Objectives:
1. To use the Binomial
Theorem and
Pascal’s Triangle to
take powers of
complex numbers
Assignment
• Homework
Supplement
Let’s say that you wanted to find (1 + i)5.
You could expand it out and then use the
distributive property until your head hurt.
1  i 
5
 1  i 1  i 1  i 1  i 1  i  = Headache
Or you could take a shortcut, which involves
something called the Binomial Theorem.
1. Notice that
 x  y  1
the
1
 x  y  x  y
expansion of
2
 x  y   x2  2 xy  y 2
a binomial is
3
symmetrical.
 x  y   x3  3x2 y  3xy 2  y3
2. There is one
4
4
3
2
2
3
4
 x  y   x  4 x y  6 x y  4 xy  y
more term
than the
power.
0
3. The sum of
 x  y  1
1
the powers
 x  y  x  y
of each term
2
2
2
 x  y   x  2 xy  y
is
the
power
3
 x  y   x3  3x2 y  3xy 2  y3
of the
4
binomial.
 x  y   x4  4 x3 y  6 x2 y 2  4 xy3  y 4
0
3+1=4
2+2=4
1+3=4
4. As the power
 x  y  1
1
of x
 x  y  x  y
decreases by
2
2
2
 x  y   x  2 xy  y
one,
the
3
 x  y   x3  3x2 y  3xy 2  y3
power of y
4
increases by
 x  y   x4  4 x3 y  6 x2 y 2  4 xy3  y 4
one.
0
The question is, how do we get those symmetrical,
binomial coefficients? Blaise has our answer.
• 1623-1662
• Also known as “Snake
Eyes” (Not really)
• Father of Probability
Theory
• Came up with an easy
way to find binomial
coefficients
(Pascal: Avid stamp collector)
Each row will begin and end with the number 1.
Power
0
1
2
3
4
5
6
7
Binomial Coefficients
Each row will begin and end with the number 1.
Power
Binomial Coefficients
0
1
2
3
4
5
6
7
1
Each row will begin and end with the number 1.
Power
Binomial Coefficients
0
1
2
3
4
5
6
7
1
1
Each row will begin and end with the number 1.
Power
Binomial Coefficients
0
1
2
3
4
5
6
7
1
1
1
Each row will begin and end with the number 1.
Power
Binomial Coefficients
0
1
2
3
4
5
6
7
1
1
1
1
To get the numbers between the ones, add the two numbers directly
above them.
Power
Binomial Coefficients
0
1
2
3
4
5
6
7
1
1
1
1
1
To get the numbers between the ones, add the two numbers directly
above them.
Power
Binomial Coefficients
0
1
2
3
4
5
6
7
1
1
1
1
2
1
To get the numbers between the ones, add the two numbers directly
above them.
Power
Binomial Coefficients
0
1
2
3
4
5
6
7
1
1
1
1
1
2
1
To get the numbers between the ones, add the two numbers directly
above them.
Power
Binomial Coefficients
0
1
2
3
4
5
6
7
1
1
1
1
1
2
1
1
To get the numbers between the ones, add the two numbers directly
above them.
Power
Binomial Coefficients
0
1
2
3
4
5
6
7
1
1
1
1
1
2
3
1
1
To get the numbers between the ones, add the two numbers directly
above them.
Power
Binomial Coefficients
0
1
2
3
4
5
6
7
1
1
1
1
1
2
3
1
1
Can you finish the triangle without clicking ahead?
Power
Binomial Coefficients
0
1
2
3
4
5
6
7
1
1
1
1
1
2
3
1
3
1
Can you finish the triangle without clicking ahead?
Power
Binomial Coefficients
0
1
2
3
4
5
6
7
1
1
1
1
1
1
2
3
4
1
3
6
1
4
1
Can you finish the triangle without clicking ahead?
Power
Binomial Coefficients
0
1
2
3
4
5
6
7
1
1
1
1
1
1
2
3
4
5
1
1
3
6
10
1
4
10
1
5
1
Can you finish the triangle without clicking ahead?
Power
Binomial Coefficients
0
1
2
3
4
5
6
7
1
1
1
1
1
1
1
2
3
4
5
6
1
3
6
10
15
1
1
4
10
20
1
5
15
1
6
1
The triangle, of course, would not stop at the 7th power. It would continue forever, or
at least until your pencil broke, your hand went dead, or you went crazy.
Power
Binomial Coefficients
0
1
2
3
4
5
6
7
1
1
1
1
1
1
1
1
7
2
3
4
5
6
1
3
6
10
15
21
1
4
10
20
35
1
1
5
15
35
1
6
21
1
7
1
This is how you use it to expand binomials:
Expand (x + 5)4.
1
 x  5 
4
4
6
4
1
(1)x 4 (4) x3 (5) (6) x 2 (52 ) (4) x(53 ) (1)(54 )
x4
20x 3
150x 2
500x
625
When you expand a difference instead of a sum,
the signs alternate, starting with +.
0
x

y

 1
1
x

y

  x y
2
x

y

  x2  2 xy  y 2
3
x

y

  x3  3x2 y  3xy 2  y3
4
x

y

  x4  4 x3 y  6 x2 y 2  4 xy3  y 4
Expand (x – 2)5.
5
x

2

 
1
5
10
10
5
1
(1)x 5 (5) x 4 (2) (10) x3 (22 ) (10) x 2 (23 ) (5) x(24 ) (1)(25 )
x5
10x 4
40x 3
80x 2
80x
32
Expand (2x – 3)3.
1
 2 x  3 
3
3
3
1
(1)(2 x)3 (3)(2 x)2 (3) (3)(2 x)(32 ) (1)(33 )
8x3
36x 2
54x
27
Expand (2 – i)4.
1
4
6
4
1
3
3
4
2
2
4

(4)(2
)(
i
)

(4)(2)(
i
)

(1)(
i
)

(6)(2
)(
i
)
2

i

(1)(2
)
 
4
16
7
32i
24i
24(1)
8( i )
1
Expand (3 + 2i)3.
1
3
2
3

(3)(3
)(2i)
3

2i

(1)(3
)


3
27
54i
9
46i
3
1
(3)(3)(2i)2
(1)(2i)3
36(1)
8( i )
Objectives:
1. To use the Binomial
Theorem and
Pascal’s Triangle to
take powers of
complex numbers
Assignment
• Homework
Supplement