Transcript complex numbers

§ 7.7
Complex Numbers
Complex Numbers
Complex Numbers
The Imaginary Unit i
The imaginary unit i is defined as
i   1, where i 2  1.
The Square Root of a Negative Number
If b is a positive real number, then
 b  (1)b   1 b  i b or
Blitzer, Intermediate Algebra, 5e – Slide #3 Section 7.7
bi.
Complex Numbers
EXAMPLE
Write as a multiple of i: (a)   300 (b) 20   5.
SOLUTION
(a)   300   3001  300 1
 100  3  1  10i 3
(b) 20   5  20  51
 20  i 5 or 20  5i
Blitzer, Intermediate Algebra, 5e – Slide #4 Section 7.7
 20  5  1
Complex Numbers
Complex Numbers & Imaginary Numbers
The set of all numbers in the form
a  bi
with real numbers a and b, and i, the imaginary unit, is called
the set of complex numbers. The real number a is called the
real part, and the real number b is called the imaginary part
of the complex number a  bi. If b  0, then the complex
number is called an imaginary number.
Blitzer, Intermediate Algebra, 5e – Slide #5 Section 7.7
Complex Numbers
Adding & Subtracting Complex Numbers
1) a  bi   c  di   a  c  b  d i
In words, this says that you add complex numbers by adding
their real parts, adding their imaginary parts, and expressing the
sum as a complex number.
2) a  bi   c  di   a  c  b  d i
In words, this says that you subtract complex numbers by
subtracting their real parts, subtracting their imaginary parts,
and expressing the difference as a complex number.
Blitzer, Intermediate Algebra, 5e – Slide #6 Section 7.7
Complex Numbers
EXAMPLE
Perform the indicated operations, writing the result in the
form a + bi: (a) (-9 + 2i) – (-17 – 6i) (b) (-2 + 6i) + (4 - i).
SOLUTION
(a) (-9 + 2i) – (-17 – 6i)
= -9 + 2i + 17 + 6i
Remove the parentheses. Change
signs of the real and imaginary
parts being subtracted.
= -9 + 17 + 2i + 6i
Group real and imaginary terms.
= (-9 + 17) + (2 + 6)i
= 8 + 8i
Add real parts and imaginary parts.
Simplify.
Blitzer, Intermediate Algebra, 5e – Slide #7 Section 7.7
Complex Numbers
CONTINUED
(b) (-2 + 6i) + (4 - i)
= -2 + 6i + 4 - i
= -2 + 4 + 6i - i
= (-2 + 4) + (6 - 1)i
= 2 + 5i
Remove the parentheses.
Group real and imaginary terms.
Add real parts and imaginary parts.
Simplify.
Blitzer, Intermediate Algebra, 5e – Slide #8 Section 7.7
Complex Numbers
EXAMPLE
Find the products: (a) -6i(3 – 5i)
(b) (-4 + 2i)(-4 - 2i).
SOLUTION
(a) -6i(3 – 5i)
 6i  3   6i  5i
 18i  30i 2
 18i  30 1
 30 18i
Distribute -6i through the
parentheses.
Multiply.
2
Replace i with -1.
Simplify and write in a + bi form.
Blitzer, Intermediate Algebra, 5e – Slide #9 Section 7.7
Complex Numbers
CONTINUED
(b) (-4 + 2i)(-4 – 2i)
 16  8i  8i  4i 2
Use the FOIL method.
 16  8i  8i  41
i 2  1
 16  4  8i  8i
Group real and imaginary terms.
 20
Combine real and imaginary
terms.
Blitzer, Intermediate Algebra, 5e – Slide #10 Section 7.7
Complex Numbers
Multiplying Complex Numbers
Because the product rule for radicals only applies to real
numbers, multiplying radicands is incorrect. When
performing operations with square roots of negative
numbers, begin by expressing all square roots in terms of
i. Then perform the indicated operation.
Blitzer, Intermediate Algebra, 5e – Slide #11 Section 7.7
Complex Numbers
EXAMPLE
Multiply:  16   4.
SOLUTION
 16   4  16  1  4  1
 16i  4i
 64i 2
Express square roots in terms of i.
16  4  64 and i  i  i 2 .
 64  1
i 2  1
 8
The square root of 64 is 8.
Blitzer, Intermediate Algebra, 5e – Slide #12 Section 7.7
DONE
Complex Numbers
In the next chapter we will study equation whose solutions involve
the square roots of negative numbers. Because the square of a real
number is never negative, there are no real number solutions to
those equations. However, there is an expanded system of
numbers in which the square root of a negative number is defined.
This set is called the set of complex numbers.
The imaginary number i is the basis of this new set.
So come… now go with us to never-never land , a place where you have
not been before…
Blitzer, Intermediate Algebra, 5e – Slide #14 Section 7.7
Complex Numbers
Complex Numbers & Imaginary Numbers
The set of all numbers in the form
a  bi
with real numbers a and b, and i, the imaginary unit, is called
the set of complex numbers. The real number a is called the
real part, and the real number b is called the imaginary part
of the complex number a  bi. If b  0, then the complex
number is called an imaginary number.
Blitzer, Intermediate Algebra, 5e – Slide #15 Section 7.7
Complex Numbers
EXAMPLE
6  3i
.
Divide and simplify to the form a + bi:
4  2i
SOLUTION
The conjugate of the denominator is 4 – 2i. Multiplication of
both the numerator and the denominator by 4 – 2i will eliminate
i from the denominator.
6  3i 6  3i 4  2i


4  2i 4  2i 4  2i
24  12i  12i  6i 2

2
4 2  2i 
Multiply by 1.
Use FOIL in the numerator and
 A  B A  B  A2  B 2 in the
denominator.
Blitzer, Intermediate Algebra, 5e – Slide #16 Section 7.7
Complex Numbers
CONTINUED
24  24i  6i 2

16  4i 2
24  24i  6 1

16  4 1
24  24i  6

16  4
18  24i

20
18 24

 i
20 20
9 6
  i
10 5
Simplify.
i 2  1
Perform the multiplications
involving -1.
Combine like terms in the
numerator and denominator.
Express answer in the form a + bi.
Simplify.
Blitzer, Intermediate Algebra, 5e – Slide #17 Section 7.7
Complex Numbers
EXAMPLE
Simplify: a  i 46
b  i 400 c  i 13.
SOLUTION
a  i
46
b i
400
   1
 i
2 23
 
 i
c i 
13
2 200
i
12
 1
23
  1
200
1
 i   i   i    16  i   1  i   i
2 6
Blitzer, Intermediate Algebra, 5e – Slide #18 Section 7.7
In Summary…
To add or subtract complex numbers, add or subtract their real parts and then add
or subtract their imaginary parts. Adding complex numbers is easy.
To multiply complex numbers, use the rule for multiplying binomials. After you
are done, remember that
i 2  1
and make the substitution. In fact, if you can only remember one thing from this
section – remember this fact, that is, when your square i, you get -1.
To divide complex numbers, multiply numerator and denominator by the
conjugate of the denominator. This gives you a real number in the denominator,
and you will know how to proceed from that point.
Blitzer, Intermediate Algebra, 5e – Slide #19 Section 7.7
Complex Numbers
Complex Numbers & Imaginary Numbers
The set of all numbers in the form
a  bi
with real numbers a and b, and i, the imaginary unit, is called
the set of complex numbers. The real number a is called the
real part, and the real number b is called the imaginary part
of the complex number a  bi. If b  0, then the complex
number is called an imaginary number.
Blitzer, Intermediate Algebra, 5e – Slide #20 Section 7.7
Complex Numbers
Simplifying Powers of i
2
1) Express the given power of i in terms of i .
2) Replace i 2 with -1 and simplify. Use the fact that -1 to
an even power is 1 and -1 to an odd power is -1.
Blitzer, Intermediate Algebra, 5e – Slide #21 Section 7.7
Complex Numbers
EXAMPLE
5i
.
Divide and simplify to the form a + bi:
 4i
SOLUTION
The conjugate of the denominator, 0 - 4i, is 0 + 4i.
Multiplication of both the numerator and the denominator by 4i
will eliminate i from the denominator.
5  i 5  i 4i


 4i  4i 4i
20i  4i 2

 16i 2
Multiply by 1.
Multiply. Use the distributive
property in the numerator.
Blitzer, Intermediate Algebra, 5e – Slide #22 Section 7.7
Complex Numbers
CONTINUED
20i  4 1

 16 1
i 2  1
20i  4

16
 4 20

 i
16 16
Perform the multiplications
involving -1.
1 5
  i
4 4
Simplify real and imaginary
parts.
Express the division in the
form a + bi.
Blitzer, Intermediate Algebra, 5e – Slide #23 Section 7.7