Complex Numbers
IE2107 – Engineering Mathematics II
A/P Su Rong
S1-B1b-59, Email: rsu@ntu.edu.sg
Complex Numbers > Learning Objectives
At the end of this lesson, you should be able to:
• Define the basics of complex numbers.
• Derive Euler’s Formula and De Moivre’s Formula.
• Derive the complex logarithm and its general power.
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EE2007: Engineering Mathematics II
Complex Numbers > Definition
A complex number 𝑧 is defined as 𝑧 = 𝑥 + 𝑖𝑦, where 𝑖 = −1. Geometrically, a complex
number is a point in the complex plane (or the Argand diagram) and can be considered as a
vector in the plane. A diagrammatic representation of the complex number is shown below.
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EE2007: Engineering Mathematics II
Complex Numbers > Definition
Here is an explanation of the equation depicted in the diagram.
𝑥 = 𝑟cos𝜃, and y = 𝑟sin𝜃
𝑟= 𝑧 = 𝑥 2 + 𝑦 2 = 𝑧ҧ = 𝑧 𝑧ҧ
𝑦
𝜃 = arg 𝑧 = arctan radians
𝑥
= Arg 𝑧 + 2𝑛π, 𝑛 = 0, ±1, ±2, …
Where, Arg 𝑧 is the principal value of arg 𝑧
and satisfies −π < Arg 𝑧 ≤ π
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EE2007: Engineering Mathematics II
Complex Numbers > Definition
Let us look at an example to understand the concept of complex numbers.
Example 1
i. Let 𝑧 = 1 + 𝑖
Then, 𝑟 = 𝑧 = 1 + 1 = 2
1
arg 𝑧 = arctan
1
π
= ± 2𝑛π, 𝑛 = 0,1,2, …
Type equation here.
4
𝜋
The principal value of the argument is .
4
−1 −π
ii. If 𝑧 = 1 − 𝑖, then arg 𝑧 = arctan = ± 2nπ, 𝑛 = 0,1,2, …
1 4
−𝜋
The principal value of the argument is .
4
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EE2007: Engineering Mathematics II
Complex Numbers > Euler’s Formula
From Euler’s formula, it can be found that:
𝑒 𝑖𝜃 = cos 𝜃 + 𝑖 sin 𝜃 and 𝑒 −𝑖𝜃 = cos 𝜃 − 𝑖 sin 𝜃
𝑒 𝑖𝜃 + 𝑒 −𝑖𝜃 𝑒 𝑖𝜃 − 𝑒 −𝑖𝜃
Thus, cos 𝜃 = and sin 𝜃 =
2 2𝑖
From Euler’s formula, 𝑒 𝑖𝜃 = cos 𝜃 + 𝑖 sin 𝜃, for any real value of 𝜃, the polar form of a
complex number can be written as 𝑧 = 𝑟𝑒 𝑖𝜃 = 𝑟∠𝜃.
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EE2007: Engineering Mathematics II
Complex Numbers > Euler’s Formula
Let us now look at some Algebraic Rules.
Let 𝑧1 = 𝑥1 + 𝑖𝑦1 = 𝑟1 ∠𝜃1 and 𝑧2 = 𝑥2 + 𝑖𝑦2 = 𝑟2 ∠𝜃2
Addition and subtraction 𝑧1 ± 𝑧2 = 𝑥1 ± 𝑥2 + 𝑖(𝑦1 ± 𝑦2 )
Multiplication 𝑧1 𝑧2 = 𝑥1 𝑥2 − 𝑦1 𝑦2 + 𝑖(𝑥1 𝑦2 + 𝑥2 𝑦1 )
𝑧1 𝑥1 𝑥2 + 𝑦1 𝑦2 𝑥2 𝑦1 − 𝑥1 𝑦2
Division = 2 2
+𝑖
𝑧2 𝑥 +𝑦 𝑥2 + 𝑦2
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EE2007: Engineering Mathematics II
Complex Numbers > Euler’s Formula
Let us now look at some Algebraic Rules.
Let 𝑧1 = 𝑥1 + 𝑖𝑦1 = 𝑟1 ∠𝜃1 and 𝑧2 = 𝑥2 + 𝑖𝑦2 = 𝑟2 ∠𝜃2
Addition
It isand more 𝑧convenient
subtraction
sometimes 1 ± 𝑧2 = 𝑥 ± 𝑥multiplication
to1 do 2 + 𝑖(𝑦1 ± 𝑦2 )
and division in the polar form.
𝑧1 𝑧2 = 𝑟1 𝑟2 ∠ 𝜃1 + 𝜃2 ,
Multiplication 𝑧1 𝑧2 = 𝑥1 𝑥2 − 𝑦1 𝑦2 + 𝑖(𝑥1 𝑦2 + 𝑥2 𝑦1 )
𝑧1 𝑟1 ∠𝜃1 𝑟1
= = ∠(𝜃1 −𝜃2 )
𝑧2 𝑟2 ∠𝜃2 𝑟2
𝑧1 𝑥1 𝑥2 + 𝑦1 𝑦2 𝑥2 𝑦1 − 𝑥1 𝑦2
Division = 2 2
+𝑖
𝑧2 𝑥 +𝑦 𝑥2 + 𝑦2
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EE2007: Engineering Mathematics II
Complex Numbers > Euler’s Formula
Let us now understand the complex conjugate of 𝑧 and its algebraic rules.
In the given equation 𝑧 = 𝑥 + 𝑖𝑦, the complex conjugate of 𝑧 is defined as 𝑧ҧ = 𝑥 − 𝑖𝑦.
Thus, it can be written as:
1 1
𝑅𝑒 𝑧 = 𝑧 + 𝑧ҧ , 𝐼𝑚 𝑧 = (𝑧 − 𝑧)ҧ
2 2𝑖
𝑧1 𝑧1 𝑧ഥ2
𝑧𝑧ҧ = 𝑥2 + 𝑦2 = 𝑧 2,
=
𝑧2 𝑧2 2
𝑧1 𝑧ഥ1
𝑧1 ± 𝑧2 = 𝑧ഥ1 ± 𝑧ഥ2 , 𝑧1 𝑧2 = 𝑧ഥ1 𝑧ഥ2 , =
𝑧2 𝑧ഥ2
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EE2007: Engineering Mathematics II
Complex Numbers > De Moivre’s Formula
Here is the derivation of the De Moivre’s formula.
Let 𝑧 = 𝑥 + 𝑖𝑦 = 𝑟 cos𝜃 + 𝑖sin𝜃 = 𝑟∠𝜃
𝑧 𝑛 = 𝑟 𝑛 cos𝜃 + 𝑖sin𝜃 𝑛
𝑧 𝑛 = 𝑧. 𝑧 … . 𝑧 = 𝑟. 𝑟 … . 𝑟∠ 𝜃 + 𝜃 + ⋯ + 𝜃 = 𝑟 𝑛 ∠(𝑛𝜃)
Then, for any integer 𝑛,
𝑛 𝑛 𝑛
= 𝑟 𝑛 cos 𝑛𝜃 + 𝑖 sin 𝑛𝜃
From the above equation, the De Moivre’s formula can be expressed as:
cos𝜃 + 𝑖sin𝜃 𝑛 = cos 𝑛𝜃 + 𝑖 sin 𝑛𝜃 which is useful in deriving certain trigonometric identities.
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EE2007: Engineering Mathematics II
Complex Numbers > De Moivre’s Formula
Let us look at a sample problem to understand the concept of complex numbers.
Sample Problem 1
Find identities for cos 2𝜃 and sin 2𝜃.
Solution:
cos 𝜃 + 𝑖 sin 𝜃 2 = cos 2 𝜃 − sin2 𝜃 + 2𝑖 cos 𝜃 sin 𝜃
Type
= cosequation
2𝜃 + 𝑖here.
sin 2𝜃
Therefore,
cos 2𝜃 = cos 2 𝜃 − sin2 𝜃 and sin 2𝜃 = 2 cos 𝜃 sin 𝜃
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EE2007: Engineering Mathematics II
Complex Numbers > De Moivre’s Formula
Let us look at another sample problem explaining the concept of complex numbers.
Sample Problem 2
Express cos 4 𝜃 in terms of multiples of 𝜃.
Solution:
Since 2 cos 𝜃 = 𝑒 𝑖𝜃 + 𝑒 −𝑖𝜃
Type equation here.
24 cos 4 𝜃 = (𝑒 𝑖𝜃 + −𝑖𝜃
𝑒 ) 4
= 𝑒 𝑖4𝜃 + 𝑒 −𝑖4𝜃 + 4 𝑒 𝑖2𝜃 + 𝑒 −𝑖2𝜃 + 6
= 2 cos 4𝜃 + 8 cos 2𝜃 + 6
1
⇒ cos 4 𝜃 = cos 4𝜃 + 4 cos 2𝜃 + 3
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EE2007: Engineering Mathematics II
Complex Numbers > Roots of Complex Numbers
Consider 𝑧 = 𝑤 𝑛 , 𝑛 = 1,2, …
For a given 𝑧 ≠ 0, the solution of 𝑤 in
the above equation is called the 𝑛𝑡ℎ
root of 𝑧 and is denoted by 𝑤 = 𝑛 𝑧.
First, 𝑧 = 𝑟∠ 𝜃 + 2𝑘𝜋 .
Next, let 𝑤 = 𝑅∠ϕ.
Then, 𝑧 = 𝑤 𝑛 gives
𝑟∠ 𝜃 + 2𝑘𝜋 = 𝑅𝑛 ∠(𝑛ϕ).
Thus, 𝑅 = 𝑛 𝑟, and
𝜃 + 2𝑘𝜋
ϕ= , k = 0,1, … , (𝑛 − 1).
𝑛
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EE2007: Engineering Mathematics II
Complex Numbers > Roots of Complex Numbers
Consider 𝑧 = 𝑤 𝑛 , 𝑛 = 1,2, …
For a given 𝑧 ≠ 0, the solution of 𝑤 in
To summarise,
the above equation is called the 𝑛𝑡ℎ
root of 𝑧 and is denoted by 𝑤 =𝑛 𝑛 𝑧. 𝑛 𝜃 + 2𝑘𝜋
𝑤𝑘 = 𝑧 = 𝑟∠ ,
𝑛
First, 𝑧 = 𝑟∠ 𝜃 + 2𝑘𝜋 .
𝑘 = 0,1, … . , (𝑛 − 1)
Next, let 𝑤 = 𝑅∠ϕ.
Geometrically, the entire set of roots lies at
Then, 𝑧 = 𝑤 𝑛 gives the vertices of a regular polygon of 𝑛 sides
𝑛
𝑛
inscribed in a circle of radius 𝑟.
𝑟∠ 𝜃 + 2𝑘𝜋 = 𝑅 ∠(𝑛ϕ).
Thus, 𝑅 = 𝑛 𝑟, and
𝜃 + 2𝑘𝜋
ϕ= , k = 0,1, … , (𝑛 − 1).
𝑛
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EE2007: Engineering Mathematics II
Complex Numbers > Roots of Complex Numbers
Let us look at an example to understand the concept of roots of complex numbers.
Example 2
1Τ 3
Let us find all values of (−8𝑖) 3 , that is, −8𝑖.
First,
−𝜋
−8𝑖 = 8∠ + 2𝑘𝜋 , 𝑘 = 0, ±1, ±2, …
2
Type equation here.
The desired roots are:
−𝜋 2𝑘𝜋
𝑤𝑘 = 2∠ + , 𝑘 = 0,1,2
6 3
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EE2007: Engineering Mathematics II
Complex Numbers > Roots of Complex Numbers
Let us look at an example to understand the concept of roots of complex numbers.
Example 2 (contd.)
The roots lie at the vertices of an
equilateral triangle, inscribed in
the circle 𝑧 = 2 and are equally
spaced around that circle every
Type equation here.
2𝜋Τ radians, starting with the
3
principal root
−𝜋
𝑤0 = 2∠ = 3 − 𝑖.
6
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EE2007: Engineering Mathematics II
Complex Numbers > Roots of Complex Numbers > Exponential Function
Let us now define the exponential function. The exponential function 𝑒 𝑧 is defined as:
∞
𝑧
1 𝑛
𝑒 = 𝑧 = 𝑒 𝑥 (cos 𝑦 + 𝑖 sin 𝑦).
𝑛!
𝑛=0
If 𝑥 = 0, then the Euler formula
becomes: 𝑒 𝑖𝑦 = cos 𝑦 + 𝑖 sin 𝑦.
Hence, the polar form of a complex
number may be written as
𝑧 = 𝑟(cos 𝜃 + 𝑖 sin 𝜃) = 𝑟𝑒 𝑖𝜃 . If 𝑧 = 𝑒 i𝑥 = cos 𝑥 + 𝑖 sin 𝑥 , then
1 1
sin 𝑥 = 𝑒 𝑖𝑥 − 𝑒 −𝑖𝑥 = (𝑧 − 𝑧),
ҧ
2𝑖 2𝑖
1 1
cos 𝑥 = 𝑒 𝑖𝑥 + 𝑒 −𝑖𝑥 = (𝑧 + 𝑧).
ҧ
2 2
It is also geometrically obvious that
𝜋
𝑒 𝑖𝜋 = −1, 𝑒 −𝑖 Τ2 = −𝑖 and 𝑒 −𝑖4𝜋 = 1.
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EE2007: Engineering Mathematics II
Complex Numbers > Complex Logarithm and General Power
The natural logarithm of 𝑧 = 𝑥 + 𝑖𝑦 is denoted by ln𝑧 and is defined as the inverse of
the exponential function.
Since, 𝑤 = ln𝑧 is defined for 𝑧 ≠ 0 by the relation 𝑒 𝑤 = 𝑧.
So, if 𝑧 = 𝑟𝑒 𝑖𝜃 , 𝑟 > 0, then ln𝑧 = ln𝑟 + 𝑖𝜃.
Note that the complex logarithm is infinitely many-valued.
The general power of a complex number, 𝑧 𝑐 , can be derived as follows:
Let 𝑦 = 𝑧 𝑐 , ⇒ ln𝑦 = 𝑐ln𝑧, ⇒ 𝑦 = 𝑧 𝑐 = 𝑒 𝑐ln𝑧 , 𝑧 ≠ 0.
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EE2007: Engineering Mathematics II
Complex Numbers > Complex Logarithm and General Power
Let us look at a sample problem to understand the concept of complex logarithm.
Sample Problem 3
i) Evaluate ln(3 − 4𝑖).
3
ii) Solve ln𝑧 = −2 − 𝑖.
2
Solution:
i) ln 3 − 4𝑖 = ln 3 − 4𝑖 + 𝑖 arg(3 − 4𝑖)
Type equation here.
= 1.609 − 𝑖 0.927 ± 2nπ , 𝑛 = 0,1, …
Principal value: When 𝑛 = 0
3 3
−2−2𝑖 3 3
ii) 𝑧 = 𝑒 = 𝑒 −2 𝑒 −𝑖2 = 𝑒 −2 cos − 𝑖 sin
2 2
= 0.010 − 𝑖 0.135
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EE2007: Engineering Mathematics II
Complex Numbers > Complex Logarithm and General Power
Here is another sample problem explaining the concept of complex logarithm.
Sample Problem 4
Find the principal value of (1 + 𝑖)𝑖 .
Solution:
Let 𝑦 = (1 + 𝑖)𝑖 . Then, ln𝑦 = 𝑖 ln(1 + 𝑖), or 𝑦 = 𝑒 𝑖 ln(1+𝑖)
Hence, (1 + 𝑖)𝑖 = 𝑒 𝑖 ln(1+𝑖)
Type equation here.
𝜋Τ +2𝑘𝜋
But, ln 1 + 𝑖 = ln 2𝑒 𝑖 4
= ln 2 + 𝑖 πൗ4 + 2𝑘π , 𝑘 = 0, ±1, …
and the principal value is when 𝑘 = 0.
𝜋
𝜋Τ ) − 4 +𝑖(ln 2)
Therefore, 𝑒 𝑖 ln(1+𝑖) = 𝑒 𝑖(ln 2+𝑖 4 =𝑒
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EE2007: Engineering Mathematics II
Summary
Complex Numbers > Summary
Key points discussed in this lesson:
• A complex number 𝑧 is defined as 𝑧 = 𝑥 + 𝑖𝑦, where 𝑖 = −1. Geometrically, a complex
number is a point in the complex plane (or the Argand diagram) and can be considered as a
vector in the plane.
• In the given complex number 𝑧 = 𝑥 + 𝑖𝑦, the complex conjugate of 𝑧 is defined as
𝑧ҧ = 𝑥 − 𝑖𝑦.
• From Euler’s Formula 𝑒 𝒊𝜃 = cos 𝜃 + 𝑖 sin 𝜃 , and 𝑒 −𝑖𝜃 = cos 𝜃 − 𝑖 sin 𝜃. Then,
𝑒 𝑖𝜃 + 𝑒 −𝑖𝜃 𝑒 𝑖𝜃 − 𝑒 −𝑖𝜃
cos 𝜃 = and sin 𝜃 = .
2 2𝑖
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EE2007: Engineering Mathematics II
Complex Numbers > Summary
Key points discussed in this lesson:
• For complex number 𝑧 = 𝑥 + 𝑖𝑦 = 𝑟 cos𝜃 + 𝑖sin𝜃 = 𝑟∠𝜃. The De Moivre’s formula is
given as: cos𝜃 + 𝑖sin𝜃 𝑛 = cos 𝑛𝜃 + 𝑖 sin 𝑛𝜃.
∞
1 𝑛
• The exponential function 𝑒𝑧 𝑧
is defined as: 𝑒 = 𝑧 = 𝑒 𝑥 (cos 𝑦 + 𝑖 sin 𝑦).
𝑛!
𝑛=0
• The natural logarithm of 𝑧 = 𝑥 + 𝑖𝑦 is denoted by ln𝑧 and is defined as the
inverse of the exponential function.
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EE2007: Engineering Mathematics II
