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2. Equations and
Inequalities
2.4 Complex Numbers
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1
Complex Numbers
Complex numbers are needed to find solutions of
equations that cannot be solved using only the set
numbers.
of real
The following chart illustrates several simple quadratic
equations and the types of numbers required for solutions.
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Complex Numbers
The solutions of the first three equations in the chart are in;
however, since squares of real numbers are never
negative, does not contain the solutions of x2 = –9.
To solve this equation, we need the complex number
system which contains both and numbers whose
squares are negative.
We begin by introducing the imaginary unit, denoted by i,
which has the following properties.
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Complex Numbers
Because its square is negative, the letter i does not
represent a real number. It is a new mathematical entity
that will enable us to obtain .
Since i, together with , is to be contained in , we must
consider products of the form bi for a real number b and
also expressions of the form a + bi for real numbers a
and b.
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Complex Numbers
The next chart provides definitions we shall use.
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Example 1 – Addition and multiplication of complex numbers
Express in the form a + bi, where a and b are real numbers:
(a) (3 + 4i) + (2 + 5i)
(b) (3 + 4i)(2 + 5i)
Solution:
(a) (3 + 4i) + (2 + 5i) = (3 + 2) + (4 + 5)i
= 5 + 9i
(b) (3 + 4i)(2 + 5i) = (3 + 4i)(2) + (3 + 4i)(5i)
= 6 + 8i + 15i + 20i2
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Example 1 – Solution
cont’d
= 6 + 23i + 20(–1)
= –14 + 23i
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Complex Numbers
The set of real numbers may be identified with the set of
complex numbers of the form a + 0i. It is also convenient to
denote the complex number 0 + bi by bi. Thus,
(a + 0i) + (0 + bi) = (a + 0) + (0 + b)i = a + bi.
Hence, we may regard a + bi as the sum of two complex
numbers a and bi (that is, a + 0i and 0 + bi).
For the complex number a + bi, we call a the real part and
b the imaginary part.
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Example 2 – Equality of complex numbers
Find the values of x and y, where x and y are real numbers:
(2x – 4) + 9i = 8 + 3yi
Solution:
We begin by equating the real parts and the imaginary
parts of each side of the equation:
2x – 4 = 8
and
9 = 3y
Since 2x – 4 = 8, 2x = 12 and x = 6. Since 9 = 3y, y = 3.
The values of x and y that make the complex numbers
equal are
x = 6 and y = 3
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Complex Numbers
With complex numbers, we are now able to solve an
equation such as x2 = –9. Specifically, since
(3i) (3i) = 32i 2 = 9(–1) = –9,
we see that one solution is 3i and another is –3i.
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Complex Numbers
In the next chart we define the difference of complex
numbers and multiplication of a complex number by a real
number.
If we are asked to write an expression in the form a + bi,
the form a – di is acceptable, since a – di = a + (–d)i.
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Example 3 – Operations with complex numbers
Express in the form a + bi, where a and b are real numbers:
(a) 4(2 + 5i) – (3 – 4i) (b) (4 – 3i)(2 + i)
(c) i(3 – 2i)2
(d) i 51
(e) i –13
Solution:
(a) 4(2 + 5i) – (3 – 4i) = 8 + 20i – 3 + 4i
= 5 + 24i
(b) (4 – 3i)(2 + i) = 8 – 6i + 4i – 3i 2
= 11 – 2i
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Example 3 – Solution
cont’d
(c) i(3 – 2i)2 = i(9 – 12i + 4i 2)
= i(5 – 12i)
= 5i – 12i 2
= 12 + 5i
(d) Taking successive powers of i, we obtain
i 1 = i, i 2 = –1, i 3 = –i, i 4 = 1,
and then the cycle starts over:
i 5 = i, i 6 = i 2 = –1, and so on.
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Example 3 – Solution
cont’d
In particular,
i51 = i48i3
= (i4)12i3
= (1)12i3
= (1)(–i)
= –i.
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Example 3 – Solution
cont’d
(e) In general, multiply i –a by i b, where a  b  a + 3 and b
is a multiple of 4 (so that i b = 1).
For i –13, choose b = 16.
i –13  i 16 = i 3
= –i
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Complex Numbers
The following concept has important uses in working with
complex numbers.
Since a – bi = a + (–bi), it follows that the conjugate of
a – bi is
a – (–bi) = a + bi.
Therefore, a + bi and a – bi are conjugates of each other.
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Complex Numbers
Illustration: Conjugates
Complex number
• 5 + 7i
• 5 – 7i
• 4i
•3
Conjugate
5 – 7i
5 + 7i
–4i
3
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Complex Numbers
The following two properties are consequences of the
definitions of the sum and the product of complex numbers.
Note that the sum and the product of a complex number
and its conjugate are real numbers.
Conjugates are useful for finding the multiplicative
inverse of a + bi, 1/(a + bi), or for simplifying the quotient
of two complex numbers.
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Complex Numbers
As illustrated in the next example, we may think of these
types of simplifications as merely rationalizing the
denominator, since we are multiplying the quotient by the
conjugate of the denominator divided by itself.
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Example 4 – Quotients of complex numbers
Express in the form a + bi, where a and b are real numbers
Solution:
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Example 4 – Solution
cont’d
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Complex Numbers
If p is a positive real number, then the equation x2 = –p has
solutions in . One solution is
, since
Similarly,
is also a solution.
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Complex Numbers
The definition of
in the next chart is motivated by
= –r for r > 0. When using this definition, take care
not to write
when
is intended.
The radical sign must be used with caution when the
radicand is negative.
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Complex Numbers
For example, the formula
, which holds for
positive real numbers, is not true when a and b are both
negative, as shown below:
But
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Complex Numbers
If only one of a or b is negative, then
In
general, we shall not apply laws of radicals if radicands are
negative.
Instead, we shall change the form of radicals before
performing any operations, as illustrated in the next
example.
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Example 5 – Working with square roots of negative numbers
Express in the form a + bi, where a and b are real numbers:
Solution:
First we use the definition
, and then we simplify:
= (5 – 3i)(–1 + 2i)
= –5 + 10i + 3i – 6i2
= –5 + 13i + 6
= 1 + 13i
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Example 7 – An equation with complex solutions
Solve the equation x3 – 1 = 0.
Solution:
Using the difference of two cubes factoring formula with
a = x and b = 1, we write x3 – 1 = 0 as
(x – 1)(x2 + x + 1) = 0.
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Example 7 – Solution
cont’d
Setting each factor equal to zero and solving the resulting
equations, we obtain the solutions
or, equivalently,
Since the number 1 is called the unit real number and the
given equation may be written as x3 = 1, we call these three
solutions the cube roots of unity.
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2.5
Other Types of Equations
Copyright © Cengage Learning. All rights reserved.
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Other Types of Equations
In applications it is often necessary to consider powers xk
with k > 2.
Some equations involve absolute values or radicals.
In this section we give examples of equations of these
types that can be solved using elementary methods.
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Example 1 – Solving an equation containing an absolute value
Solve the equation |x – 5| = 3.
Solution:
If a and b are real numbers with b > 0, then |a| = b if and
only if a = b or a = –b.
Hence, if |x – 5| = 3, then either
x–5=3
or
x – 5 = –3.
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Example 1 – Solution
cont’d
Solving for x gives us
x=5+3=8
or
x = 5 – 3 = 2.
Thus, the given equation has two solutions, 8 and 2.
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Other Types of Equations
If an equation is in factored form with zero on one side,
then we may obtain solutions by setting each factor equal
to zero.
For example, if p, q, and r are expressions in x and if
pqr = 0, then either p = 0, q = 0, or r = 0.
In the next example we factor by grouping terms.
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Example 2 – Solving an equation using grouping
Solve the equation x3 + 2x2 – x – 2 = 0.
Solution:
x3 + 2x2 – x – 2 = 0
x2(x + 2) – 1(x + 2) = 0
(x2 – 1)(x + 2) = 0
(x + 1)(x – 1)(x + 2) = 0
Given
group terms
factor out x + 2
factor x2 – 1
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Example 2 – Solution
x + 1 = 0,
x = – 1,
x – 1 = 0,
x = 1,
cont’d
x+2=0
x = –2
Zero factor theorem
solve for x
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Example 3 – Solving an equation containing rational exponents
Solve the equation x3/2 = x1/2.
Solution:
x3/2 = x1/2
x3/2 – x1/2 = 0
x1/2(x – 1) = 0
x1/2 = 0,
x–1=0
x = 0,
x=1
given
subtract x1/2
factor out x1/2
Zero factor theorem
solve for x
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Other Types of Equations
In Example 3 it would have been incorrect to divide both
sides of the equation x3/2 = x1/2 by x1/2, obtaining x = 1,
since the solution x = 0 would be lost.
In general, avoid dividing both sides of an equation by an
expression that contains variables—always factor instead.
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Example 4 – Solving an equation containing a radical
Solve the equation
Solution:
given
cube both sides
x2 – 1 = 8
x2 = 9
property of
add 1
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Example 4 – Solution
x=
cont’d
take the square root
Thus, the given equation has two solutions, 3 and –3.
Except to detect algebraic errors, a check is unnecessary,
since we raised both sides to an odd power.
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Other Types of Equations
In the last solution we used the phrase cube both sides of
In general, for the equation xm/n = a, where x is a real
number, we raise both sides to the power n/m (the
reciprocal of m/n) to solve for x.
If m is odd, we obtain x = an/m, but if m is even, we have
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Other Types of Equations
If n is even, extraneous solutions may occur—for example,
if x3/2 = –8, then
x = (–8)2/3 =
= (–2)2 = 4.
However, 4 is not a solution of x3/2 = –8 since 43/2 = 8
not –8.
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Other Types of Equations
Illustration: Solving xm/n = a, m odd, x real
Equation
Solution
• x3/1 = 64
x = 641/3 =
• x3/2 = 64
x = 642/3 =
=4
= 42 = 16
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Other Types of Equations
Illustration: Solving xm/n = a, m even, x real
Equation
Solution
• x4/1 = 16
• x2/3 = 16
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Other Types of Equations
In the next example, before we raise both sides of the
equation to a power, we isolate a radical—that is, we
consider an equivalent equation in which only the radical
appears on one side.
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Example 5 – Solving an equation containing a radical
Solve the equation
Solution:
3+
=x
given
=x–3
isolate the radical
= (x – 3)2
square both the sides
3x + 1 = x2 – 6x + 9
simplify
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Example 5 – Solution
x2 – 9x + 8 = 0
(x – 1)(x – 8) = 0
x – 1 = 0, x – 8 = 0
x = 1,
x=8
cont’d
subtract 3x + 1
factor
Zero factor theorem
solve for x
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Example 5 – Solution
cont’d
We raised both sides to an even power, so checks are
required.
Check:
x = 1 LS: 3 +
=3+
=3+2
=5
RS: 1
Since 5 ≠ 1, x = 1 is not a solution.
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Example 5 – Solution
Check:
x = 8 LS: 3 +
cont’d
=3+
=3+5
=8
RS: 8
Since 8 = 8 is a true statement, x = 8 is a solution.
Hence, the given equation has one solution, x = 8.
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Other Types of Equations
An equation is of quadratic type if it can be written in the
form
au2 + bu + c = 0.
where a ≠ 0 and u is an expression in some variable.
If we find the solutions in terms of u, then the solutions of
the given equation can be obtained by referring to the
specific form of u.
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Example 7 – Solving an equation of quadratic type
Solve the equation x2/3 + x1/3 – 6 = 0.
Solution:
Since x2/3 = (x1/3)2, the form of the equation suggests that
we let u = x1/3, as in the second line below:
x2/3 + x1/3 – 6 = 0
u2 + u – 6 = 0
(u + 3)(u – 2) = 0
given
let u = x1/3
factor
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Example 7 – Solution
u + 3 = 0,
u–2=0
u = –3,
u=2
x1/3 = –3,
x1/3 = 2
x = –27,
x=8
cont’d
Zero factor theorem
solve for u
u = x1/3
cube both sides
A check is unnecessary, since we did not raise both sides
to an even power. Hence, the given equation has two
solutions, –27 and 8.
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Example 7 – Solution
cont’d
An alternative method is to factor the left side of the given
equation as follows:
x2/3 + x1/3 – 6 = (x1/3 + 3)(x1/3 – 2)
By setting each factor equal to 0, we obtain the solutions.
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