1. MOUNTAIN TOP UNIVERSITY
COURSE CODE: MTH 101
11/19/2018
COURSE TITLE: ALGEBRA
COURSE LECTURER: Mr. TAIWO A.
MTH 101/L8
          ALGEBRA by Mr. Taiwo A. is licensed under a Creative Commons Attribution-Non Commercial 4.0 International License

2. PRINCIPLE OF MATHEMATICAL INDUCTION PERMUTATIONS AND COMBINATIONS
SEQUENCES AND SERIES
PRINCIPLE OF MATHEMATICAL INDUCTION
PERMUTATIONS AND COMBINATIONS
THE BINOMIAL THEOREM
COMPLEX NUMBERS
11/19/2018
MTH 101/L8

3. SEQUENCES AND SERIES
11/19/2018
MTU/MTH 101/WK 7
MTH 101
(ALGEBRA)

4. SEQUENCES
11/19/2018
A sequence is an ordered list of numbers. Examples are given below.
(i) 1, 3, 5, 7,
(ii) 3, 2, , ,
(iii) 1, -1, 1, -1, 1,
(iv) 3, 10, 21, 36,
MTU/MTH 101/WK 7

5. SEQUENCES
11/19/2018
The numbers in the ordered list are called ‘elements’ or ‘terms’. Each term is named according to the position in the list. E.g. in (i), 1 is the first term, 3 is the second term, 5 is the third term and so on.
The nth term is denoted as 𝑎 n .
Sequence may be finite or infinite. (i) above is an example of a finite sequence while (ii) – (iv) are infinite.
MTU/MTH 101/WK 7

6. SEQUENCES
11/19/2018
Each sequence has a formula for obtaining its terms. For instance, the formula for the nth term of (iv) is 𝑎 n =n(2n+1).
Example Write the first four terms of the sequence whose nth term is 𝑎 𝑛 = (−1) 𝑛 (2𝑛) .
MTU/MTH 101/WK 7

7. The first four terms of the sequence are as follows:
Solution
The first four terms of the sequence are as follows:
𝑎 1 = (−1) 1 2(1) =− 1 2
𝑎 2 = (−1) 2 2(2) = 1 4
𝑎 3 = (−1) 3 2(3) =− 1 6
𝑎 4 = (−1) 4 2(4) = 1 8
11/19/2018
MTU/MTH 101/WK 7

8. Arithmetic Progression or Sequence
11/19/2018
Each term after the first is formed by adding a fixed amount, called the common difference, to the preceding term. If 𝑎 is the first term, 𝑑 the common difference and 𝑛 the number of terms of an A. P, then the terms are
𝑎, 𝑎+𝑑, 𝑎+2𝑑, 𝑎+3𝑑, 𝑎+(𝑛−1)𝑑
nth term 𝑎 n =𝑎+(n−1)𝑑 (6.1)
MTU/MTH 101/WK 7

9. Example Find the 14th term of the progression 2, 7, 12, 17, . . .
Solution
The sequence is an A. P.
𝑎=2, 𝑑=5, n=14
a 14 =𝑎+13𝑑=2+13 5
=2+65=67
11/19/2018
MTU/MTH 101/WK 7

10. Arithmetic Progression or Sequence
Example The 4th and 10th terms of an A. P. are -13 and -37 respectively. Find the sum of the 9th and 20th terms.
Solution
𝑎 4 =𝑎+3𝑑=−13 ⋯ (i)
𝑎 10 =𝑎+9𝑑=−37 ⋯ (ii)
ii − i : 𝑑=−24
𝑑=−4
Arithmetic Progression or Sequence
11/19/2018
MTU/MTH 101/WK 7

11. Arithmetic Progression or Sequence
Put the value of 𝑑 in (i): 𝑎−12=−13
𝑎=−1
𝑎 9 =𝑎+8𝑑=−1+8 −4 =−1−32=−33
∴ 𝑎 20 =𝑎+19𝑑= −1 +19 −4 =−77
𝑎 9 + 𝑎 20 =−33−77=−110
Arithmetic Progression or Sequence
11/19/2018
MTU/MTH 101/WK 7

12. Arithmetic Means The arithmetic mean of two numbers 𝑥 and 𝑧 is the number 𝑦 such that 𝑥, 𝑦, 𝑧 are in A. P. This implies that 𝑦−𝑥=𝑧−𝑦 2𝑦=𝑥+𝑧 𝑦= 𝑥+𝑧 2 For example, the arithmetic mean of -3 and 15 is 6.
11/19/2018
MTU/MTH 101/WK 7

13. Example Insert five arithmetic means between 4 and 22. Solution
We are to find 5 numbers 𝑣, 𝑤, 𝑥, 𝑦, 𝑧 such that 4, 𝑣, 𝑤, 𝑥, 𝑦, 𝑧, 22are in A. P.
So,
𝑎 1 =𝑎=4
𝑎 7 =𝑎+6𝑑=22
⇒ 𝑑=3
∴𝑣=7, 𝑤=10, 𝑥=13, 𝑦=16, 𝑧=19
11/19/2018
MTU/MTH 101/WK 7

14. Geometric Progression
11/19/2018
In a Geometric Progression, the ratio of a term and the immediately preceding term is always a constant.
This constant term is called the common ratio (r).
Examples are:
(a) 2, 4, 8, 16, . . .
MTU/MTH 101/WK 7

15. If 𝑎 is the first term and 𝑟 is the common ratio, then we have 𝑎 1 =𝑎
(b) 𝑥, 𝑥 3 , 𝑥 5 , 𝑥 7 ,
(c) 12, −8, , −3 5 9 , . . .
If 𝑎 is the first term and 𝑟 is the common ratio, then we have
𝑎 1 =𝑎
𝑎 2 =𝑎×𝑟
𝑎 3 =𝑎× 𝑟 2 ⋮
𝑎 n =𝑎× 𝑟 n−1
11/19/2018
MTU/MTH 101/WK 7

16. The nth term of a G.P. is 𝑎 n =𝑎 𝑟 n−1 .
Example Find the seventh term of the sequence 5, -10, 20, -40,⋯
Solution
𝑎=5, 𝑟=−2 , 𝑎 7 =?
𝑎 7 =5 (−2) 7−1
=5 (2) 6
=320
11/19/2018
MTU/MTH 101/WK 7

17. 𝑛-th term = 𝑎 𝑟 𝑛−1 =4 3 𝑛−1 or 4 (−3) 𝑛−1
Example Find the 𝑛-th term of a G.P. whose 3rd term is 36 and whose 5th term is 324.
Solution
𝑎 3 =𝑎 𝑟 2 =36⋯ i
𝑎 5 =𝑎 𝑟 4 =324⋯ ii
𝑎 5 𝑎 3 = 𝑎 𝑟 4 𝑎 𝑟 2 = ⇒ 𝑟 2 =9
Hence 𝑟=±3, 𝑎= 36 9 =4
𝑛-th term = 𝑎 𝑟 𝑛−1 =4 3 𝑛−1 or 4 (−3) 𝑛−1
11/19/2018
MTU/MTH 101/WK 7

18. Example Find the possible values of 𝑥 if 𝑥−3, 𝑥+1 , 4𝑥−2 are in G. P
Example Find the possible values of 𝑥 if 𝑥−3, 𝑥+1 , 4𝑥−2 are in G.P. Solution Since 𝑥−3, 𝑥+1, 4𝑥−2 are in G.P., 𝑥+1 𝑥−3 = 4𝑥−2 𝑥+1 ⇒ 𝑥+1 𝑥+1 =(𝑥−3)(4𝑥−2) 𝑥 2 +2𝑥+1=4 𝑥 2 −14𝑥+6
11/19/2018
MTU/MTH 101/WK 7

19. 3 𝑥 2 −16𝑥+5=0
𝑥−5 3𝑥−1 =0
𝑥=5 or 1 3
11/19/2018
MTU/MTH 101/WK 7

20. 𝑛-th term = 𝑎 𝑟 𝑛−1 =4 3 𝑛−1 or 4 (−3) 𝑛−1
Example Find the 𝑛-th term of a G.P. whose 3rd term is 36 and whose 5th term is 324.
Solution
𝑎 3 =𝑎 𝑟 2 =36⋯ i
𝑎 5 =𝑎 𝑟 4 =324⋯ ii
𝑎 5 𝑎 3 = 𝑎 𝑟 4 𝑎 𝑟 2 = ⇒ 𝑟 2 =9
Hence 𝑟=±3, 𝑎= 36 9 =4
𝑛-th term = 𝑎 𝑟 𝑛−1 =4 3 𝑛−1 or 4 (−3) 𝑛−1
11/19/2018
MTU/MTH 101/WK 7

21. Example A ball is dropped from a height of 125 feet
Example A ball is dropped from a height of 125 feet. If it rebounds of the height from which it falls every time it hits the ground, how high will it bounce after it strikes the ground for the fifth time?
Solution
The first rebound= =75 feet. The second rebound= =45 feet. The heights of the rebounds form a geometric sequence with first term75 and common ratio Thus the fifth term is
𝑎 5 = = = = feet
11/19/2018
MTU/MTH 101/WK 7

22. Geometric Mean
11/19/2018
If 𝑥, 𝑦, 𝑧 are 3 consecutive terms of a G.P., then 𝑦 is called the geometric mean of 𝑥 and 𝑧. It is given as 𝑦= 𝑥𝑧 Proof: 𝑦 𝑥 = 𝑧 𝑦 ⇒ 𝑦 2 =𝑥𝑧 𝑦=± 𝑥𝑧
MTU/MTH 101/WK 7

23. Harmonic Progression
11/19/2018
A sequence of the form
1 𝑎 , 1 𝑎+𝑑 , 1 𝑎+2𝑑 ,
is called a Harmonic Progression (H. P.).
Note that the terms of an H.P. are reciprocals of the terms of an A.P.
The 𝑛-th term of an H.P. is 𝑎 𝑛 = 1 𝑎+(𝑛−1)𝑑
Example is the sequence 1 2 , 1 7 , , ,⋯
MTU/MTH 101/WK 7

24. Harmonic Mean 1 7 is the harmonic mean of 1 2 and 1 12
11/19/2018
If 𝑥, 𝑦, 𝑧 are 3 consecutive terms of an H.P., then y= 2𝑥𝑧 𝑥+𝑧 ,
is the Harmonic Mean of 𝑥 and 𝑧.
1 7 is the harmonic mean of and 1 12
MTU/MTH 101/WK 7

25. Series
11/19/2018
A series is a sequence of partial sum. Given a sequence: 𝑎 1 , 𝑎 2 , 𝑎 3 , , the corresponding series is 𝑎 1 + 𝑎 2 + 𝑎 3 + ∙ ∙ ∙ If we denote the sum of the first n terms by S n , then S 1 = 𝑎 1 S 2 = 𝑎 1 + 𝑎 2 S 3 = 𝑎 1 + 𝑎 2 + 𝑎 3 ⋮
MTU/MTH 101/WK 7

26. S n = 𝑎 1 + 𝑎 2 + 𝑎 3 + ∙ ∙ ∙ + 𝑎 n−1 S n−1 + 𝑎 𝑛
∴ 𝑎 n = S n − S n−1 ∙ ∙ ∙ ∙ ∙ ∙ ∙ (6.2)
(6.2) can be used to find the nth term given the sum of the first nth term of the sequence.
Note: Another way of writing 𝑆 𝑛 is
𝑖=1 𝑛 𝑎 𝑖
11/19/2018
MTU/MTH 101/WK 7

27. SERIES
11/19/2018
Example The sum of the first n terms of a sequence is given by S n = n 2 (n+1) Find the 5th term of the sequence. Solution Using (6.2), 𝑎 5 = S 5 − S 4 = 5 2 × − 4 2 × = 225−100=125
MTU/MTH 101/WK 7

28. Sum of an Arithmetic Progression
11/19/2018
The sum of the first 𝑛 terms of an A.P. is called an Arithmetic Series. If this is denoted as 𝑆 𝑛 , then
𝑆 𝑛 = 𝑎 1 + 𝑎 2 +⋯+ 𝑎 𝑛
The sum of the first 𝑛 terms of an A.P. is 𝑆 𝑛 = 𝑛 2 2𝑎+ 𝑛−1 𝑑 2𝑎+ 𝑛−1 𝑑 = 𝑛 2 [ 𝑎 1 + 𝑎 𝑛 ]
MTU/MTH 101/WK 7

29. Example Find the sum of the first 21 terms of the series: 3.5, 4.1, 4.7, 5.3, ∙ ∙ ∙
Solution
The sequence is an A.P.
𝑎=3.5, 𝑑=0.6, 𝑛=21
𝑆 21 = =
= 21 2 ×19
= 199.5
11/19/2018
MTU/MTH 101/WK 7

30. The series 3, 6, 9, 12, ⋯, 207 is an A.P. 𝑎=3, 𝑑=3, 𝑛=?, 𝑎 𝑛 =207
Example Find the sum of all the numbers between 1 and 208 which are exactly divisible by 3.
Solution
The series 3, 6, 9, 12, ⋯, 207 is an A.P. 𝑎=3, 𝑑=3, 𝑛=?, 𝑎 𝑛 =207
From (6.1),
207=3+ 𝑛−1 3=3𝑛
⇒𝑛=69
∴ 𝑆 69 = =7245
11/19/2018
MTU/MTH 101/WK 7

31. Example Three numbers are in A. P
Example Three numbers are in A.P. Their sum is 15 and their product is 80. Determine the three numbers.
Solution
Let the numbers be 𝑎−𝑑, 𝑎 and 𝑎+𝑑.
Their sum: 𝑎−𝑑+𝑎+𝑎+𝑑=3𝑎=15
∴𝑎=5
Their product: 𝑎−𝑑 𝑎 𝑎+𝑑 =80
𝑎 𝑎 2 − 𝑑 2 =80
11/19/2018
MTU/MTH 101/WK 7

32. Arithmetic Series 5 25− 𝑑 2 =80 25− 𝑑 2 =16 𝑑 2 =9 𝑑=±3
11/19/2018
5 25− 𝑑 2 =80
25− 𝑑 2 =16
𝑑 2 =9
𝑑=±3
The numbers are 2, 5 and 8.
MTU/MTH 101/WK 7

33. Sum of a Geometric Progression
11/19/2018
The sum 𝑆 𝑛 of the first 𝑛 terms of a G.P. is called a Geometric Series. Thus 𝑆 𝑛 =𝑎+𝑎𝑟+𝑎 𝑟 2 +𝑎 𝑟 3 + ∙ ∙ ∙+𝑎 𝑟 𝑛−1 ⇒𝑟 𝑆 𝑛 =𝑎𝑟+𝑎 𝑟 2 +𝑎 𝑟 3 + ∙ ∙ ∙+𝑎 𝑟 𝑛−1 +𝑎 𝑟 𝑛 Subtract: 1−𝑟 𝑆 𝑛 =𝑎−𝑎 𝑟 𝑛 ⇒ 𝑆 𝑛 = 𝑎(1− 𝑟 𝑛 ) 1−𝑟 , 𝑟≠1
MTU/MTH 101/WK 7

34. 𝑆 5 = 4(1− (−3) 5 ) 1−(−3) = 4(1−(−243)) 4 =244
Example
(a) Find the sum of the first five terms of the geometric progression with first term 4 and common ratio -3.
(b) Find the sum of the first six terms of the geometric sequence 1 4 , 1 8 , , ⋯
Solution
(a) 𝑆 5 =?, 𝑎=4, 𝑟=−3
𝑆 5 = 4(1− (−3) 5 ) 1−(−3) = 4(1−(−243)) 4 =244
11/19/2018
MTU/MTH 101/WK 7

35. (b) 𝑆 6 =?, 𝑎= 1 4 , 𝑟= 1 2 .
𝑆 6 = − − 1 2 = −
= (1− 1 64 ) 2
= 64−1 128 =
11/19/2018
MTU/MTH 101/WK 7

36. Geometric Series
11/19/2018
Example Evaluate 𝑘=2 𝑛 2 𝑘+3 3 𝑘 Solution 𝑘=2 𝑛 2 𝑘+3 3 𝑘 = ∙ ∙ ∙+ 2 𝑛+3 3 𝑛 This is a G.P. containing (𝑛−1) terms with 𝑎= 1 9 , 𝑟= 1 3 .
MTU/MTH 101/WK 7

37. ⇒ 𝑆 𝑛−1 = 𝑘=2 𝑛 2 𝑘+3 3 𝑘 =𝑎 1− 𝑟 𝑛−1 1−𝑟 = 2 5 3 2 1− 2 3 𝑛−1 1− 2 3 = 2 5 3 1− 2 3 𝑛−1
11/19/2018
MTU/MTH 101/WK 7

38. Sum to Infinity
11/19/2018
The infinite sum 𝑆 ∞ = 𝑖=1 ∞ 𝑎 𝑛 = 𝑎 1 + 𝑎 2 + ∙ ∙ ∙ + 𝑎 𝑛 + ∙ ∙ ∙ is called an infinite series of 𝑎 𝑛 . If 𝑆 ∞ is defined and finite, we say that the series converges or is convergent. We can also say that 𝑆 ∞ = lim 𝑛→∞ 𝑆 𝑛
MTU/MTH 101/WK 7

39. Sum to infinity
11/19/2018
For a G.P. with |r|<1, 𝑆 𝑛 = 𝑎(1− 𝑟 𝑛 ) 1−𝑟 . The value of 𝑟 𝑛 approaches zero as n tends to infinity. The sum of the first n terms as n approaches infinity is called the sum to infinity ( 𝑆 ∞ ).
𝑆 ∞ = 𝑎 1−𝑟
MTU/MTH 101/WK 7

40. Example Find the sum to infinity of the sequence 1,− 1 4 , 1 16 ,− 1 64 , . . .
Solution
𝑎=1, 𝑟=− 1 4
𝑆 ∞ = 𝑎 1−𝑟 =
= = 4 5
11/19/2018
MTU/MTH 101/WK 7

41. Example Find the sum to infinity 𝑖=1 ∞ 2 𝑖+1 3 𝑖
𝑖=1 ∞ 2 𝑖 𝑖
Solution
The series is a G.P. with 𝑎= 4 3 and 𝑟= 2 3
𝑆 ∞ = − 2 3
= 4 3 × 3 1 =4
11/19/2018
MTU/MTH 101/WK 7

42. Telescoping Series
11/19/2018
A telescoping series is a series whose partial sums eventually only have a fixed number of terms after cancellation.
Example Evaluate
𝑎 𝑘=1 𝑛 1 𝑘(𝑘+1) 𝑏 𝑘=1 ∞ 1 𝑘(𝑘+1)
Solution
Resolve into partial fractions
MTU/MTH 101/WK 7

43. = 1− 1 2 + 1 2 − 1 3 + 1 3 − 1 4 + ∙ ∙ ∙+ 1 𝑛−1 − 1 𝑛 + 1 𝑛 − 1 𝑛+1
𝑘=1 𝑛 1 𝑘(𝑘+1) = 𝑘=1 𝑛 1 𝑘 − 1 𝑘+1
= 1− − − ∙ ∙ ∙+ 1 𝑛−1 − 1 𝑛 𝑛 − 1 𝑛+1
=1− 1 𝑛+1
= 𝑛 𝑛+1
11/19/2018
MTU/MTH 101/WK 7

44. Telescoping Series 𝑏 𝑘=1 ∞ 1 𝑘(𝑘+1) = lim 𝑛→∞ 𝑛 𝑛+1 = lim 𝑛→∞ 1 1+ 1 𝑛
11/19/2018
𝑏 𝑘=1 ∞ 1 𝑘(𝑘+1) = lim 𝑛→∞ 𝑛 𝑛+1
= lim 𝑛→∞ 𝑛 =1
MTU/MTH 101/WK 7

45. Harmonic Progression
11/19/2018
A sequence of the form
1 𝑎 , 1 𝑎+𝑑 , 1 𝑎+2𝑑 ,
is called a Harmonic Progression (H. P.).
Example is the sequence 1 2 , 1 7 , , ,⋯
The 𝑛-th term of an H.P. is 𝑎 𝑛 = 1 𝑎+(𝑛−1)𝑑
MTU/MTH 101/WK 7

46. Harmonic Progression
11/19/2018
1. The reciprocals of the terms are in Arithmetic Progression.
2. If 𝑥, 𝑦 and 𝑧 are three consecutive terms,
then
𝑦= 2𝑥𝑧 𝑥+𝑧
𝑦 is called the harmonic mean of 𝑥 and 𝑧.
Another example is the sequence 12, 6, 4, 3, . . .
MTU/MTH 101/WK 7

47. Harmonic Progression
11/19/2018
Example: Show that 12, 6, 4, 3, is a harmonic progression. Find the next two terms and state the formula for the nth term.
Solution
1 12 , 1 6 , 1 4 , 1 3 , is in A.P. Therefore the original sequence is a H.P.
MTU/MTH 101/WK 7

48. MATHEMATICAL INDUCTION
11/19/2018
Mathematical Induction is used to prove a basic arithmetic fact about the natural numbers.
For instance, we can prove that for all 𝑛∈ℕ,
(i) 9 divides 2 2𝑛 −3𝑛−1
ii ∙ ∙ ∙ +𝑛= 𝑛(𝑛+1) 2
iii 𝑛(𝑛+1)(𝑛+2) is divisible by 6.
MTU/MTH 101/ CHAPTER 7

49. MATHEMATICAL INDUCTION
11/19/2018
iv 𝑘=1 𝑛 𝑘 2 = 1 6 𝑛 𝑛+1 2𝑛+1
v 2 𝑛 >𝑛
MTU/MTH 101/ CHAPTER 7

50. Principle of Mathematical Induction
11/19/2018
Let P(𝑛) be a statement concerning natural number 𝑛.
If 𝑃(1) is true and
the hypothesis that 𝑃(𝑘) is true for a particular 𝑘 is sufficient to ensure that it is also true for 𝑃(𝑘+1) then 𝑃(𝑛) is true for every positive integer 𝑛 .
MTU/MTH 101/ CHAPTER 7

51. Implication of the principle
11/19/2018
Show that P(1) is true.
Show that if P(k) holds, then also P(k+1) holds.
Example Prove by Mathematical Induction that ∙ ∙ ∙ +𝑛= 𝑛(𝑛+1) 2
Solution
The statement P(n) is that
∙ ∙ ∙ +𝑛= 𝑛(𝑛+1) 2
MTU/MTH 101/ CHAPTER 7

52. MATHEMATICAL INDUCTION
11/19/2018
Show that P(1) is true.
Consider 𝑛=1
L. H. S=1
R. H. S= 1(1+1) 2 =1
Since the L.H.S = R.H.S the statement is true for 𝑛=1.
MTU/MTH 101/ CHAPTER 7

53. MATHEMATICAL INDUCTION
11/19/2018
Assume the statement is true for 𝑛=𝑘
∙ ∙ ∙ +𝑘= 𝑘 𝑘+1 2
We then prove that 𝑃(𝑘+1) is also true.
For 𝑛=𝑘+1
L. H. S= ∙ ∙ ∙ +𝑘 + 𝑘+1 = 𝑘 𝑘 (𝑘+1)
MTU/MTH 101/ CHAPTER 7

54. = (𝑘+1)(𝑘+2) 2 R. H.S= (𝑘+1)(𝑘+1+1) 2 We have that L.H.S = R.H.S
Hence by the principle of mathematical induction, 𝑃(𝑛) is true for all 𝑛∈ ℕ.
11/19/2018
MTU/MTH 101/ CHAPTER 7

55. MATHEMATICAL INDUCTION
11/19/2018
Example Prove that
1∙2+2∙3+3∙4+ ∙ ∙ ∙+𝑛 𝑛+1 = 𝑛(𝑛+1)(𝑛+2) 3
Solution
Consider 𝑛=1.
L. H. S=1 1+1 =2
MTU/MTH 101/ CHAPTER 7

56. MATHEMATICAL INDUCTION
11/19/2018
R. H.S= 1(1+1)(1+2) 3 =2=L. H.S
So it is true for 𝑛=1.
Assume it is true for 𝑛=𝑘.
1∙2+2∙3+3∙4+ ∙ ∙ ∙+𝑘 𝑘+1 = 𝑘 𝑘+1 𝑘 ⋯(1)
MTU/MTH 101/ CHAPTER 7

57. MATHEMATICAL INDUCTION
11/19/2018
We then prove it for 𝑛=𝑘+1.
L. H. S=
1∙2+2∙3+3∙4+ ∙ ∙ ∙+𝑘 𝑘+1 +(𝑘+1)(𝑘+2)
= 𝑘(𝑘+1)(𝑘+2) 3 +(𝑘+1)(𝑘+2)
= (𝑘+1)(𝑘+2)(𝑘+3) 3
MTU/MTH 101/ CHAPTER 7

58. MATHEMATICAL INDUCTION
11/19/2018
R. H. S= (𝑘+1)(𝑘+2)(𝑘+3) 3 =L. H. S
So 𝑃(𝑘+1) is true whenever 𝑃(𝑘) is true. Therefore, the statement is true for all 𝑛∈ℕ by the principle of mathematical induction.
MTU/MTH 101/ CHAPTER 7

59. Exercise Use mathematical induction to prove that
11/19/2018
Use mathematical induction to prove that
∙ ∙ ∙+2𝑛−1= 𝑛 2
1 1∙ ∙ ∙ ∙ ∙+ 1 𝑛 𝑛+1 = 𝑛 𝑛+1
Prove that for all integers 𝑛:
15 is a factor of 2 4𝑛 −1.
MTU/MTH 101/ CHAPTER 7

60. MATHEMATICAL INDUCTION
11/19/2018
Example Prove that 7 2𝑛 +2 is a multiple of 3 for all natural numbers 𝑛. Solution Let 𝑃(𝑛) be the statement that 7 2𝑛 +2 is a multiple of 3. Consider 𝑛=1. 𝑃 1 =49+2=51 =3(17) ∴𝑃 1 is true
MTU/MTH 101/ CHAPTER 7

61. MATHEMATICAL INDUCTION
11/19/2018
Assume 𝑃 𝑘 is true.
Then there exists 𝑚∈ℤ such that
7 2𝑘 +2=3𝑚⋯(2)
Now for 𝑛=𝑘+1
7 2𝑘+2 +2= 7 2𝑘 ∙
=49(7 2𝑘 +2)−98+2
=49 3𝑚 −96
MTU/MTH 101/ CHAPTER 7

62. MATHEMATICAL INDUCTION
11/19/2018
=147𝑚−96 =3(49𝑚−32) =3𝑡 Where 𝑡=49𝑚−32 ∈ℤ. Hence it is true for 𝑛=𝑘+1. By the principle of mathematical induction 𝑃(𝑛) is true for all 𝑛∈ℕ
MTU/MTH 101/ CHAPTER 7

63. DEFINITION OF TERMS
11/19/2018
Event: An event here refers to a task or an experiment to perform. For example, below is the event of different arrangements of the letters of the word ‘FISH’
FISH FIHS FSIH FSHI FHIS FHSI
SFIH SFHI SHIF SHFI SIFH SIHF
ISFH ISHF IHSF IHFS IFSH IFHS
HISF HIFS HFIS HFSI HSIF HSFI
MTH 101/L8

64. Different arrangements of three letters using the F, I, S,H without repetitions are FIS FIH FHI FHS FSI FSH SIF SIH SFI SFH SHI SHF IFS IFH ISF ISH HIS IHF HIS HIF HSI HSF HFI HFS Different arrangements using two out of the four letters are: FI FS FH IF IS IH SI SF SH HI HS HF
11/19/2018
MTH 101/L8

65. Basic Counting Principles
11/19/2018
Suppose an event 𝐸 1 can occur in 𝑛 1 ways, a second event 𝐸 2 can occur in 𝑛 2 ways, and, following 𝐸 2 ; a third event 𝐸 3 can occur in 𝑛 3 ways, and soon. Then:
Sum Rule: If no two events can occur at the same time, then one of the events can occur in:
𝑛 1 + 𝑛 2 + 𝑛 3 + ∙ ∙ ∙𝑤𝑎𝑦𝑠
MTH 101/L8

66. SUM RULE
11/19/2018
Example A college has 3 different history courses, 4 different literature courses, and 2 different sociology courses. How many ways can a student choose one of the courses?
Solution
The number n of ways a student can choose just one of the courses =
= 9
MTH 101/L8

67. PRODUCT RULE
11/19/2018
Product Rule: If the events occur one after the other, then all the events can occur in the order indicated in:
𝑛 1 × 𝑛 2 × 𝑛 3 × 𝑤𝑎𝑦𝑠
Example A college has 3 different history courses, 4 different literature courses, and 2 different sociology courses. How many ways can a student choose one of each kind of courses?
MTH 101/L8

68. Product Rule
11/19/2018
Solution
The number of ways a student can choose one of each kind of courses
=3×4×2
=24
MTH 101/L8

69. FACTORIAL FUNCTION n!
11/19/2018
n! (n factorial) is the product of the positive integers from 1 to n inclusive.
𝑛!=1∙2∙3∙ 𝑛−2 𝑛−1 𝑛 OR
𝑛!=𝑛 𝑛−1 𝑛−2 ∙ ∙3∙2∙1
5!=5∙4∙3∙2∙1=120
1!=1
MTH 101/L8

70. FACTORIAL FUNCTION n! We also define 0!=1
11/19/2018
We also define 0!=1
Example: Evaluate 7!×6! 8!×2!
Solution
7!×6! 8!×2! = 7!×6×5×4×3×2×1 8×7!×2
=45
MTH 101/L8

71. PERMUTATIONS
11/19/2018
Any arrangement of a set of n objects in a given order is called a permutation of the objects.
TYPES OF PERMUTATIONS
Permutation of n different objects taken all at a time.
Permutation of n different objects taken r ≤𝑛 at a time.
MTH 101/L8

72. PERMUTATIONS 3. Permutation involving indistinguishable objects.
11/19/2018
3. Permutation involving indistinguishable objects.
Permutation with repetitions allowed.
Circular Arrangement
Conditional Permutations
MTH 101/L8

73. PERMUTATIONS Permutation of n different objects taken all at a time.
11/19/2018
Permutation of n different objects taken all at a time.
The number of ways n different objects can be arranged taken all at a time =𝑛!
Example How many ways can five students be arranged on a line?
Solution
The number of ways =5!=120
MTH 101/L8

74. PERMUTATIONS
11/19/2018
2. Permutation of n different objects taken r ≤𝑛 at a time.
The number of permutations of n objects taken r at a time will be denoted by
𝑛 𝑃 𝑟 , P(n, r) or nPr,
The following theorem applies.
MTH 101/L8

75. PERMUTATIONS Theorem 8.1 𝑃 𝑛, 𝑟 = 𝑛! 𝑛−𝑟 ! =𝑛 𝑛−1 𝑛−2 ∙ ∙ ∙(𝑛−𝑟+1)
11/19/2018
Theorem 8.1
𝑃 𝑛, 𝑟 = 𝑛! 𝑛−𝑟 !
=𝑛 𝑛−1 𝑛−2 ∙ ∙ ∙(𝑛−𝑟+1)
Example Find the number m of permutations of six objects, say, A, B, C, D, E, F, taking three at a time.
MTH 101/L8

76. PERMUTATIONS
11/19/2018
Solution The number of permutations of 6 different objects taken 3 at a time =𝑃 6,3 = 6! (6−3)! = 6! 3! = 6×5×4×3×21 3×2×1 =120
MTH 101/L8

77. PERMUTATIONS
11/19/2018
Example In how many ways can the 1st, 2nd and 3rd positions be awarded to a group of 10 students?
Solution
𝑃 10, 3 = 10! 7!
=10×9×8
=720
MTH 101/L8

78. PERMUTATIONS Permutation involving indistinguishable objects. Let
11/19/2018
Permutation involving indistinguishable objects.
Let
𝑃(𝑛; 𝑛 1 , 𝑛 2 , , 𝑛 𝑟 )
denote the number of permutations of objects(multi-set) of which 𝑛 1 are alike, 𝑛 2 are alike, . . ., 𝑛 𝑟 are alike.
MTH 101/L8

79. There are 5 letters with 3 letters alike.
PERMUTATIONS
11/19/2018
Theorem 8.3
𝑃 𝑛; 𝑛 1 , 𝑛 2 , , 𝑛 𝑟 = 𝑛! 𝑛 1 !× 𝑛 2 !×⋯× 𝑛 𝑟 !
Example How many five-letter words can be formed using letters from the word BABBY?
Solution
There are 5 letters with 3 letters alike.
MTH 101/L8

80. 𝑃 5;3 = 5! 3! =20 Example Find the number m of seven-letter words that can be formed using the letters of the word “BENZENE.” Solution There 7 objects of which 3 are alike (the three E’s), and 2 are alike (the two N’s). By Theorem 8.3, 𝑚=𝑃 7;3, 2 = 7! 3!2! = 7×6×5×4 2×1 =420
11/19/2018
MTH 101/L8

81. 4 Permutations with repetition allowed or Sampling with replacement
11/19/2018
4 Permutations with repetition allowed or Sampling with replacement
Here the element is replaced in the set S before the next element is chosen. Thus, each time there are n ways to choose an element (repetitions are allowed). The Product rule tells us that the number of such samples is:
MTH 101/L8

82. PERMUTATIONS 𝑛×𝑛×𝑛×∙ ∙ ∙ ×𝑛 = 𝑛 𝑟 𝑟
11/19/2018
𝑛×𝑛×𝑛×∙ ∙ ∙ ×𝑛 = 𝑛 𝑟 𝑟
Example How many 4-letter code words can be formed using the letters A, B, C, D, E and allowing repetition of letters?
Solution
The number is =5 4 =625
MTH 101/L8

83. PERMUTATIONS 5. Circular Arrangements
11/19/2018
5. Circular Arrangements
The number of ways of arranging 𝑛 different objects in a circle is 𝑛−1 !
If the circular object can be turned over e.g a ring or bead, the number of different arrangements is
𝑛−1 ! 2
MTH 101/L8

84. Example A family of 5 are to seat round a circular table for dinner
Example A family of 5 are to seat round a circular table for dinner. In how many ways can the family be seated?
Solution
The number of ways 5−1 !=4!=24
11/19/2018
MTH 101/L8

85. Example In how many ways can 10 boys take positions around a circular table if two particular boys must not be seated side by side?
Solution
Represent the total permutations without any restriction be P, the number of permutations when the boys seated together by 𝑃 𝑇 and the number of permutations when they are not together by 𝑃 𝑁 . Then
𝑃 𝑇 + 𝑃 𝑁 =𝑃
11/19/2018
MTH 101/L8

86. 𝑃=9! 𝑃 𝑇 =2×8! ∴ 𝑃 𝑁 =9!−2×8!=7×8!
11/19/2018
MTH 101/L8

87. PERMUTATIONS 6 Conditional Permutation
11/19/2018
6 Conditional Permutation
Example In how many ways can the letters of the word SHALLOW be arranged if
(a) the two L’s must be together;
(b) the two L’s must not be together;
(c) S must start the arrangement?
MTH 101/L8

88. The number of arrangements =6!=720.
Solution
(a) If the L’s must be together, we can lump the two L’s together as one; we need to arrange SHA LL OW.
The number of arrangements =6!=720.
(b) If the two L’s must not be together each of the L’s can occupy any of the starred positions in ∗S∗H∗A∗O∗W∗
11/19/2018
MTH 101/L8

89. PERMUTATIONS
11/19/2018
The first L can occupy any of the starred positions in 6 ways and the other L can occupy any of the remaining in 5 ways.
The letters S H A O W can be arranged in 5! ways. Therefore, the number of possible arrangements
6×5×5! 2! =1800
MTH 101/L8

90. (c) If S must start the arrangement, then the first letter can be chosen in 1 way(there is only one S), and the remaining six letters in 6! 2! ways(note that the two L’s are indistinguishable and the accounts for the 2!). Therefore, the number of arrangement
=1× 6! 2! =360
11/19/2018
MTH 101/L8

91. Q1 How many numbers greater than can be formed using the digits 4, 5, 6, 7 and 8 if no digit is repeated?
Q2 How many even 5-digit numbers can be formed using the digits 1, 2, 3, 4, 5, 6, and 7 if there is no repetition?
Q3 How many 3-digit numbers greater than 300 can be formed with the digits 1,2 ,3 , 4, 5, 6, if no digit is repeated in any number?
Q4 How many rearrangements of the word MATHEMATICS contain the object MAT exactly once?
11/19/2018
MTH 101/L8

92. (a) all the six flags are to be used?
Q5 6 flags, consisting one blue, 3 identical red and two identical green flags, are to be arranged in a row. In how many ways can this be done if
(a) all the six flags are to be used?
(b) at least 5 of the flags are to be used?
11/19/2018
MTH 101/L8

93. Partitions of 𝒏 Distinct Objects
11/19/2018
Theorem 5 The number of partitions of n distinct objects into classes, where the first class contains 𝑟 1 objects, the second 𝑟 2 and so on with the last class containing 𝑟 𝑘 objects, where 𝑟 𝑟 2 + ∙ ∙ ∙+ 𝑟 𝑘 =𝑛 is
𝑛! 𝑟 1 ! 𝑟 2 !∙ ∙∙ 𝑟 𝑘 !
MTH 101/L8

94. The number of partitions = 9! 3!4!2! = 9×8×7×6×5 3×2×1×2 =1260
Example In how many ways can 9 students be divided into three groups of 3, 4 and 2?
Solution
The number of partitions
= 9! 3!4!2! = 9×8×7×6×5 3×2×1×2 =1260
11/19/2018
MTH 101/L8

95. COMBINATION Combination deals with selection.
11/19/2018
Combination deals with selection.
In selections, the order in which the objects are chosen does not matter. A combination/selection of 𝑟 objects from a set of 𝑛 different objects will be denoted 𝑛 𝐶 𝑟 or 𝐶 𝑛, 𝑟 or 𝑛 𝑟 .
Theorem 𝐶 𝑛, 𝑟 = 𝑃(𝑛, 𝑟) 𝑟! = 𝑛! 𝑛−𝑟 !𝑟!
MTH 101/L8

96. EXAMPLE Find the number of ways of selecting 3 out of the 4 objects, A, B, C, D.
ABC : ABC, ACB, BAC, BCA, CAB, CBA
ABD: ABD, ADB, BAD, BDA, DAB, DBA
ACD: ACD, ADC, CAD, CDA, DAC, DCA
BCD: BDC, BDC, CBD, CDB, DBC, DCB
Thus the number of combinations multiplied by 3! gives us the number of permutations; that is,
𝐶 4, 3 ×3!=𝑃 4, 3 or 𝐶 4, 3 = 𝑃(4, 3) 3!
11/19/2018
MTH 101/L8

97. COMBINATION
11/19/2018
Example From a shelf containing 12 different toys, a child is permitted to select 3. In how many ways can this be done?
Solution
The required number is
𝐶 12, 3 = 12! 12−3 !3! = 12∙11∙10 3∙2∙1 =220
MTH 101/L8

98. 6 men can be chosen from 10 men in 10 6 = 10(9)(8)7 4(3)(2)1 =210 ways
Example In how many ways can a committee of 6 men and 6 women be chosen from a council of 10 men and 12 women?
Solution
6 men can be chosen from 10 men in
10 6 = 10(9)(8)7 4(3)(2)1 =210 ways
11/19/2018
MTH 101/L8

99. 12 6 = 12(11)(10)(9)(8)7 6(5)(4)(3)(2)1 =924 ways
6 women can be chosen from 12 women in
12 6 = 12(11)(10)(9)(8)7 6(5)(4)(3)(2)1 =924 ways
Therefore, the committee can be chosen in
210×924= ways.
Example How many committees of five can be formed from 8 men and 4 women if each committee is to have at least 2 women?
11/19/2018
MTH 101/L8

100. COMBINATION
11/19/2018
Solution: There are three types of committee which can be formed:
3 men and 2 women
2 men and 3 women
1 man and 4 women
The number of committees
=𝐶 8, 3 ×𝐶 4, 2 =56 6 =336
MTH 101/L8

101. (ii) The number of committees =𝐶 8, 2 ×𝐶 4, 3 =28 4 =112 (iii) The number of committees =𝐶 8, 1 ×𝐶 4, 4 =8 1 =8 Hence the total number of committees = =456
11/19/2018
MTH 101/L8

102. (a) The six guests may be selected in 𝐶 10, 6 = 10! 4!6! =210 ways.
Example A lady gives a dinner party for six guests. (a) In how many ways may they be selected from among ten friends? (b) In how many ways if two of her friends will not attend the party together?
Solution
(a) The six guests may be selected in 𝐶 10, 6 = 10! 4!6! =210 ways.
11/19/2018
MTH 101/L8

103. COMBINATION
11/19/2018
(b) If the two friends attend together, the number of ways the six guests may be selected
=𝐶 8, 4 = 8! 4!4! =70
So if the two friends will not attend the party together, the number of ways
=210−70=140.
MTH 101/L8

104. Prove that
𝑛 𝑟 + 𝑛 𝑟+1 = 𝑛+1 𝑟+1
𝑛 𝑟 = 𝑛 𝑛−𝑟
11/19/2018
MTH 101/L8

105. Theorem (The Binomial Theorem)
11/19/2018
If 𝑎 and 𝑏 are any real numbers and 𝑛 is a positive integer, then
(𝑎+𝑏) 𝑛
=𝑎 𝑛 +𝑛 𝑎 𝑛−1 𝑏+ 𝑛(𝑛−1) 2! 𝑎 𝑛−2 𝑏 ∙ ∙ ∙+𝑛𝑎 𝑏 𝑛−1 + 𝑏 𝑛
MTH 101/L8

106. =𝑎 𝑛 + 𝑛 1 𝑎 𝑛−1 𝑏+ 𝑛 2 𝑎 𝑛−2 𝑏 2 + ∙ ∙ ∙+ 𝑛 𝑛−1 𝑎 𝑏 𝑛−1 + 𝑏 𝑛 = 𝑟=0 𝑛 𝑛 𝑟 𝑎 𝑛−𝑟 𝑏 𝑟 Example: Expand 𝑎+𝑏 4 . Example: Expand 3𝑥+2 𝑦 2 5 and simplify term by term. Example: Expand 𝑥− 1 𝑥 6
11/19/2018
MTH 101/L8

107. Example: Find the third term in the expansion of 2−3𝑥 7 in ascending power of x. Example: Obtain the first four terms of the expansion of 1+2𝑥 12 in the ascending power of 𝑥 and use the expansion to find the value of correct to 4 decimal places.
11/19/2018
MTH 101/L8

108. The Binomial Theorem for 𝑛ϵℝ and |𝑏|<|𝑎|
11/19/2018
(𝑎+𝑏) 𝑛 = 𝑎 𝑛 +𝑛 𝑎 𝑛−1 𝑏+ 𝑛(𝑛−1) 2! 𝑎 𝑛−2 𝑏 2 + ∙ ∙ ∙ +𝑛𝑎 𝑏 𝑛−1 + 𝑏 𝑛
Example: Write the first five terms of 𝑎+𝑏 −4 , |𝑏|<|𝑎|.
Example: Obtain the first five terms of 1−3𝑥 −3
Example: Write down the first 4 terms and simplify term by term 1− 𝑥 3 − 2 3 , |𝑥|<1.
MTH 101/L8

109. Complex Numbers Solve the equation 𝑥 2 −2𝑥+5=0 Solution
11/19/2018
Solve the equation
𝑥 2 −2𝑥+5=0
Solution
𝑥= −𝑏± 𝑏 2 −4𝑎𝑐 2𝑎
⇒𝑥= 2± −2 2 −4(1)(5) 2(1)
𝑥= 2± 4−20 2
MTU/100L/MTH 101/CHAPTER TEN

110. 𝑥= 2± −16 2 𝑥= 2±4 −1 2 𝑥= 2±4𝑖 2 , where 𝑖= −1 𝑥=1±2𝑖
11/19/2018
MTU/100L/MTH 101/CHAPTER TEN

111. With the symbol 𝑖= −1 , it then implies that −5 = 5 × −1 = 5 𝑖 −16 = 16 × −1 =4𝑖 Properties of the symbol 𝑖 2 =−1, and for higher integral powers we have 𝑖 3 = 𝑖 2 ×𝑖=−1×𝑖=−𝑖 𝑖 4 = 𝑖 2 × 𝑖 2 =−1×−1=1 𝑖 5 = 𝑖 4 ×𝑖=𝑖
11/19/2018
MTU/100L/MTH 101/CHAPTER TEN

112. Definition
Complex Number (ℂ) : a number of the form 𝑎+𝑏𝑖, where 𝑎 and 𝑏 are real numbers and 𝑖 2 =−1.
The first term 𝑎 is called the real part and the second term 𝑏 is called the imaginary part.
Two complex numbers 𝑎+𝑏𝑖 and 𝑐+𝑑𝑖 are said to be equal if and only if 𝑎=𝑐 and 𝑏=𝑑.
11/19/2018
MTU/100L/MTH 101/CHAPTER TEN

113. Algebra of Complex Numbers
Evaluate the following:
3+2𝑖 2 − 1−𝑖 2
5−4𝑖 3+2𝑖
Solution
3+2𝑖 2 =(3+2𝑖)(3+2𝑖)
=9+6𝑖+6𝑖+2 𝑖 2
=9+12𝑖−2
=7+12𝑖
11/19/2018
MTU/100L/MTH 101/CHAPTER TEN

114. Example: Solve the equation 𝑧−3𝑖 4+𝑖 =5𝑖+2𝑧
5−4𝑖 3+2𝑖 = 5−4𝑖 3+2𝑖 × 3−2𝑖 3−2𝑖
= 15−10𝑖−12𝑖−8 9+4
= 7−22𝑖 13
Example: Solve the equation
𝑧−3𝑖 4+𝑖 =5𝑖+2𝑧
11/19/2018
MTU/100L/MTH 101/CHAPTER TEN

115. Example Find the quadratic equation whose roots are 3+𝑖 and 3−𝑖.
Solution
Sum of the roots = 3+𝑖 + 3−𝑖 =6
Product of the roots = 3+𝑖 3−𝑖 =9+1=10
The equation is 𝑥 2 −6𝑥+10=0
11/19/2018
MTU/100L/MTH 101/CHAPTER TEN

116. The Argand Diagram
11/19/2018
Argand diagram is a pictorial representation of complex numbers.
The complex number 𝑧=𝑥+𝑦𝑖 may be represented graphically by the point P whose rectangular coordinates are (𝑥, 𝑦).
MTU/100L/MTH 101/CHAPTER TEN

117. 11/19/2018
MTU/100L/MTH 101/CHAPTER TEN

118. 11/19/2018
MTU/100L/MTH 101/CHAPTER TEN

119. The Argand Diagram
11/19/2018
The length OP = 𝑟 is called the modulus or absolute value of the complex number 𝑧. It is also written as |𝑧|. By Pythagoras’ theorem,
𝑟= 𝑧 = 𝑥 2 + 𝑦 2
Note that
𝑧∙ 𝑧 = 𝑥+𝑖𝑦 𝑥−𝑖𝑦 = 𝑥 2 + 𝑦 2 = |𝑧| 2
MTU/100L/MTH 101/CHAPTER TEN

120. Polar or Trigonometric Form of Complex Numbers
Example The modulus of 3+4𝑖
= = 25 =5
Polar or Trigonometric Form of Complex Numbers
11/19/2018
MTU/100L/MTH 101/CHAPTER TEN

121. Therefore 𝑧=𝑥+𝑖𝑦=𝑟 cos 𝜃 +𝑖𝑟 sin 𝜃 =𝑟( cos 𝜃 +𝑖 sin 𝜃 )
The angle 𝜃 is called the amplitude or argument of 𝑧.
From the diagram, we can see that
𝑥=𝑟 cos 𝜃 ;𝑦=𝑟 sin 𝜃
tan 𝜃 = 𝑦 𝑥 so that 𝜃= tan −1 𝑦 𝑥
Therefore
𝑧=𝑥+𝑖𝑦=𝑟 cos 𝜃 +𝑖𝑟 sin 𝜃
=𝑟( cos 𝜃 +𝑖 sin 𝜃 )
11/19/2018
MTU/100L/MTH 101/CHAPTER TEN

122. 𝑧=𝑟( cos 𝜃 +𝑖 sin 𝜃 ) is called the polar form of 𝑧.
Example Write −4 3 −4𝑖 in polar form.
Solution
𝑟= − (−4) 2 = = 64 =8
11/19/2018
MTU/100L/MTH 101/CHAPTER TEN

123. Note that the complex number is in the third quadrant.
So 𝜃= tan −1 −4 − = tan − =210°
Therefore, the polar form is
𝑧=8( cos 210°+𝑖 sin 210° )
11/19/2018
MTU/100L/MTH 101/CHAPTER TEN

124. Example Write 8( cos 210°+𝑖 sin 210° ) in rectangular form
Example Write 8( cos 210°+𝑖 sin 210° ) in rectangular form. Solution 8 cos 210°+𝑖 sin 210° =8 − 3 2 − 1 2 𝑖 =−4 3 −4𝑖
11/19/2018
MTU/100L/MTH 101/CHAPTER TEN

125. PRINCIPAL argument
11/19/2018
The principal argument of a complex number, denoted by 𝐴𝑟𝑔(𝑧) is the value of the argument 𝜃 in the range (−𝜋, 𝜋] i.e. −𝜋<𝐴𝑟𝑔(𝑧)≤𝜋
Example The principal argument of −4 3 −4𝑖 is −150°= − 5 6 𝜋
MTU/100L/MTH 101/CHAPTER TEN

126. Multiplication and Division in Polar Form
11/19/2018
Let 𝑧 1 = 𝑟 1 ( cos 𝜃 1 +𝑖 sin 𝜃 1 ) and 𝑧 2 = 𝑟 2 cos 𝜃 2 +𝑖 sin 𝜃 2 1. Multiplication 𝑧 1 ∙ 𝑧 2 = 𝑟 1 𝑟 2 cos 𝜃 1 + 𝜃 2 +𝑖 sin 𝜃 1 + 𝜃 2 So, 𝑧 1 𝑧 2 = 𝑟 1 𝑟 2 and 𝑎𝑟𝑔 𝑧 1 𝑧 2 = 𝜃 1 + 𝜃 2 = 𝑎𝑟𝑔 𝑧 1 +𝑎𝑟𝑔⁡( 𝑧 2 )
MTU/100L/MTH 101/CHAPTER TEN

127. Example 3 cos 30° +𝑖 sin 30° ∙5 cos 70° +𝑖 sin 70° =15( cos 100° +𝑖 sin 100° ) 2. Division 𝑧 1 𝑧 2 = 𝑟 1 ( cos 𝜃 1 +𝑖 sin 𝜃 1 ) 𝑟 2 ( cos 𝜃 2 + 𝑖 sin 𝜃 2 ) = 𝑟 1 𝑟 2 [ cos 𝜃 1 − 𝜃 2 +𝑖 sin ( 𝜃 1 − 𝜃 2 ) ]
11/19/2018
MTU/100L/MTH 101/CHAPTER TEN

128. So, 𝑧 1 𝑧 2 = 𝑟 1 𝑟 2 , 𝑎𝑟𝑔 𝑧 1 𝑧 2 = 𝜃 1 − 𝜃 2 = 𝑎𝑟𝑔 𝑧 1 −𝑎𝑟𝑔⁡( 𝑧 2 )
11/19/2018
MTU/100L/MTH 101/CHAPTER TEN

129. Applications of De Moivre’s Theorem
11/19/2018
If 𝑛 is any rational number and
𝑧=𝑟( cos 𝜃 +𝑖 sin 𝜃 ), then
𝑧 𝑛 = 𝑟( cos 𝜃 +𝑖 sin 𝜃 ) 𝑛 = 𝑟 𝑛 ( cos 𝑛𝜃 +𝑖 sin 𝑛𝜃 )
Applications of De Moivre’s Theorem
Example Evaluate −𝑖 10
MTU/100L/MTH 101/CHAPTER TEN

130. Let 𝑧= 3 −𝑖. 𝑧 = = 3+1 =2 arg 𝑧 = tan −1 − 1 3 =330° 𝑧=2( cos 330° +𝑖 sin 330° ) ∴ 𝑧 10 = 2 10 cos ° +𝑖 sin ° =1024( cos 𝑖 sin 3300 )
11/19/2018
MTU/100L/MTH 101/CHAPTER TEN

131. =1024( cos 60° +𝑖 sin 60° )
= 𝑖 = 𝑖
Example: Show that if 𝑧= cos 𝜃 +𝑖 sin 𝜃 and 𝑚 is a positive integer then
1. sin 2𝜃 =2 sin 𝜃 cos 𝜃
𝑧 𝑚 + 1 𝑧 𝑚 =2 cos 𝑚𝜃
𝑧 𝑚 − 1 𝑧 𝑚 =2𝑖 sin 𝑚𝜃
11/19/2018
MTU/100L/MTH 101/CHAPTER TEN

132. ROOTS OF COMPLEX NUMBERS
11/19/2018
Let 𝑧 𝑛 =𝜆 , where λ is a complex number. The equation has exactly 𝑛 distinct roots which is given as 𝑧= 𝜆 1 𝑛 = 𝑟 1 𝑛 cos 𝜃+2𝑘𝜋 𝑛 +𝑖 sin 𝜃+2𝑘𝜋 𝑛 𝑘=0, 1, 2, ⋯,𝑛−1, 𝜃 is the argument and 𝑟 the modulusof 𝜆.
MTU/100L/MTH 101/CHAPTER TEN

133. Example Solve the equation 𝑧 5 −4=−4𝑖. Solution 𝑧 5 =4−4𝑖 𝑧= 5 4−4𝑖
arg 4−4𝑖 = tan −1 − 4 4 =315°= 7 4 𝜋
𝑟= = 32 =4 2
11/19/2018
MTU/100L/MTH 101/CHAPTER TEN

134. 𝑘=0, 1, 2, 3, 4 The five roots are 𝑧 0 = 2 cos 63°+𝑖 sin 63°
The fifth roots are given as
𝑧 𝑛 = cos 315°+360°𝑘 5 +𝑖 sin 315°+360°𝑘 5
= 2 cos 63°+72°𝑘 +𝑖 sin (63°+72°𝑘)
𝑘=0, 1, 2, 3, 4
The five roots are
𝑧 0 = 2 cos 63°+𝑖 sin 63°
𝑧 1 = 2 cos 135°+𝑖 sin 135°
11/19/2018
MTU/100L/MTH 101/CHAPTER TEN

135. Roots of Unity 𝑧 2 = 2 cos 207°+𝑖 sin 207° 𝑧 3 = 2 cos 279°+𝑖 sin 279°
The case when λ = 1 is called n-th roots of unity.
Note that 1= cos 0 +𝑖 sin 0 .
So arg 1 =0.
11/19/2018
MTU/100L/MTH 101/CHAPTER TEN

136. Roots of Unity
11/19/2018
If 𝑧 𝑛 =1, the 𝑛−th roots are given as 𝑧 𝑛 = 1 1 𝑛 = cos 2𝑘𝜋 𝑛 +𝑖 sin 2𝑘𝜋 𝑛 𝑘=0, 1, 2, ⋯,𝑛−1.
MTU/100L/MTH 101/CHAPTER TEN

137. Example The cube roots of unity are 𝑧 0 =1
𝑧 1 = cos 120°+𝑖 sin 120° =− 1 2 +𝑖
𝑧 2 = cos 240°+𝑖 sin 240° =− 1 2 −𝑖
11/19/2018
MTU/100L/MTH 101/CHAPTER TEN

138. Example Find all the fourth roots of −16. Solution Modulus of −16=16
−16=−16+0𝑖
arg −16 = tan − −16 =𝜋
4 −16 = cos 𝜋+2𝑘𝜋 4 +𝑖 sin 𝜋+2𝑘𝜋 4
𝑘=0, 1, 2, 3
11/19/2018
MTU/100L/MTH 101/CHAPTER TEN

139. 𝑧 1 =2 cos 3𝜋 4 +𝑖 sin 3𝜋 4 =2 − 2 2 +𝑖 2 2 =− 2 +𝑖 2
The four roots are
𝑧 0 =2 cos 𝜋 4 +𝑖 sin 𝜋 4 = 𝑖 = 2 +𝑖 2
𝑧 1 =2 cos 3𝜋 4 +𝑖 sin 3𝜋 4 =2 − 𝑖 =− 2 +𝑖 2
11/19/2018
MTU/100L/MTH 101/CHAPTER TEN

140. 𝑧 2 =2 cos 5𝜋 4 +𝑖 sin 5𝜋 4 =2 − 2 2 −𝑖 2 2 =− 2 −𝑖 2
11/19/2018
MTU/100L/MTH 101/CHAPTER TEN
          ALGEBRA by Mr. Taiwo A. is licensed under a Creative Commons Attribution-Non Commercial 4.0 International License

141. 11/19/2018
ALL THE BEST
MTU/100L/MTH 101/CHAPTER TEN