form 𝑧 = 𝑟𝑐𝑖𝑠𝜃 where 𝑟 is the
modulus/length of z , is the argument of z and cis 𝜃 = cos 𝜃 + 𝑖 sin 𝜃
To find the value of and r , we use this formula: tanθ =
𝑦
𝑥
→ 𝜃 = tan
−
𝑦
𝑥
2
2
Example 9 :
Write in polar form: (a) 2 𝑖 (b) 1 + 𝑖
(a) 𝑟 =
𝜋
2
(b) 𝑟 =
2
2
𝜃 = tan
− 1
1
1
Example 1 0 : Convert
2 𝜋
3
) to Cartesian form.
2.6 DeMOIVRE’S THEOREM
𝑛
𝑛
Prove:
Let 𝑧 = 𝑟𝑐𝑖𝑠𝜃
2
= 𝑟𝑐𝑖𝑠𝜃 × 𝑟𝑐𝑖𝑠𝜃
2
2
Now 𝑧
3
2
𝑐𝑖𝑠2𝜃 × 𝑟𝑐𝑖𝑠𝜃
3
3
The generalization of this process is (𝑟𝑐𝑖𝑠𝜃)
𝑛
𝑛
Example 14 :
Find the exact value of ( √
8
using De Moivre’s theorem.
√ 3 + 𝑖 has length, 𝑟 =
2
2
Argument of √
3 + 𝑖 = tan
− 1
1
√
3
𝜋
6
8
8
8
4 𝜋
3
8
1
2
√
3
2
Example 15:
By considering cos 2 𝛽 + 𝑖 sin 2 𝛽, deduce the duplication formulae for cos 2 𝛽 and sin2 𝛽.
cos 2 𝛽 + 𝑖 sin 2 𝛽 = 𝑐𝑖𝑠 2 𝛽
2
2
2
2
Equating real parts, 𝑐𝑜𝑠 2 𝛽 = 𝑐𝑜𝑠
2
2
Equating imaginary parts, 𝑠𝑖𝑛 2 𝛽 = 2 𝑠𝑖𝑛𝛽𝑐𝑜𝑠𝛽
2.7 SOLVING EQUATIONS AND FINDING ROOTS OF COMPLEX NUMBERS
DEFINITION: The nth roots of complex number C are the n solutions of 𝒛
𝒏
For example, the 4
th
roots of 2𝑖 are the four solutions of 𝑧
4
nth roots may be found by:
factorization,
using the ‘ nth root method’ of complex numbers.
Example 1 6 :
Find the three cube roots of 1 using:
(a) factorization (b) the ‘ nth roots method’
(a) By factorisation
3
3
2
By using quadratic formula,
(b) By ‘ nth roots method’
3
3
= 1 𝑐𝑖𝑠( 0 + 𝑘 2 𝜋) {polar form}
1
3 𝑐𝑖𝑠(
𝑘 2 𝜋
3
2 𝜋
3
4 𝜋
3
∴ 𝑧 = (cos 0 + 𝑖𝑠𝑖𝑛 0 ) , ( 𝑐𝑜𝑠
2 𝜋
3
2 𝜋
3
(cos
4 𝜋
3
4 𝜋
3
1
2
√
3
2
1
2
√
3
2
Example 1 7 :
Find the cube roots of 8i.
The cube roots of 8i are the solutions of 𝑧
3
3
1
3 𝑐𝑖𝑠 (
𝜋+𝑘 4 𝜋
6
Hence, 𝑧 = 2 𝑐𝑖𝑠 (
𝜋
6
5 𝜋
6
9 𝜋
6
