# Objectives: Assignment To use the Binomial Homework

```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
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
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
```