Download 7.7 - History of Complex Roots

§ 7.7
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, 4e – Slide #94
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  10 3i
(b) 20   5  20  51  20  5 1  20  5i
Blitzer, Intermediate Algebra, 4e – Slide #95
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, 4e – Slide #96
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, 4e – Slide #97
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, 4e – Slide #98
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, 4e – Slide #99
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.
(b) (-4 + 2i)(-4 – 2i)
 16  8i  8i  4i 2
Use the FOIL method.
Blitzer, Intermediate Algebra, 4e – Slide #100
Complex Numbers
CONTINUED
 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, 4e – Slide #101
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, 4e – Slide #102
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, 4e – Slide #103
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
Multiply by 1.
24  12i  12i  6i 2

2
4 2  2i 
Use FOIL in the numerator and
 A  B  A  B   A2  B 2 in the
denominator.
Blitzer, Intermediate Algebra, 4e – Slide #104
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, 4e – Slide #105
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, 4e – Slide #106
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, 4e – Slide #107
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, 4e – Slide #108
Complex Numbers
EXAMPLE
Simplify: a  i 46
b  i 400 c i 13.
SOLUTION
 i
a  i
46
b i
400
 i
c i 
13
  1
2 23

23
2 200
i
12
 1
  1
200
1
 i   i   i    16  i   1  i   i
2 6
Blitzer, Intermediate Algebra, 4e – Slide #109