Complex numbers
Introduction
We started the study of number systems with the set of natural numbers ℕ, and then the
number zero was included to form the system of whole numbers. Thereafter, negative
numbers were defined. This extended our number system to integers ℤ. To solve the
problems of the type 𝑞𝑞𝑞𝑞 = 𝑝𝑝 we included rational numbers ℚ in the system of integers.
The system of rational numbers was extended further to irrational numbers as all
lengths cannot be measured in terms of lengths expressed in rational numbers to solve
problems like 𝑥𝑥 2 = 2 (has no rational solutions but it has real solutions). The rational
and irrational numbers taken together are termed real numbers ℝ. These are numbers
that exist and can be represented on the number line, but the system of real numbers is
not sufficient to solve all algebraic equations. There are no real numbers which satisfy
the equation 𝑥𝑥 2 + 1 = 0 as you might know (no solution). In order to solve such
equations, i.e., to find square roots of negative numbers, we need to extend the system
of real numbers to a new system of numbers known as complex numbers ℂ.
The equation 𝑥𝑥 2 + 1 = 0 has no real solutions, because for any real number 𝑥𝑥 the
square 𝑥𝑥 2 is non-negative, and so 𝑥𝑥 2 + 1 can never be less than 1. In other words, we
can say that there is no real numbers whose square is −1. In order to solve
such equations, let us 𝑖𝑖magine that there exist a number 𝑖𝑖 which is equal to √−1
(√−1 = 𝑖𝑖). In 1748, a great mathematician, Leonard Euler named a number 𝑖𝑖 as 𝐼𝐼ota
whose square is −1. This Iota or 𝑖𝑖 is defined as an imaginary unit. With the introduction
of the new symbol 𝑖𝑖, we can interpret the square root of a negative number as a
product of a real number with 𝑖𝑖. Therefore, we can denote the solution of 𝑥𝑥 2 + 1 = 0 as
𝑥𝑥 = ± √−1 = ±𝑖𝑖.
From high school in SA, using the discriminant, we identify the three different cases of
quadratic equations as follows:
1. Positive discriminant: 𝑏𝑏 − 4𝑎𝑎𝑎𝑎 > 0, two real roots, eg 𝑥𝑥 2 + 3𝑥𝑥 + 2 = 0
2. Zero discriminant: 𝑏𝑏 − 4𝑎𝑎𝑎𝑎 = 0, one repeated real root, eg 𝑥𝑥 2 + 4𝑥𝑥 + 4 = 0.
1
3. Negative discriminant: 𝑏𝑏 2 − 4𝑎𝑎𝑎𝑎 < 0, complex conjugate roots. 𝑎𝑎 − 𝑏𝑏𝑏𝑏 and 𝑎𝑎 + 𝑏𝑏𝑏𝑏
are conjugate pairs.
It very useful to assume that there is a number 𝑖𝑖 for which one has 𝑖𝑖 2 = −1. Any
complex number is then an expression of the form 𝑎𝑎 + 𝑏𝑏𝑏𝑏, where 𝑎𝑎 and 𝑏𝑏 are real
2
2
numbers. For example, 2 = +2 = = �√2� = 2 + 0𝑖𝑖 and 2𝑖𝑖 = 0 + 2𝑖𝑖. The number 𝑎𝑎 is
1
called the real part of 𝑎𝑎 + 𝑏𝑏𝑏𝑏, and 𝑏𝑏 is called the imaginary part. Traditionally the letters 𝑧𝑧
and 𝑤𝑤 are used to stand for complex numbers. Since any complex number is specified
by two real numbers, one can visualise them by plotting a point with coordinates (𝑎𝑎, 𝑏𝑏)
in the plane for a complex number 𝑎𝑎 + 𝑏𝑏𝑏𝑏. The plane in which one plot these complex
numbers is called the Complex plane, or Argand plane (named after Swiss
mathematician). NB: For real numbers, we have the Cartesian plane from Rene
Descartes who was French.
Some real polynomials, like 𝑧𝑧 2 = −9, cannot be factored completely into degree 1
real polynomials, but do factor into degree 1 complex polynomials:
(𝑧𝑧 + 3𝑖𝑖)(𝑧𝑧 − 3𝑖𝑖) = 0 where 𝑧𝑧 = ±√−9 = ±√9 × −1 = ±√9 × √−1 = 0 ± 3𝑖𝑖.
(for 𝑥𝑥 2 − 9 = 0; 𝑤𝑤𝑤𝑤 𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘 (𝑥𝑥 − 3)(𝑥𝑥 + 3) = 0 𝑡𝑡ℎ𝑒𝑒𝑒𝑒 𝑥𝑥 = ±3 + 0𝑖𝑖) In fact, every complex
polynomial factors completely into degree 1 complex polynomials. This implies the
Fundamental theorem of algebra:
Every degree 𝑛𝑛 complex polynomial 𝑓𝑓(𝑧𝑧) has exactly 𝑛𝑛 complex roots, if counted
with multiplicity.
For real polynomials, the theorem sometimes does not hold eg, 𝑥𝑥 3 − 8 = 0. What is 𝑥𝑥?
𝑥𝑥 = 2. But according to the theorem, we expect to hve THREE REAL ROOTS. For real
polynomials, the non-real roots can be paired off with their complex conjugates.
Complex conjugate root theorem states that if 𝑃𝑃 is a polynomial in one variable with
real coefficients, and 𝑎𝑎 + 𝑏𝑏𝑏𝑏 is a root of 𝑃𝑃 with 𝑎𝑎 and 𝑏𝑏 real numbers, then its complex
conjugate 𝑎𝑎 − 𝑏𝑏𝑏𝑏 is also a root of 𝑃𝑃. 1 − 𝑖𝑖 also 1 + 𝑖𝑖 is a root. Also difference of two
squares: (𝑥𝑥 − 2)(𝑥𝑥 + 2) = 𝑥𝑥 2 + 2𝑥𝑥 − 2𝑥𝑥 − 4 = 𝑥𝑥 2 − 4
2
The simplest explanation is that complex roots come in conjugate pairs (by the way this
is only true for real polynomials, not those with complex coefficients) is that these pairs
are needed to ensure we don’t get complex values for our coefficients/constants of our
polynomial when we multiply through. E.g. (𝑎𝑎 + 𝑖𝑖𝑖𝑖)(𝑎𝑎 − 𝑖𝑖𝑖𝑖) = 𝑎𝑎2 + 𝑎𝑎𝑎𝑎𝑎𝑎 − 𝑎𝑎𝑎𝑎𝑎𝑎 − 𝑖𝑖 2 𝑏𝑏 2 =
𝑎𝑎2 + 𝑏𝑏 2 .
It follows from this (and the fundamental theorem of algebra) that, if the degree of a real
polynomial is odd, it must have at least one real root.
This can be proved as follows.
•
Since non-real complex roots come in conjugate pairs, there are an even number
of them;
•
But a polynomial of odd degree has an odd number of roots;
•
Therefore, some of them must be real.
For example, 𝑥𝑥 3 − 8 = 0 has one real root (𝑥𝑥 = 2) and two complex roots to give a total
of three roots/solutions. Also 𝑥𝑥 3 + 8 = 0 has at least one real solution, which is 𝑥𝑥 = −2.
The other two are non-real (complex).
Consider another quadratic equation:
By the quadratic formula, a=1; b=-6 and c=13
𝑥𝑥 =
−(−6) ± √36 − 52 6 ± √−16 6 ± √−1 × 16 6 ± 4√−1
=
=
=
= 3 + 2𝑖𝑖 𝑜𝑜𝑜𝑜 3 − 2𝑖𝑖
2
2
2
2
3
It has TWO non-real (complex) roots shown above because it is even. But, if 𝑎𝑎 = 0 and
𝑏𝑏 ≠ 0, then the root becomes 𝑏𝑏𝑏𝑏 which is a purely imaginary complex number, 2𝑖𝑖.
If 𝑎𝑎 ≠ 0 and 𝑏𝑏 = 0 then the root becomes 𝑎𝑎 which is a real number, 3. Hence the
real numbers are particular cases of complex numbers. ℝ ⊆ ℂ.
We know that the roots of 𝑥𝑥 2 − 6𝑥𝑥 + 25 = 0 are 3 ± 4𝑖𝑖. This can be written as 3 + 4𝑖𝑖 or
3 − 4𝑖𝑖. Can you show it?
The polynomial 𝑥𝑥 3 − 7𝑥𝑥 2 + 41𝑥𝑥 − 87 has roots 3, 2 + 5𝑖𝑖 , 2 − 5𝑖𝑖. Can you show it?
Thus can be factored as ( 𝑥𝑥 − 3 ) ( 𝑥𝑥 − 2 − 5 𝑖𝑖 ) ( 𝑥𝑥 − 2 + 5 𝑖𝑖 ) .
In computing the product of the last two factors, the imaginary parts cancel, and we get
2
( 𝑥𝑥 − 3 ) ( 𝑥𝑥 2 − 4 𝑥𝑥 + 29 ) . 5𝑖𝑖 × 5𝑖𝑖 = 25𝑖𝑖 2 = 25�√−1� = 25(−1) = −25
Also, if 𝑧𝑧 = 𝑎𝑎 − 𝑏𝑏𝑏𝑏, then the complex conjugate is 𝑧𝑧̅ = 𝑎𝑎 + 𝑏𝑏𝑏𝑏. Now 𝑧𝑧̿ = 𝑎𝑎 − 𝑏𝑏𝑏𝑏.
Verify it.
Moreover,
��������
EG 𝑤𝑤 = 2 + 2𝑖𝑖 𝑎𝑎𝑎𝑎𝑎𝑎 𝑧𝑧 = 3 − 3𝑖𝑖 . Now 𝑤𝑤 + 𝑧𝑧 = 2 + 2𝑖𝑖 + 3 − 3𝑖𝑖 = 5 − 𝑖𝑖 𝑡𝑡ℎ𝑒𝑒𝑒𝑒 𝑤𝑤
+ 𝑧𝑧 = 5 + 𝑖𝑖
Also 𝑤𝑤
� = 2 − 2𝑖𝑖 𝑎𝑎𝑎𝑎𝑎𝑎 𝑧𝑧̅ = 3 + 3𝑖𝑖. 𝑁𝑁𝑁𝑁𝑁𝑁 𝑤𝑤
� + 𝑧𝑧̅ = 5 + 𝑖𝑖
The first one just says that if we conjugate twice we get back to what we started with
originally and hopefully this makes some sense. The remaining three just say we can
break up sum, differences, products and quotients into the individual pieces and then
conjugate.
4
Properties of complex numbers
2
Also 𝑧𝑧𝑧𝑧̅ = |𝑧𝑧|2 = �√𝑎𝑎2 + 𝑏𝑏 2 � = 𝑎𝑎2 + 𝑏𝑏 2 shown below.
Task
1
1
1
𝑖𝑖
(iv) + 𝑖𝑖 then the conjugate is − .
2
2
2
2
Task
If 𝑧𝑧 = 2 + 3𝑖𝑖 and 𝑤𝑤 = 4 − 9𝑖𝑖, find 𝑧𝑧𝑧𝑧.
Division of complex numbers
5
(rationalising the denominator)
(a) 60-18i
(b) 60 + 78𝑖𝑖 + 24𝑖𝑖 2 = 36 + 76𝑖𝑖
(c)
(d)
5𝑖𝑖(1+7𝑖𝑖)
(1−7𝑖𝑖)(1+7𝑖𝑖)
=
5𝑖𝑖−35
1+49
=
−35+5𝑖𝑖
50
=−
7
10
+
1
10
6
𝑖𝑖.
Higher powers of 𝑖𝑖
We know that 𝑖𝑖 = √−1
Thus, we find that any higher powers of 𝑖𝑖 can be expressed in terms of one of four
values 𝑖𝑖, – 1, – 𝑖𝑖, 1.
Example
7
Example
Equal complex numbers
If two complex numbers are equal, then the real and imaginary parts are also equal.
We call this equating like parts.
Task
Solutions
1 − 6𝑖𝑖 − 9 − 5𝑥𝑥 − 5𝑦𝑦𝑦𝑦 = 𝑥𝑥 + 𝑦𝑦𝑦𝑦
4
Comparing Real parts: 1 − 9 − 5𝑥𝑥 = 𝑥𝑥 𝑡𝑡ℎ𝑒𝑒𝑒𝑒 8 = 6𝑥𝑥 which gives 𝑥𝑥 = .
3
Imaginary parts (terms in 𝑖𝑖): −6𝑖𝑖 − 5𝑦𝑦𝑖𝑖 = 𝑦𝑦𝑖𝑖 𝑡𝑡ℎ𝑒𝑒𝑒𝑒 − 6 = 6𝑦𝑦 𝑤𝑤ℎ𝑖𝑖𝑖𝑖ℎ 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑦𝑦 = −1.
Task 2
Task
where 𝑧𝑧 = 𝑥𝑥 + 𝑦𝑦𝑦𝑦.
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Answer:
𝑧𝑧 2 − 2𝑧𝑧 + 3 − 4𝑖𝑖 − 6 − 3𝑖𝑖 = 0
𝑧𝑧 2 − 2𝑧𝑧 − (3 + 7𝑖𝑖) = 0
Use the quadratic formula where 𝑎𝑎 = 1, 𝑏𝑏 = −2 𝑎𝑎𝑎𝑎𝑎𝑎 𝑐𝑐 = −(3 + 7𝑖𝑖)
𝑧𝑧 =
To complete this later!!!
2 ± �4 + 4 × 1 × (3 + 7𝑖𝑖)
2
𝑧𝑧 =
2 ± √16 + 28𝑖𝑖
2
𝑧𝑧 =
2 − 𝑖𝑖 3 − 3𝑖𝑖
×
3 + 3𝑖𝑖 3 − 3𝑖𝑖
Task
Answer:
6 − 6𝑖𝑖 − 3𝑖𝑖 + 3𝑖𝑖 2
𝑧𝑧 =
9 − 9𝑖𝑖 2
𝑧𝑧 =
Task
𝑧𝑧 =
6 − 9𝑖𝑖 − 3
9+9
1
1
3 − 9𝑖𝑖 1 1
= − 𝑖𝑖. 𝑁𝑁𝑁𝑁𝑁𝑁 𝑖𝑖𝑖𝑖 𝑦𝑦𝑦𝑦𝑦𝑦 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 𝑥𝑥 = 𝑎𝑎𝑎𝑎𝑎𝑎 𝑦𝑦 = − .
18
6 2
6
2
9
Solution
Substitute the root: (1 + 2𝑖𝑖)2 + (𝑝𝑝 + 5𝑖𝑖)(1 + 2𝑖𝑖) + 𝑞𝑞(2 − 𝑖𝑖) = 0
1 + 4𝑖𝑖 − 4 + 𝑝𝑝 + 2𝑝𝑝𝑝𝑝 + 5𝑖𝑖 − 10 + 2𝑞𝑞 − 𝑞𝑞𝑞𝑞 = 0 + 0𝑖𝑖
Equating real parts: −13 + 𝑝𝑝 + 2𝑞𝑞 = 0 and imaginary parts: 9 + 2𝑝𝑝 − 𝑞𝑞 = 0.
𝑝𝑝 = 13 − 2𝑞𝑞. Substitute in (2): 9 + 2(13 − 2𝑞𝑞) − 𝑞𝑞 = 0; −14 − 4𝑞𝑞 − 𝑞𝑞 = −9
−5𝑞𝑞 = −9 + 14
𝑞𝑞 = −1. Subst to get 𝑝𝑝: 𝑝𝑝 = 13 − 2(−1) = 15
(b) 𝑧𝑧 = 1 − 2𝑖𝑖 is also a solution.
Task
Answer: working in reverse for the solution of a quadratic equation, we get:
if 𝑧𝑧 = 3 + √6𝑖𝑖 is a root and also 𝑧𝑧 = 3 − √6𝑖𝑖.
Thus 𝑧𝑧 − (3 + √6𝑖𝑖) = 0 and 𝑧𝑧 − (3 − √6𝑖𝑖) = 0
Then �𝑧𝑧 − �3 + √6𝑖𝑖�� � 𝑧𝑧 − �3 − √6𝑖𝑖�� = 0
Use FOIL 𝑧𝑧 2 − 3𝑧𝑧 + √6𝑖𝑖𝑖𝑖 − 3𝑧𝑧 − √6𝑖𝑖𝑖𝑖 + 9 + 6 = 0
𝑧𝑧 2 − 6𝑧𝑧 + 15 = 0.
GEOMETRICAL REPRESENTATION OF COMPLEX NUMBERS
Argand diagram is what we call the plane that will allow us to plot complex numbers. It
is named after the Swiss mathematician Jean Argand (1768 –1822). Using the 𝑥𝑥-axis as
the real axis, and the y-axis as the imaginary axis, the ordered pairs (𝑎𝑎,b) reflect
complex numbers of the form 𝑎𝑎 + 𝑏𝑏𝑏𝑏.
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The plane of two axes representing complex numbers as points is called the complex
plane or Argand Plane. The diagram which represents complex number in the Argand
Plane is called Argand Diagram.
If 𝑧𝑧 = 𝑎𝑎 + 𝑏𝑏𝑏𝑏, then 𝑅𝑅𝑅𝑅(𝑧𝑧) = 𝑎𝑎 and 𝐼𝐼𝐼𝐼(𝑧𝑧) = 𝑏𝑏.
Absolute value
The absolute value (or magnitude or modulus) |𝑧𝑧| of a complex number 𝑧𝑧 = 𝑥𝑥 + 𝑖𝑖𝑖𝑖 is its
distance to the origin:
This is shown below:
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From the image above, we can see that |𝑧𝑧| = √𝑎𝑎2 + 𝑏𝑏 2 is nothing more than the length
of the line segment (vector) that we are using to represent the complex number 𝑧𝑧 = 𝑎𝑎 +
𝑏𝑏𝑏𝑏. |𝑧𝑧| = �(−1)2 + 1 = √2. This interpretation also tells us that the inequality |𝑧𝑧1 | < |𝑧𝑧2 |
means that 𝑧𝑧1 is closer to the origin (in the complex plane) than 𝑧𝑧2 is. But this done not
give us an indication of the angle of inclination of the 𝑧𝑧1 and 𝑧𝑧2 . We now turn to the
angle 𝜃𝜃, called the argument.
The angle 𝜃𝜃 is called the argument of the complex number 𝑧𝑧. Notation: arg 𝑧𝑧 = 𝜃𝜃. The
argument is defined in an ambiguous way: it is only defined up to a multiple of 2𝜋𝜋 (360o)
radians. For example, the argument of −1 + 0𝑖𝑖 could be 𝜋𝜋 (1800), or −𝜋𝜋, or 3𝜋𝜋, etc. In
general one says arg(−1) = tan−1 (−1) = 𝜋𝜋 + 𝑘𝑘𝑘𝑘, where 𝑘𝑘 may be any integer.
Now let’s look at just one point in detail 𝑃𝑃(𝑥𝑥, 𝑦𝑦) = 𝑥𝑥 + 𝑖𝑖𝑖𝑖.
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𝑧𝑧1 = 𝑥𝑥 + 𝑖𝑖𝑖𝑖 = 𝑟𝑟 cos 𝜃𝜃 + 𝑟𝑟 sin 𝜃𝜃 = 𝑟𝑟( cos 𝜃𝜃 + 𝑖𝑖 sin 𝜃𝜃). In the 4th quadrant, we know cos is
+ve and sin is –ve, so 𝑧𝑧2 = 𝑟𝑟(cos 𝜃𝜃 − 𝑖𝑖 sin 𝜃𝜃). Hence z1 and z2 are conjugate pairs and
we write z2 as 𝑧𝑧̅ = 𝑟𝑟(cos 𝜃𝜃 − 𝑖𝑖 sin 𝜃𝜃).
|𝑧𝑧| = 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠(𝑟𝑟 2 (cos 𝑥𝑥)^2 + 𝑟𝑟^2 (sin 𝑥𝑥)^2 ) = 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 (𝑟𝑟 2 (cos2 𝑥𝑥 + sin2 𝑥𝑥)) = �𝑟𝑟 2 = 𝑟𝑟
𝑥𝑥
cos 𝜃𝜃 = then 𝑥𝑥 = 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝜃𝜃 and 𝑦𝑦 = 𝑟𝑟 sin 𝜃𝜃
𝑟𝑟
The Greek letters 𝜃𝜃 (theta) and 𝜑𝜑 (phi) are often used for angles. In the calculus it is
convenient to measure angles in radians instead of degrees. An angle 𝜃𝜃 in radians is
defined as the length of the arc of the angle on a circle of radius one. Since a circle of
radius one has circumference 2𝜋𝜋. Thus 360 degrees = 2𝜋𝜋 radians.
The radian measure
The convention for Complex numbers is to use radians as the measure for angles. But
what are the radians?
An arc of a circle is a "portion" of the circumference of the circle. The length of an arc
is simply the length of its "portion" of the circumference. Actually, the circumference
itself can be considered an arc length.
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The length of an arc (or arc length) is traditionally symbolized by 𝑠𝑠. In the diagram
above, it can be said that arc AB subtends angle 𝜃𝜃.
If the length of an arc of a circle, 𝑠𝑠, (think of straightening it out), is the same as the
length of the circle's radius, 𝑟𝑟, a unique situation occurs. The angle, 𝜃𝜃, created by this
situation is called a radian.
A radian is the measure of an angle 𝜃𝜃 that, when drawn as a central angle, subtends
an arc whose length equals the length of the radius of the circle.
The measure of 𝜃𝜃 = 1 radian.
When radius r = arc length r, the angle 𝜃𝜃 measures 1 radian.
Radian measure is another way of expressing the measure (size) of an angle. It is
considered to be a "pure" measure since it is based upon the radius of the circle being
wrapped along the circumference. While at first it may seem easier to work with
degrees, you will find that there is a mathematical simplicity to the use of radians. So,
now we have two methods of expressing the measure of an angle: in degrees and in
radians.
𝑠𝑠
If the arc length 𝑠𝑠 is twice the radius 𝑟𝑟, the 𝜃𝜃 = 2 radians, and so on. Therefore, 𝜃𝜃 = .
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𝑟𝑟
NB: Radians, being a ratio of two lengths, are not units OR do not have units of
measurement.
Example
Convert 50 degrees to radians.
Answer
Example
Convert
Answer
𝜋𝜋
6
to degrees.
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Also, we know that the arc length, 𝑠𝑠 = 𝑟𝑟𝑟𝑟. When the arc length equals an entire
circumference (2𝜋𝜋𝜋𝜋), we can use 𝑠𝑠 = 𝑟𝑟𝑟𝑟 to get
2𝜋𝜋𝜋𝜋 = 𝑟𝑟𝑟𝑟
2𝜋𝜋 = 𝜃𝜃
This verifies what we already know to be true, that 2𝜋𝜋 𝑟𝑟𝑟𝑟𝑟𝑟 = 3600 .
Task
How long is the arc subtended by an angle of 3150 on a circle of radius 20 cm?
Task
Answers
NOTE: By convention if the units are not specified they are RADIANS.
NB: Radians are real numbers which do not carry units, hence we can freely do
𝜋𝜋
operations with radian measure. For example, × sin 𝜋𝜋 is valid, while 450 × sin 1800 is
√2
2
4
𝜋𝜋
ridiculous. From now, cos −1 � � = .
Polar form
4
Let’s now take a look at the first alternate form for a complex number. If we think of the
non-zero complex number 𝑧𝑧 = 𝑎𝑎 + 𝑏𝑏𝑏𝑏 as the point (𝑎𝑎, 𝑏𝑏) in the 𝑥𝑥𝑥𝑥-plane we also know
that we can represent this point by the polar coordinates (𝒓𝒓, 𝜽𝜽), where 𝑟𝑟 is the distance
of the point from the origin and 𝜃𝜃 is the angle, in radians, from the positive 𝑥𝑥-axis to the
ray (line segment) connecting the origin to the point (anticlockwise).
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Task
For a given complex number 𝑧𝑧 pick any of the possible values of the argument, say 𝜃𝜃. If
you now increase the value of 𝜃𝜃, which is really just increasing the angle that the point
makes with the positive 𝑥𝑥 −axis, you are rotating the point about the origin in a counter-
clockwise manner. Since it takes 2𝜋𝜋 radians to make one complete revolution you will
be back at your initial starting point when you reach 𝜃𝜃 + 2𝜋𝜋 and so have a new value of
the argument. See the figure below.
If you keep increasing the angle you will again be back at the starting point when you
reach 𝜃𝜃 + 4𝜋𝜋, which is again a new value of the argument. Continuing in this fashion we
can see that every time we reach a new value of the argument we will simply be adding
multiples of 2𝜋𝜋 onto the original value of the argument.
Likewise, if you start at 𝜃𝜃 and decrease the angle you will be rotating the point about the
origin in a clockwise manner and will return to your original starting point when you
reach 𝜃𝜃 − 2𝜋𝜋. Continuing in this fashion and we can again see that each new value of
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the argument will be found by subtracting a multiple of 2𝜋𝜋 from the original value of the
argument.
Task
Show that the conjugate is 𝑧𝑧̅ = 𝑟𝑟 (cos 𝜃𝜃 − sin 𝜃𝜃).
You can see here that a conjugate is a reflection of the point on the 𝑥𝑥-axis in the
complex plane.
Multiplication of two complex numbers (in polar)
Solutions
Task
Show division of complex numbers in polar form
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Task
Task
Calculate the modulus of
CIRCLES
2+𝑖𝑖
3−𝑖𝑖
in two ways. Also find the argument 𝜃𝜃.
Since |𝑧𝑧| defines the size of 𝑧𝑧 from the origin (modulus/magnitude), then
{𝑧𝑧: |𝑧𝑧 + 0 + 0𝑖𝑖| = 𝑟𝑟 } defines the collection of points that are all distance 𝑟𝑟 from the
origin. This describes a circle. Also {𝑧𝑧: |𝑧𝑧 + 𝑎𝑎 + 𝑏𝑏𝑏𝑏| = 𝑟𝑟, where 𝑟𝑟, 𝑎𝑎, 𝑏𝑏 ∈ 𝑅𝑅} is a circle,
radius 𝑟𝑟 with centre (−𝑎𝑎 − 𝑏𝑏𝑏𝑏). The centre has been translated from the origin – 𝑎𝑎 units
in the real direction, and – 𝑏𝑏 in the imaginary direction. For a circle 𝑥𝑥 2 + 𝑦𝑦 2 = 4 with
radius 2 and the origin is the centre. (𝑥𝑥 + 1)2 + (𝑦𝑦 + 2)2 = 1 this is a circle of radius 1
and centre is (-1,-2).
Example
Sketch |𝑧𝑧| = 1.
Answer
We know if 𝑧𝑧 = 𝑥𝑥 + 𝑖𝑖𝑖𝑖, then we remember from last lecture |𝑧𝑧| = �𝑥𝑥 2 + 𝑦𝑦 2 .
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Hence |𝑧𝑧| = 1 is the same as �𝑥𝑥 2 + 𝑦𝑦 2 = 1. Which becomes 𝑥𝑥 2 + 𝑦𝑦 2 = 12 . This is a
circle of radius 1 and centred at (0,0) as shown below.
Example
Sketch |𝑧𝑧 − 3 + 0𝑖𝑖| = 2.
Answer
|𝑥𝑥 + 𝑖𝑖𝑖𝑖 − 3| = |(𝑥𝑥 − 3) + 𝑖𝑖𝑖𝑖|
Thus (𝑥𝑥 − 3)2 + 𝑦𝑦 2 = (2)2 which is circle of centre (3,0) and radius 2.
Example
Sketch |𝑧𝑧 − 𝑖𝑖| = 1.
Answer
|𝑥𝑥 + 𝑖𝑖𝑖𝑖 − 𝑖𝑖| = 1
|𝑥𝑥 + (𝑦𝑦 − 1)𝑖𝑖| = 1
�𝑥𝑥 2 + (𝑦𝑦 − 1)2 = 1
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𝑥𝑥 2 + (𝑦𝑦 − 1)2 = 12
Task
Sketch the following
(a) |𝑧𝑧 − 2 + 3𝑖𝑖| = 3.
(b) |𝑧𝑧| < 3 all points inside the circle centre (0,0) and radius 3. Points on the
circumference are not part of the solution. Show it as a dotted line.
(c) 2 < |𝑧𝑧| < 3 represents the region between circles of radii 2 and 3.
(d) arg 𝑧𝑧 <
𝜋𝜋
2
SQUARE ROOTS OF COMPLEX NUMBERS
We can use complex numbers to compute square roots.
Example
Find the square root of −8 + 6𝑖𝑖.
Answer
Squaring both sides;
�−8 + 6𝑖𝑖 = 𝑎𝑎 + 𝑏𝑏𝑏𝑏
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After comparing real parts and imaginary parts.
𝑎𝑎2 = −9
𝑎𝑎 = ±√−9 = ±3𝑖𝑖
In other 𝑎𝑎 = 3𝑖𝑖 𝑜𝑜𝑜𝑜 𝑎𝑎 = −3𝑖𝑖. But there is a problem.
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The solutions are 1 + 3𝑖𝑖 and −1 − 3𝑖𝑖.
NB: √4 = 2 but 𝑥𝑥 2 − 4 = 0 gives us 𝑥𝑥 = ±2.
Task
If 𝑧𝑧 2 is known to be 3 − 4𝑖𝑖 what is 𝑧𝑧?
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