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COMPLEX NUMBERS
Assumed Knowledge from Extension 1
Make sure you are familiar with the following before proceeding
• Solving Quadratic Equations
• Absolute Values
• Preliminary Trigonometry
• Locus
• Circle Geometry
Where are we up to?
• Complex Numbers and Quadratic Equations
• The Argand Diagram
• Modulus-Argument Form
• Vector Representation
• Locus Problems
• Roots of Unity
1. Introduction to complex numbers
1.1 Number Systems
Throughout your progression in primary to high school mathematics you were introduced to
different number systems which eventually led to the commonly used real number system:
• Natural Numbers N = {0, 1, 2, 3, ....} - the problem with this system is that the operation
of say 1 − 2 gives no solution in the natural number set, so we introduce the concept of
negative numbers
• Integers Z = {...., −2, −1, 0, 1, 2, ....} - the problem with this system is that the operation
of say 1 ÷ 2 gives no solution in the integer set, so we introduce the concept of fractions
o
nm
: m ∈ Z, n ∈ Z - the problem with this system is that some
• Rational Numbers Q =
n
numbers cannot be written as a fraction (e.g. π), so we introduce the concept of irrational
numbers
• Real Numbers R = Q ∪ Q - the problem with this system is that the operation of say
√
−1 has no solution in the real number set, so we introduce the concept of IMAGINARY
NUMBERS
• Complex Numbers C - is the union of real numbers and imaginary numbers. We will now
look at this number system in detail
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1.2 Imaginary numbers
These are based on the following definition of the imaginary unit i where:
i2 = −1
In other words we are proposing there exists this unusual ’thing’ called i such that when it is
squared we get negative one. All imaginary numbers exist in the form iy where y is a real number. In other words, any real number multiplied by i becomes an imaginary number e.g. 2i, 2i , 7i.
They have no real physical meaning, as they are ’imaginary’.
1.3 Complex numbers
These are the UNION (combination) of real numbers and imaginary numbers. Therefore, all
complex numbers (commonly called z) can be written in the form
z = x + iy
where:
x is the REAL PART of z or Re(z)
y is the IMAGINARY PART z or Im(z)
If Re(z) = 0, then we say that z is purely imaginary (real part ’does not exist’)
If Im(z) = 0, then we say that z is purely real (imaginary part ’does not exist’)
The complex number system has no ordering. In other words, we cannot say one complex number
is greater than or less than another complex number.
1.4 Arithmetic involving complex numbers
Addition, subtraction and multiplication on the set of complex numbers are calculated using
the standard arithmetic procedures we are used to doing in real numbers and algebra.
e.g. Addition: (1 + 2i) + (2 + 6i) = ..............................
e.g. Subtraction: (1 + 2i) − (2 + 6i) = ..............................
e.g. Multiplication: (1 + 2i)(2 + 6i) = ............................................................................................
Notice that the final answer is often written with their respective real and imaginary parts in
the form x + iy. Division of complex numbers involves a bit more effort to get it in the form x + iy,
so before we look at division we need to first look at complex conjugates.
1.5 Complex conjugates
If z = x + iy, then we define the complex CONJUGATE of z as z = x − iy. This allows us
to derive a very useful property. Note that:
zz = (x + iy)(x − iy)
= ............................
= ............................
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Note that this number is purely real (i.e. no imaginary parts) and is always non-negative.
e.g. If z = 3 + 2i then find z, hence find the value of zz.
1.6 Division involving complex numbers
The idea of complex conjugates is needed to divide two complex numbers and simplify them
to the form x + iy. The first step is to make the denominator REAL (sometimes called ’realising’ the denominator). In order to do this we needed to multiply the numerator and denominator
by the conjugate of the denominator, because this converts the denominator into a real number
(as shown in the previous part) rather than a non-real number. It is very similar to the technique
of rationalising the denominator.
e.g. Simplify
3 + 2i
in the form x + iy.
1 + 4i
3 + 2i 1 − 4i
3 + 2i
=
×
1 + 4i
1 + 4i 1 − 4i
= ......................................................
= ......................................................
= ......................................................
BRIEF EXERCISES - express the following in the form x + iy
1. (3 + 2i) + (4 − 5i)
2. (2 + 3i) − (3 − 2i)
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3. (1 + i)(2 + i)
4.
3 − 4i
1 + 2i
2. Quadratic Equations Involving Complex Numbers
2.1 Equality of complex numbers
Suppose z = a + ib and say ω = c + id. If z = ω then we can say that a = c and b = d.
In other words Re(z) = Re(ω) (the real parts are equal) and Im(z) = Im(ω) (the imaginary parts
are equal).
e.g. If (2 + i)2 = x + iy, find the values of x and y.
e.g. If
1+i
= x + iy, find the values of x and y.
1−i
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√
3
1
i = cos θ + i sin θ for 0 < θ < π2 , find the value of θ.
e.g. If +
2
2
(Note that we are using radian measures of angles rather than degrees. The conversion from
degrees to radians uses π radians = 180 degrees. You will be shown the reason for this in more
detail in the latter part of the Extension 1 course, but for now just accept the result.)
2.2 Square root of complex numbers
√
Brief Exercise - Suppose a + ib = x + iy, find an expression for a and find another expression for b in terms of x and y.
√
Suppose we want to find say 3 + 4i in the form x + iy (keep in mind that x and y are real
numbers). Using the same method employed above (here a = 3 and b = 4), notice that we have
a set of simultaneous equations to solve for x and y. Since one of the equations
√ is quadratic, we
expect TWO pairs of solutions which means there are TWO ways to write 3 + 4i in the form
x + iy. We solve this problem as follows:
√
3 + 4i = x + iy
⇒ 3 + 4i = (x + iy)2
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BRIEF EXERCISES - express the following in the form x + iy
√
1. 24 + 10i
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2.
√
15 + 8i
3.
√
−21 + 20i
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4.
√
i
2.3 Solving quadratic equations with real coefficients
Many quadratic equations you may have encountered may have had no real solution because
you end up square rooting negative numbers. This does not necessarily mean that it is impossible
to have a solution. It just means that the solutions that exist are not in the real number system.
Using the knowledge of complex numbers, it is now possible to solve EVERY quadratic equation.
In other words, whilst the solutions of a quadratic equation may not exist in the real number
system R, they will always exist in the complex number system C.
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e.g. Solve x2 + 2 = 0 if
a) x ∈ R
x2 = −2 gives no real solutions for x
b) x ∈ C
x2 = −2
√
x = ± −2
√
x=i± 2
Note that there are always two solutions in the complex number system. We can now solve equations which were otherwise unsolvable in the real number system.
e.g. Solve x2 + 2x + 5 = 0 for x ∈ C
e.g. Solve 2z 2 − 2z + 1 = 0 for z ∈ C
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2.4 Solving quadratic equations with non-real coefficients
The complex number system also allows us to invent quadratic equations with complex coefficients.
These are solvable using the same methods as solving any real quadratic equation. However, they
often require a bit more effort to simplify to the form x + iy because we need to evaluate the
square root of a complex number.
e.g. Solve 3z 2 − 3z − i = 0 for z ∈ C
p
(−3)2 − 4(3)(−i)
z=
2(3)
√
3 ± 9 + 12i
=
6
3±
√
The 9 + 12i needs to be simplified so that z can be written in the form x + iy. We employ the
standard procedure for finding the square root of the complex number.
Let
√
9 + 12i = x + iy
⇒ 9 + 12i = (x + iy)2
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BRIEF EXERCISES - solve the following quadratic equations expressing your solutions in
the form x + iy
1. z 2 + 1 = 0
2. u2 − 2u + 3 = 0
3. 3y 2 + 3y + 1 = 0
4. x2 − (4 + i)x + 9 + 7i = 0
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5. a2 − (1 − 4i)a − 5 + i = 0
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