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L10-Mon-26-Sep-2016-Sec-A-7-Complex-Num-HW08-A-10-Radicals-HW09-Q09, page 86
L10-Mon-26-Sep-2016-Sec-A-7-Complex-Num-HW08-A-10-Radicals-HW09-Q09
L10-Mon-26-Sep-2016-Sec-A-7-Complex-Num-HW08-A-10-Radicals-HW09-Q09, page 87
We use exponents to indicate the number of factors of a base that are to be multiplied. For example,
the exponent in 32 tells us to multiply two factors of the base, 3 , to get 9 .
In some situations we may have to do the reverse; that is, we may be given the final product and
asked to find the base. For example, find numbers whose square is 16 . We call those numbers the
square roots of 16 and we use the radical symbol
to indicate the square root.
Every positive real number has two square roots, which are opposites of each other. The square root
that is positive is called the principal square root; the square root that is negative is called the
negative square root.
 principal square root of 16 is 4 since 42 is 16. We indicate the principal square root of 16 as
16
 negative square root of 16 is -4 since  4  is 16. We write the negative square root of 16 as 
2
 Square root of 16 is 4 and -4. Write the square root of 16 as 
An expression that contains a radical symbol,
16
16 .
is called a radical expression or simply a radical.
Expression under radical symbol is the radicand. E.g.,,
16 is a radical with a radicand of 16.
L10-Mon-26-Sep-2016-Sec-A-7-Complex-Num-HW08-A-10-Radicals-HW09-Q09, page 88
20x 3 
27x 5 y
12xy 3
2  2  5xxx  2 5 x

27x 5 y

12xy 3
x  2x
3  3  3xxxxxy

2  2  3xyyyy
5x
3  3xxxx
3x 2

2  2 yyyy
2y 2
L10-Mon-26-Sep-2016-Sec-A-7-Complex-Num-HW08-A-10-Radicals-HW09-Q09, page 89
250x 9 y 16 
2  5  5  5  xxxxxxxxxyyyyyyyyyyyyyyyy
5  xxxx
2 5

x  yyyyyyyy
 5x 4 y 8 10x
3
250x 9 y 16

3
2  5  5  5  xxxxxxxxxyyyyyyyyyyyyyyyy

3
2 5x3  y5 3 y1
 5x 3 y 5  3 2 y
A quick way to simplify a variable part is to divide the exponent by the index. E.g., in the above:
 For x 9 divide 9 / 3  3 , to get the final exponent for x outside the radical.
 For y 16 divide 16 / 3  5 with remainder 1. The quotient is the final exponent for y outside
the radical and the remainder is the final exponent for y inside the radical.
3
1 7

3
1 7

1 7
1 7


3 1  7
12 
7 

7 
49


3 1  7
1 7
  3 1  7   1 
Another example:
3 12  2 3  5 18  3 
223 2
 32
 6
3 2
3 2
 4 3  15 2
Another example:
2
3 6

2
3 6

6
6

2 6
6

6
3
3 5
3 53
3  15 2
233
2
6
2
7
L10-Mon-26-Sep-2016-Sec-A-7-Complex-Num-HW08-A-10-Radicals-HW09-Q09, page 90
Two ways of solving:
Correct way:
3 2•
x 4 5
2•
x 4 2
x 4 1

x 4

2
 1
2
x 4 1
x  3
or
Incorrect way (it works
by pure luck):
3 2•
x 4 5
5•
x 4 5
x 4 1

x 4

2
 1
2
x 4 1
x  3
12  x  x

12  x

2
 x2
12  x  x 2
0  x 2  x  12
0   x  3  x  4 
x 3
or
x  4
Be sure solutions work! Be sure solution is in domain of original equation.
12  x  0
Since the radicand must be nonnegative, we have
x  12 . Also, since the left side of the
x  12
equation is positive (it is the principal square root) we also have x  0 . Thus the domain is 0,12  .
So, X = 3 is the only solution.
L10-Mon-26-Sep-2016-Sec-A-7-Complex-Num-HW08-A-10-Radicals-HW09-Q09, page 91
x2 
3
Simplify:
3
4
x
x3
4
x2 
x
3
x
x 2 3  x12

 x 2 3  x 1 2  x 3 4  x 8 12  x 6 12  x 9 12  x 5 12 
3 4
x
Another example:
 4x y 
x y 
1
1
13
23
3 2
2 x y 

x y 
2
13
1
3 2
1

23
24 3 x 2 3 y 2 9
x
3 2
 24 3 x
y
3 2
4 6   9 6 
 24 3 x
2 3   3 2 
y 2 93 2
y 4 18  27 18  24 3 x 13 6 y 23 18 
24 3 x 13 6
y 23 18
12
x5
L10-Mon-26-Sep-2016-Sec-A-7-Complex-Num-HW08-A-10-Radicals-HW09-Q09, page 92
A.7 Complex Numbers
When mathematicians tried to solve the simple quadratic equation
x 2  1  0 they were stumped because x 2 is always a positive
number. So, showing their usual pluck, the solved it this way:
x2 1 0
x 2  1
x2  
1
x 
1
Since
 1 does not exist, they defined a
new number called the imaginary unit.
i 
1 .
Although ancient Greeks are known to have observed these numbers,
imaginary numbers were defined in 1572 AD by Rafael Bombelli.
L10-Mon-26-Sep-2016-Sec-A-7-Complex-Num-HW08-A-10-Radicals-HW09-Q09, page 93
However, imaginary numbers have concrete applications in sciences such as signal processing,
control theory, electromagnetism, fluid dynamics, quantum mechanics, cartography, and vibration
analysis. Other topics utilizing imaginary numbers include the investigation of electrical current,
wavelength, liquid flow in relation to obstacles, analysis of stress on beams, the movement of shock
absorbers in cars, the study of resonance of structures, the design of dynamos and electric motors,
and the manipulation of large matrices used in modeling. For example, the mathematical models that
describe how AC current flows through wires use imaginary numbers.
From quantum mechanics, here is a model of the propagation of a plane wave along the x-axis as a
function of time (Merzbacher, p17):
  x ,t  
1
 2


2    x

 2
 i 
e



x
x  t 
d

2

x
L10-Mon-26-Sep-2016-Sec-A-7-Complex-Num-HW08-A-10-Radicals-HW09-Q09, page 94
 
Divide 37 by 4 to get 9 with 1 left over. Thus, i 37  i 4
9
 i 1  19  i  i
L10-Mon-26-Sep-2016-Sec-A-7-Complex-Num-HW08-A-10-Radicals-HW09-Q09, page 95
5   4  7i  



5
4  7i

4  7i 4  7i
5  4  7i 
16  49i 2
5  4  7i 
16  49
5  4  7i 
65
4  7i

13
4  7 


i
13  13 
 2  3i  4  i   8  2i  12i  3i
 8  10i  3  1
 8  10i  3
 11  10i
Write in complex form:
4
4 i


3i
3i i
4i

3i 2
4i

3  1
4
i
3
 4
 0   i
 3

4
3i
2
L10-Mon-26-Sep-2016-Sec-A-7-Complex-Num-HW08-A-10-Radicals-HW09-Q09, page 96
Simplify:
2
 18
2
 18 
2 i 
18  i

36  i 2

36   1
 6  1
 6
Note that:
2
 18 
36  6
Simplify:  9 
 16 
 9 
 25  3  4i  5i
 16 
 25
 3  i
L10-Mon-26-Sep-2016-Sec-A-7-Complex-Num-HW08-A-10-Radicals-HW09-Q09, page 97
Let’s prove that false by proving that 1 = -1.
Start with this identity 1 
1
1 
 1 1

1
1
 1  i  i  i 2  1 Q .E .D .
Be careful with the properties!
L10-Mon-26-Sep-2016-Sec-A-7-Complex-Num-HW08-A-10-Radicals-HW09-Q09, page 98
Square root of i
If i 
 1 then what is
i ? Do we need to invent a new number for this? No. Mathematicians
have proved that for all polynomial expressions all we need is the set of complex numbers. So,
i
is a complex number and can be written in the form a  bi . Let’s see what this is:
i  a  bi

i

2
  a  bi 
2
i  a 2  2abi  b 2i 2
i  a 2  b 2  2abi
In order for these to be equal, the imaginary parts must be equal and the real parts must be equal. So
we can write:
0  a2  b2
0   a  b  a  b 
0  a b
b a
or
0  a b
or  b  a
i  2abi
1  2ab
and
1
a
2b
When a = b we have
1
b
2b
1
 b2
2
1

b
2
2

b
2
When a = -b we have:
1
 b
2b
1
 b 2
2
This is a contradiction since a positive number cannot equal a negative number, so a cannot equal –b.
So, we have
i  a  bi 
2
2

2
2
i
L10-Mon-26-Sep-2016-Sec-A-7-Complex-Num-HW08-A-10-Radicals-HW09-Q09, page 99
We can check to see if squaring
 2


 2

2
2
2

2
2
i yields i.

 2
2  2
2 


i 
i 
i

 2
 2

2
2
2




2
2
1 i  


1  i 
2
2
2
1 i  i  i 2

4
1
 1  2i  1
2
1
  2i 
2
i
2

