Transcript College Algebra Fifth Edition James Stewart  Lothar Redlin  Saleem Watson Equations and Inequalities.

College Algebra
Fifth Edition
James Stewart  Lothar Redlin

Saleem Watson
1
Equations and
Inequalities
Introduction
In Section 1.3, we saw that, if the
discriminant of a quadratic equation is
negative, the equation has no real solution.
• For example, the equation
x2 + 4 = 0
has no real solution.
Introduction
If we try to solve this equation, we get:
x2 = –4
So, x   4.
• However, this is impossible—since the square
of any real number is positive.
• For example, (–2)2 = 4, a positive number.
• Thus, negative numbers don’t have real
square roots.
1.4
Complex Numbers
Complex Number System
To make it possible to solve all quadratic
equations, mathematicians invented
an expanded number system—called
the complex number system.
Complex Number
First, they defined the new number
i  1
• This means i 2 = –1.
• A complex number is then a number
of the form a + bi, where a and b are
real numbers.
Complex Number—Definition
A complex number is an expression
of the form
a + bi
where:
• a and b are real numbers.
• i 2 = –1.
Real and Imaginary Parts
The real part of this complex number is a.
The imaginary part is b.
• Two complex numbers are equal if and only if
their real parts are equal and their imaginary parts
are equal.
Real and Imaginary Parts
Note that both the real and
imaginary parts of a complex
number are real numbers.
E.g. 1—Complex Numbers
Here are examples of complex numbers.
3 + 4i
Real part 3, imaginary part 4
1/2 – 2/3i Real part 1/2, imaginary part –2/3
6i
Real part 0, imaginary part 6
–7
Real part –7, imaginary part 0
Pure Imaginary Number
A number such as 6i, which has
real part 0, is called:
• A pure imaginary number.
Complex Numbers
A real number like –7 can be
thought of as:
• A complex number with imaginary part 0.
Complex Numbers
In the complex number system, every
quadratic equation has solutions.
• The numbers 2i and –2i are solutions of
x2 = –4 because:
(2i)2 = 22i2 = 4(–1) = –4
and
(–2i)2 = (–2)2i2 = 4(–1) = –4
Complex Numbers
Though we use the term imaginary here,
imaginary numbers should not be thought of
as any less “real”—in the ordinary rather than
the mathematical sense of that word—than
negative numbers or irrational numbers.
• All numbers (except possibly the positive integers)
are creations of the human mind—the numbers –1
and 2 as well as the number i.
Complex Numbers
We study complex numbers as they
complete—in a useful and elegant fashion—
our study of the solutions of equations.
• In fact, imaginary numbers are useful not only in
algebra and mathematics, but in the other sciences
too.
• To give just one example, in electrical theory,
the reactance of a circuit is a quantity whose
measure is an imaginary number.
Arithmetic Operations
on Complex Numbers
Arithmetic Operations on Complex Numbers
Complex numbers are added, subtracted,
multiplied, and divided just as we would
any number of the form a + b c .
• The only difference we need to keep in mind
is that i2 = –1.
Arithmetic Operations on Complex Numbers
Thus, the following calculations are valid.
(a + bi)(c + di)
= ac + (ad + bc)i + bdi2
(Multiply and collect
all terms)
= ac + (ad + bc)i + bd(–1)
= (ac – bd) + (ad + bc)i
(i2 = –1)
(Combine real and
imaginary parts)
Arithmetic Operations on Complex Numbers
We therefore define the sum,
difference, and product of complex
numbers as follows.
Adding Complex Numbers
(a + bi) + (c + di)
= (a + c) + (b + d)i
• To add complex numbers, add
the real parts and the imaginary parts.
Subtracting Complex Numbers
(a + bi) – (c + di)
= (a – c) + (b – d)i
• To subtract complex numbers, subtract
the real parts and the imaginary parts.
Multiplying Complex Numbers
(a + bi) . (c + di)
= (ac – bd) + (ad + bc)i
• Multiply complex numbers like binomials,
using i 2 = –1.
E.g. 2—Adding, Subtracting, and Multiplying
Express the following in the form a + bi.
(a) (3 + 5i) + (4 – 2i)
(b) (3 + 5i) – (4 – 2i)
(c) (3 + 5i)(4 – 2i)
(d) i 23
E.g. 2—Adding
Example (a)
According to the definition, we add the real
parts and we add the imaginary parts.
(3 + 5i) + (4 – 2i)
= (3 + 4) + (5 – 2)i
= 7 + 3i
E.g. 2—Subtracting
(3 + 5i) – (4 – 2i)
= (3 – 4) + [5 – (– 2)]i
= –1 + 7i
Example (b)
E.g. 2—Multiplying
Example (c)
(3 + 5i)(4 – 2i)
= [3 . 4 – 5(– 2)] + [3(–2) + 5 . 4]i
= 22 + 14i
E.g. 2—Power
23
i
=
22
+
1
i
Example (d)
2
11
(i ) i
=
= (–1)11i
= (–1)i
= –i
Dividing Complex Numbers
Division of complex numbers is much like
rationalizing the denominator of a radical
expression—which we considered in
Section P.8.
Complex Conjugates
For the complex number z = a + bi,
we define its complex conjugate
to be:
z  a  bi
Dividing Complex Numbers
Note that:
z  z  a  bi a  bi   a  b
2
2
• So, the product of a complex number and its
conjugate is always a nonnegative real number.
• We use this property to divide complex
numbers.
Dividing Complex Numbers—Formula
a  bi
c  di
multiply the numerator and the denominator
by the complex conjugate of the denominator:
To simplify the quotient
a  bi  a  bi  c  di 



c  di  c  di  c  di 
 ac  bd    bc  ad  i

2
2
c d
Dividing Complex Numbers
Rather than memorize this entire formula,
it’s easier to just remember the first step
and then multiply out the numerator and
the denominator as usual.
E.g. 3—Dividing Complex Numbers
Express the following in the form a + bi.
3  5i
a
1  2i
7  3i
b 
4i
• We multiply both the numerator and denominator
by the complex conjugate of the denominator to
make the new denominator a real number.
E.g. 3—Dividing Complex Numbers Example (a)
The complex conjugate of 1 – 2i is:
1  2i  1  2i
3  5i  3  5i   1  2i 



1  2i  1  2i   1  2i 
7  11i
7 11

  i
5
5 5
E.g. 3—Dividing Complex Numbers Example (b)
The complex conjugate of 4i is –4i.
7  3i  7  3i  4i 



4i
 4i  4i 
12  28i 3 7

  i
16
4 4
Square Roots of
Negative Numbers
Square Roots of Negative Numbers
Just as every positive real number r has two
square roots r and  r , every negative
number has two square roots as well.


• If -r is a negative number, then its square roots
are i r , because:

and

i r

i r

2
2
 i 2r  r
 i 2r  r
Square Roots of Negative Numbers
If –r is negative, then the principal
square root of –r is:
r  i r
• The two square roots of –r are: i r
• We usually write i b instead of bi
to avoid confusion with bi .
and  i r
E.g. 4—Square Roots of Negative Numbers
 a  1  i 1  i
 b  16  i 16  4i
 c  3  i 3
Square Roots of Negative Numbers
Special care must be taken when
performing calculations involving square
roots of negative numbers.
• Although a  b  ab when a and b
are positive, this is not true when both
are negative.
Square Roots of Negative Numbers
For example,
2  3  i 2  i 3  i
However,
• Thus,
2
6  6
 2 3   6

2  3   2 3
Square Roots of Negative Numbers
When multiplying radicals of negative
numbers, express them first in the form
i r
(where r > 0)
to avoid possible errors of this type.
E.g. 5—Using Square Roots of Negative Numbers
Evaluate  12  3  3  4 
and express in the form a + bi.



  12  i 3  3  i 4 
  2 3  i 3   3  2i 
 6 3  2 3   i 2  2 3  3 3 
12  3 3  4
8 3 i 3
Complex Solutions of
Quadratic Equations
Complex Roots of Quadratic Equations
We have already seen that, if a ≠ 0,
then the solutions of the quadratic
equation ax2 + bx + c = 0
are:
b  b  4ac
x
2a
2
Complex Roots of Quadratic Equations
If b2 – 4ac < 0, the equation has no real
solution.
However, in the complex number system,
the equation will always have solutions.
• This is because negative numbers have
square roots in this expanded setting.
E.g. 6—Quadratic Equations with Complex Solutions
Solve each equation.
(a) x2 + 9 = 0
(b) x2 + 4x + 5 = 0
E.g. 6—Complex Solutions
Example (a)
The equation x2 + 9 = 0 means x2 = –9.
So,
x   9  i 9  3i
• The solutions are therefore 3i and –3i.
E.g. 6—Complex Solutions
Example (b)
By the quadratic formula, we have:
4  4  4  5 4  4
x

2
2
4  2i

2
2  2  i 

 2  i
2
2
• So, the solutions are
–2 + i and –2 – i.
E.g. 7—Complex Conjugates as Solutions of a Quadratic
Show that the solutions of the equation
4x2 – 24x + 37 = 0
are complex conjugates of each other.
E.g. 7—Complex Conjugates as Solutions of a Quadratic
We use the quadratic formula to get:
24   24   4  4  37 
x
2  4
2
24  16 24  4i
1


3 i
8
8
2
• So, the solutions are 3 + ½i and 3 – ½i.
• These are complex conjugates.