Advanced Elementary Algebra - UCSD Mathematics

Advanced Elementary Algebra

Guershon Harel & Sara-Jane Harel ©2011

Advanced Elementary Algebra Guershon Harel University of California, San Diego With Sara-Jane Harel

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Advanced Elementary Algebra The unit consists of three parts, which mark three evolving stages in the development of its content. Your work in this unit revolves around four kinds of activities: 1. In-Class Problems (ICP) 2. Questions (Q) and Answers (A) 3. Probes (P) 4. Homework Problems (HP) ICP: The ICPs serve two purposes. Some ICPs serve as a starting point for a new idea introduced in a lesson; their goal is to bridge new knowledge to be learned with the knowledge you have already learned. Other ICPs aim at enhancing your ability to read and understand mathematical text. In these problems, you are asked to study a solution to a problem or a proof of an assertion. • To study a mathematical text means to understand the underlying ideas in the text and the mathematical reason for each claim made in the text. • To test your understanding, reproduce from memory the complete solution or the proof, and, in doing so, use your own words and choose your own symbols (different from those used in the text). P: Often, as you read a mathematical text, you will be asked to respond to probes (marked by P). They appear in the form of queries, such as “why,” “how,” and “explain.” It is necessary that you respond to these queries before continuing reading. This will help you better understand the text and, for a long run, improve your ability to read mathematical texts. Q&A: The aim of the Questions is to motivate new concepts and ideas, by helping you see that new knowledge always come about out of a need to solve a problem, resolve a puzzle, explain a phenomenon, etc. Each Question is then followed by an Answer, which often contain probes. HP: Last, but not least, are the Homework Problems (HP) on each of the lesson. To master a mathematical subject, one must practice the reasoning particular to that subject. The ultimate goal of these problems is to make you a better mathematical reasoner, by gradually improving your ability to prove assertion, solve problems, think in general terms, and be fluent in computing with understanding. Some of the HPs aim at extending your knowledge beyond what is presented in a particular lesson, to prepare you for subsequent lessons.

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Part I Lesson 1 ICP 1: The sum of the volumes of two cubes is 16, and the product of the side of one cube by the side of the other cube is 4. What are their dimensions? 27 9 and the product of the sides is ICP 2: Repeat Problem 1, where the sum of the volumes is 4 4 ICP 3: We can create as many problems like Problems 1 and 2 as we want by varying the volumes of the cubes and the product of their sides. a. Create two such new problems and solve them. b. Is it possible that among such problems there are ones with no solutions? ICP 4: Study the following solution to Problem 3b, and compare it to your own solution. Solution Let u and v be the sides of the two cubes. Let P and Q be two positive numbers, so that uv = P (*) u 3 + v3 = Q  P To solve this system, we use the substitution, v = . We get the equation u 3

P u +  = Q. u 3

Multiplying both sides of this equation by u 3 , we get

(u )

3 2

− Qu 3 + P 3 = 0.

Solving for u 3 , we get 2

Q Q u3 = ±   − P3 . 2 2 By symmetry, 2

Q Q v = ±   − P3 . 2 2 3

Since u 3 + v3 =, Q 2

Q Q u3 = +   − P3 2 2

3



2

Q Q v = −   − P3 2 2 3

Hence: 2

Q Q 3 u= +   − P3 2 2 2

Q Q v = −   − P3 2 2 For u and v to be meaningful numbers, P and Q must satisfy the condition 3

2

Q 3   −P ≥0 2 Note that with this condition, u and v are positive numbers (P: why?). In summary, the answer to ICP 3b is: 1. We can create as many problems like Problems 1 and 2 by varying P and Q over the positive numbers. 2. The sides of the two cubes, u and v , must of course be positive numbers. This is 2

Q insured by the condition   − P 3 ≥ 0 . 2 2

Q 3. Hence, if either P < 0 , Q < 0 , or   − P 3 < 0 , the problem does not have a 2 solution.  Homework Problems on Lesson 1 System (*) in this lesson belongs to a family of systems involving products and cubes of unknowns. Systems (a)-(f) below belong to the same family. The goal of this homework is to learn how to solve this kind of systems. 1. Solve the following systems. If you encounter a difficulty in solving any of these systems, describe the difficulty. If you can, discuss the difficulty with a classmate. 1  uv = 32 a.  1 u 3 + v3 =  16 4uv = 25 b.  3 3 125 4u + 4v = 4



2uv = 1 c.  3 3 −7 8u − 8v = 1  uv = 27 d.  2 u 3 + v3 = −  9 3 −2uv = e.  3 3 170 u + 8v =  uv = −2  f.  u + v u 3 + v3 = 32 

5



Lesson 2 Q:

A:

It is likely that you solved systems (a)-(e) by substitution, as in the solution to ICP 3b. When we apply the same method to system (f), we get a polynomial equation of degree 6. We know of no technique for solving this type of equation. So, what do we do? As often happens, when our solution approach fails, we return to the problem and try to gather more cues, as you will now see.

Solution To avoid confusion with other systems in future HPs, we first rename system (f) as system (**)  uv = −2  (**) u + v u 3 + v3 = 32  A close look at this system reveals that it contains three expressions, u 3 + v3 , u + v , and uv , which make up the well-known identity: (u + v)3 − u 3 − 3u 2 v − 3vu 2 − v 3 = 0 These expressions can be seen clearly when we factor out uv : (1) (u + v)3 − 3uv(u + v) − (u 3 + v3 ) = 0 st From the system’s 1 equation, we have (2) uv = −2(u + v) . nd And from the system’s 2 equation, we have (3) u 3 + v3 =. 32 Substituting (2) and (3) in (1), we get an equation of a single variable, u + v : (u + v)3 − 3 ( −2(u + v) ) (u + v) − 32 = 0 Simplifying, we get (u + v)3 + 6(u + v) 2 − 32 = 0 For convenience, let u + v = x . Equation (4) is then: (4)

x 3 + 6 x 2 − 32 = 0 The last equation can easily be solved. By trying a few values or by using the Rational Root Theorem, we find that x = 2 is a root of this equation. To find the other roots we divide x 3 + 6 x 2 − 32 into x − 2 . The division yields the polynomial x 2 + 8 x + 16 . x = −4 is the only root of this polynomial. Hence, we have two possible values for u + v : u + v = 2 or u + v =−4 . Case 1: u + v = 2. In this case, uv = −4 (by equation (2)), and so we have the system 6



2 u + v =  uv = −4 Solving by substitution, we get the quadratic equation u 2 − 2u − 4 =. 0 Its roots are: u = 1 + 5 and u = 1 − 5 . By substitution (or by symmetry) we get v = 1 − 5 and v = 1 + 5 , respectively. Case 2: u + v =−4 . In this case, uv = 8 , and so we have the system u + v =−4  uv = 8 Solving by substitution, we get the quadratic equation u 2 + 4u + 8 =, 0 with no roots. Conclusion: System (**) has two solutions: u= 1 + 5, v = 1− 5 or u= 1 − 5, v = 1+ 5 .  The solution to system (**) we have just found is based on identity (1). For this reason we call this solution method, completing the cube. Homework Problems on Lesson 2 1: Solve systems (a)-(f). If you encounter difficulty in solving any of these systems, describe the difficulty. If you can, discuss the difficulty with a classmate. 1  10uv  5u − 6v = 3 a.  34  1 u 3 − 27 v 3 = −  8 125 5 uv + u − v =−20 b.  3 3 100 u − v = uv + u − v =−2 c.  3 3 −8 u − v = uv   2 − 3(u + v) = 1 d.  u 3 + v3 − 2uv = 50 

7



8 uv(u + v) = e.  3 3 −31 u + v − 2u − 2v = f.

uv   2 − 3(u + v) = −2  u 3 + v3 − 2uv = −72 

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Lesson 3 Your attempt to solve system (e) in the HP on Lesson 2 likely resembled the following. [Again, we rename the system as system (***), just to avoid confusion with other systems in the unit.] 8  uv = (***) u+v  u 3 + v3 − 2(u + v) = −31  (u + v)3 − (u 3 + v 3 ) − 3uv(u + v) = 0 8 (u + v)3 − 2(u + v) + 31 − 3 ⋅ (u + v) = 0 u+v (u + v)3 − 2(u + v) + 31 − 24 = 0 (u + v)3 − 2(u + v) + 7 = 0

x= u + v

x − 2x + 7 = 0. Attempts to guess a solution to this equation or to apply the Rational Root Theorem, as we successfully did for the earlier systems, would most likely fail. This raises the following question: 3

Q: A:

Is there a formula for solving the general cubic equation, ax 3 + bx 2 + cx + d = 0? The answer to this question begins here and will be completed in Lesson 6. Recall how we developed the quadratic formula. We arrived at this formula by the method of Completing the Square. We transformed the quadratic equation ax 2 + bx + c = 0 into an b , easily solvable equation of the form, A( x + T ) 2 + L = 0 , where A = a , T = 2a 4ac − b 2 . The solution to this equation is the quadratic formula: L= 4a L b 2 − 4ac b −b ± b 2 − 4ac . x =± − − T =± − = A 4a 2 2a 2a It is tempting to try a similar approach by transforming the cubic equation

ax3 + bx 2 + cx + d = 0 into an equation of the form A( x + T )3 + L = 0 , which is easy to solve. Unfortunately, this cannot always be done. The reason is simply because such A , T , and L do not always exist. Here is why: Expanding the left-hand side of the last equation, we get A( x + T )3 + L= Ax 3 + 3 ATx 2 + 3 AT 2 x + AT 3 + L . In order for A( x + T )3 + L to be equal to ax3 + bx 2 + cx + d for all values of x , it is necessary that 9



A=a 3AT = b

3AT 2 = c AT 3 + L = d Such T does not always exist. For example, for a = 1 , b = 0 , and c = 1 there is no such T (P: why?). In what follows, we will learn a different approach. The idea behind the new approach is this: the Completing the Cube Method, which we used to solve system (**), when combined with the substitution method, which we used to solve system (*), leads to an interesting and important observation. We start with an example: ICP 5: Understand the solution below to the cubic equation, x3 − 2 x + 7 = 0. Solution The idea of the solution is to turn this equation into a system of equations like system (*), which we know how solve. We do so by two key steps we are familiar with by now: (a) partially reversing the process that led us to this equation (see the opening of this lesson) and (b) using the identity, (u + v)3 − u 3 − 3u 2 v − 3vu 2 − v 3 = 0. Here is how: Let (1) x= u + v . Cube both sides of this equation: 3 x= (u + v)3 . Expand

x3 = u 3 + v3 + 3u 2 v + 3v 2u Factor uv x 3 = u 3 + v 3 + 3uv(u + v) .

Substitute (1) in the last equation

x3 = u 3 + v3 + 3uvx Rearrange (2)

x 3 − 3uvx − ( u 3 + v3 ) = 0.

Before proceeding, pause and explain why for any choice of u and v , the sum, x= u + v , is a solution to equation (2). Now attend to equation (2) and our equation, x3 − 2 x + 7 = 0 . These equations are identical for u and v that satisfy the system 2  uv = 3  u 3 + v3 = −7  10



Before proceeding, pause again and explain why for any u and v that solve this system,

0. x= u + v is a solution to our equation, x3 − 2 x + 7 = This system has the same form as system (*), which we are familiar with from Lesson 1. Substitution v =

2 from the first equation into the second equation, we get a quadratic 3u

equation in u 3 : 3

 2  u +  = −7  3u  3

33 ( u 3 ) + 7 ⋅ 33 u 3 + 23 = 0 2

Its solutions are:

u=

3

−63 − 3873 −63 + 3873 and v = 3 18 18

u=

3

−63 + 3873 −63 − 3873 and v = 3 18 18

Or

Hence x =

3

−63 + 3873 3 −63 − 3873 is a solution to our cubic equation. + 18 18

 The approach we have just used to solve a particular cubic equation includes key ideas for developing a general formula for cubic equations, as you will see in the Homework Problems on Lessons 3-5. Homework Problems on Lesson 3 1: Find at least one root to the following cubic equations and compare the root you found with the root provided (in parentheses). For equation where you cannot find a root, describe the difficulty you encountered and share it with a classmate if possible. a.

x3 + 6 x − 1 =0

(x=

3

( x = 2 3)

b. x 3 − 9 x − 6 3 = 0 c. 2 x3 + 9 x + 3 = 0

1 + 33 + 3 1 − 33 ) 3 2

(x=

11

3

3 7 −3 − 3 3 7 +3 ) 3 4



d. 4 x3 − 3 x + 2 = 0

(x=

e. 2 x3 + 4 x − 1 =0

(x=

f.

3x3 − 3x − 7 = 0

(x=

g.

x3 − 3 3x − 3 − 3 = 0

(= x

h. x 3 − 3 35 x − 37 = 0

(x=

i.

x3 − 3 2 x − 2 − 4 = 0

j.

(= x x 3 + 3 35 x + 5 5 − 7 7 = 0

(x =

3

3−2 − 3 3+2 ) 2

3

9 + 465 + 3 9 − 465 ) 3 36

3

21 + 429 + 3 21 − 429 ) 3 18

6

3+ 3 3)

3

3

37 + 1 + 3 37 − 1 ) 3 2 2 2 +2 + 3 2− 2 ) 7− 5)

3 0 2: Find a general solution to the cubic equation, x + Ax + B =

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Lesson 4 ICP 6: In the last HP, you were asked to solve the general cubic equation, x 3 + Ax + B = 0. Compare your solution to the following solution: Solution We set (1) x= u + v Cubing both sides and rearranging, we get: 3 (u + v)3 x=

x3 = u 3 + v3 + 3u 2 v + 3v 2u x 3 = u 3 + v 3 + 3uv(u + v)

x3 = u 3 + v3 + 3uvx

x3 − 3uvx − ( u 3 + v3 ) = 0 By setting (2) (3)

−3uv = A 3 3 B −(u + v ) =

we get our cubic equation, x 3 + Ax + B = 0. This implies that for any u and v that satisfy conditions (2) and (3) , x= u + v is a solution to the cubic equation x 3 + Ax + B = 0. Condition (2) and (3) form the system uv = P  3 3 Q u + v = where A and Q = − B . 3 We use substitution and solve for u and v , as we did for system (*) (in Lesson 1). We get P= −

Q −B B Q  −B   − A   B  A u= +   − P3 = 3 +   −  =3 − +   +   2 2 2 2  2   3  2 3 2

2

3

2

3

3

Q −B B Q  −B   − A   B  A v= −   − P3 = 3 −   −  =3 − −   +   2 2 2 2  2   3  2 3 2

2

3

Hence, a solution to the cubic equation x 3 + Ax + B = 0 is

13

3

2

3



2

3

2

3

B B  B  A  B  A x =u + v = − +   +   + 3 − −   +   2 2 2 3 2 3 . 3

 Homework Problems on Lesson 4 1: Find at least one root for the following cubic equations: As before, if you encounter a difficulty in solving any of these systems, describe the difficulty. a.

(x =

x3 + 3x − 4 6 = 0

b. x3 − 6 6 2 x + 6 2 = 0

3

2 6 −5 + 3 2 6 +5 )

( x =3 4 − 4 2 − 3 2 2 + 4 ) (x =

3

18 − 2 )

d. x3 + 15 x + 20 = 0

(= x

3

5 − 3 25 )

e.

x3 + 12 x − 30 = 0

(x =

3

32 − 3 2 )

f.

6 x3 + 15 x + 5 = 0

c.

x 3 + 6 3 18 x − 10 = 0

( x =3 −

5 5 13 3 5 5 13 ) + − + 12 12 3 12 12 3

g. 2 x3 + 9 x = 14

(x=

3

7 5 5 37 5 5 ) + + − 2 2 2 2 2 2

h. x3 + x + 10 = 0

(x=

3

−5 +

i.

x3 − 1 =−15 x + 15

26 26 ) − 3 5+ 3 3 3 3

( x = 3 8 + 3 21 + 3 8 − 3 21 )

28 620 5 (x= ) x− = 0 3 3 27 2: Use a calculator to find the value of the roots of equations (h) and (i). The result is surprising. Can you prove it?

j.

x3 + x 2 +

14



Lesson 5 We now know a formula for solving any cubic equation of the form (1) x3 + Ax + B = 0 This formula is not applicable to equations with a second term, as you have seen in Problem (j) in the previous homework. We need, therefore, to develop a formula for the general cubic equation, (2) ax3 + bx 2 + cx + d = 0. Since a ≠ 0 (because otherwise the equation is not cubic), we can save one parameter by dividing by a and then rewrite equation (2) as: (3) x3 + Ex 2 + Fx + G = 0. Therefore, it is sufficient to deal with equations of form (3). To this end, we revisit the quadratic equation for a possible insight. The idea behind the Completing the Square method is essentially to reduce a quadratic equation with a first term into one without it. Under this method, the equation x 2 + Bx + C = 0 is reduced

B  − B 2 + 4C −B  into:  x +  + , = 0 . Put it in different words, the change of variable, x= y + 2 4 2  2

reduces the equation, x 2 + Bx + C = 0 , with a first term, into the equation, y 2 +

− B 2 + 4C = 0, 4

without a first term. This insight suggests that we try a similar variable change for the cubic equation. The change of variable, x= y + R , where R is a fixed number, transforms equation (3) into

( y + R)

3

+ E ( y + R) + F ( y + R) + G = 0. 2

We are looking R that would turn this equation into one without second term variable. Expanding and adding like terms, we get y 3 + ( 3R + E ) y 2 + ( 3R 2 + 2 ER + F ) y + R 3 + ER 2 + FR + G = 0.

(4)

We see that if we take R = −

E , equation (4) becomes 3

3 2   − E 2   −E   −E   −E   −E  (5) y +  3 + 2E  + F  y+ + E +F 0.     +G =   3    3   3   3   3    Simplifying, we get 3

3   E2  E E  + − + y3 +  F − y G F 2 0      = 3  3 3      We have, thus, successfully reduced the general cubic equation (3) into an equation of form (1).

(6)

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Homework Problems on Lesson 5 1. Find at least one root for the following cubic equations: a.

x3 + 3x 2 +

3

27 17 x = x+ 0= 2 2

6 + 19 2 + 3 6 − 19 2 −1 3 4

b. x3 − 6 x 2 + 24 x − 40 = 0

x = 3 4−4 5 + 3 4+4 5 +2

c. 125 x3 + 75 x 2 − 60 x + 64 = 0

x=

d. x3 + 9 x 2 + 21x + 15 = 0

x= −3 4 − 3 2 −3

3

−39 + 2 349 + 3 −39 − 2 349 − 1 5

1 2 2. Two objects, A and B, are moving in space. At some point of time, it was discovered that the objects change course and they are now moving according to the following patterns: At time

e. 2 x3 + 3 x 2 − 3 x − 9 = 0

x =1

t , Object A is at the coordinates (t 3 + t 2 − 3 6, 6 6 2t − 9 2, 0) , whereas Object B at (−3t + t 2 + 6, t 3 − 3 2, 0) .

a. Will the objects ever collide? b. The objects’ paths cannot be changed, but their speeds can be controlled. If the objects are to collide, can you change their speeds to avoid the collision? c. Repeat Assignments 1 and 2 for objects moving according to the following patterns: At 2 3

time t Satellite A is at (4t − 7t + 1, t − 3 2 − 2 17 t + 3(2 − 2 17) t + 2 17 − 2, 0) , 0) 3

3

2

3

and Satellite B is at (3t 3 + 5t + 5, 2 + 2 17, 0) 3 2 0. 3. Find a solution to the cubic equation, x + Ax + Bx + C =

4. Compare the formula you obtained in Problem 3 to the following formula 3

= x

3

3

3

3

2

2

2

3

 A  B   A  C  B  1 A  B  C   A   C   A  B  C      −   − +   −     +   + 2     − 2     +  3  2   3  2  3  3 3   2   2   3   2   3  2  2  3

2

2

2

3

 A  B   A  C  B  1 A  B  C   A   C   A  B  C  A     −   − −   −     +   + 2     − 2     − .  3  2   3  2  3  3 3   2   2   3   2   3  2  2  3

0 . Show (again) that the change of variable, 5. Take the quadratic equation, x + Bx + C =  B x = y +−   2  results in a quadratic equation without the 1st term. Give a geometric 2

explanation to this fact.

16



 B x = y +−   3  in the cubic equation, 6. We have seen that the change of variable,

x3 + Bx 2 + Cx + D = 0 , results in a cubic equation without the second term. This and problem  a  x = y +  − n −1   n  in the 5 suggests the following conjecture: “The change of variable, n n −1 0 results in an n -degree polynomial equation polynomial equation x + an −1 x +  + a0 =

n −1 without the x term.” Check this conjecture for n = 1 . Is this conjecture true for any natural number n ? (Hint: use the Binomial Theorem; see Unit X) 4 3 2 0. 7. Solve the equation 8 x − 8 x + 8 x − 3 x − 8 =

17



Part II Lesson 6 The cubic formula holds a few surprises. Surprise 1: The first surprise is that the cubic formula, unlike the quadratic formula, does not seem to give all the solutions to the equation. Consider, for example, the equation,

x3 − 12 x + 16 = 0 . It is easy to see that x = 2 is a solution. On the other hand, using the cubic formula, we get x = −4 is a solution: 16 16  16   −12   16   −12  x = − +   +  + 3 − −   +  = 2 2  2  3   2  3  2

3

2

3

3

3

−8 +

(8) + ( −4 ) 2

3

+ 3 −8 −

(8) + ( −4 ) 2

3

=3 −8 + 3 −8 =−4

This raises the question, how many roots does a cubic equation have? We will return to this question in Lesson 12. ICP 7: Construct two cubic equations, each with more than one root. Which of these roots are obtained by the cubic formula? Surprise 2: Sometimes simple roots appear as complicated expressions. For example, you have seen in Problem 1viii in the Homework on Lesson 4 that for the equation x3 + x + 10 = 0 the

26 26 , but when we evaluate this root by a − 3 5+ 3 3 3 3 calculator, we find it to be −2 . Similarly, the Cubic Formula produces the root cubic formula produces the root x =

3

−5 +

3 x = 3 8 + 3 21 + 3 8 − 3 21 for the equation x − 1 =−15 x + 15 , and when this root is evaluated by a calculator we find it to be 1. These results are not due to a rounding off error; rather, they are the actual values of the roots. Here is why.

Consider the equation, x3 + x + 10 = 0 . It is easy to see that x = −2 is a root of this equation. Since, x3 and x are monotonic functions, their sum, x 2 + x , is also a monotonic function (why?). The function x3 + x + 10 is a shift upward of a monotonic function; hence, it too is a monotonic function (See Figure 1). This shows that our equation has at most one root which proves that x = −2 is the only root it has. Hence,

3

surprising result.

18

−5 +

26 26 − 3 5+ =−2 , a pleasantly 3 3 3 3



12

f 2 = x 3 + x + 10 10

8

6

4

f1 = x 3 + x

2

-5

5 -2

-4

Figure 1

We will now show another proof for the equality

26 26 − 3 5+ =−2 . 3 3 3 3 You may deem this proof to be less elegant than the one we have just shown. However, the proof is important because it is based on an idea which will serve us in the near future. Here the proof: The first expression in the equality can be written as follows: 3

3

−5 +

−5 +

3 −15 3 + 26 26 == 3 3 3 3 3

−15 3 + 26 = 3 33

3

−15 3 + 26 = 6 3 3

3

3

−15 3 + 26 . 3

Now set 3

−15 3 + 26 =a + b 3

Cubing both sides we get: (1)

−15 3 + 26 = (a + b 3)3

Our goal is to find a and b satisfying equation (1). Expanding the right-hand side of equation (1), we get: −15 3 + 26 = a 3 + 3a 2b 3 + 3b 2

( 3)

2

a + b3

( 3)

3

a 3 + 3a 2b 3 + 9ab 2 + 3 3b3 = ( a3 + 9ab2 ) + ( 3a 2b + 3b3 ) 3 We have: 19



−15 3 + 26 = ( a 3 + 9ab 2 ) + ( 3a 2b + 3b3 ) 3 Equating separately the rational and irrational expressions, we get:

26 = a 3 + 9ab 2

(2)

−15 3 = ( 3a 2b + 3b3 ) 3

(3)

We try to find desirable a and b by trial and error. Let’s try b = 1 . Then equations (2) and (3) become

26 = a 3 + 9a

(4)

−15 3 =( 3a 2 + 3) 3

(5)

Unfortunately, equation (5) does not have a solution, and so b = 1 is not suitable. Let’s try b = −1 . Then equations (2) and (3) become

26 = a 3 + 9a

(6)

−15 3 = ( −3a 2 − 3) 3

(7) From equation (6), we get:

3a 2 = 12 This gives us two values for a : a = 2 , a = −2 . We see that a = 2 satisfies equation (6), but a = −2 does not. Thus, we found desirable values for a and b : a = 2 and b = −1 satisfies equation (1) Let’s check: ?

−15 3 + 26 = (2 + (−1) 3)3 . Expanding the right-hand side, we get that it indeed equals the right-hand side: (2 − 3)3 =− 8 12 3 + 18 − 3 3 = −15 3 + 26 . We, thus, have , 3

(8)

−15 3 + 26 2 − 3 = = 3 3

2 − 1. 3

In a similar way, we find: 3

−15 3 − 26 2 + 3 = = 3 3

2 +1 . 3

Hence: 3

3 26 26 −15 3 + 26 3 −15 3 − 26  2   2  3 −5 + − 5+ = − = − 1 −  − 1 =−2 3 3 3 3 3 3  3   3 



20



Surprise 3: Consider the equation, x3 − 15 x − 4 = 0 . By inspection, or by the Rational Root Theorem, we can determine that the equation has exactly three roots: x =−2 ± 3 and x = 4 . On the other hand, using the Cubic Formula, we find its root to be: −4 −4  −4   −15   −4   −15  +   + −   +  +3−  = 2 2  2   3   2   3  2

x =3 − 3

The expression,

3

3

2

3

2 + 4 − 125 + 3 2 − 4 − 125 = 3 2 + 11 −1 + 3 2 − 11 −1

2 + 11 −1 + 3 2 − 11 −1 , produced by the Cubic Formula as a root to our

equation seems meaningless since the expression −1 is meaningless. This raises the question: How is it possible that the cubic formula fails to produce any of the correct roots to our equation? These surprises will lead us to new knowledge. Homework Problems on Lesson 6 1. Prove that the root produced by the cubic formula to the equation, x3 + x − 2 = 0 is exactly 1. Provide three proofs. Two of the proofs should be like the one you learn in the lesson. For the third proof, use the following hint: divide x3 + x − 2 by x − 1 . 2. The roots of each of the following equations are rational. Find them by using the Rational Root Theorem or by inspection. Does Cubic Formula give any of these roots?

x3 − 28 x + 48 = 0 b. 4 x 3 − 43 x + 21 = 0 a.

c.

x3 − 13x + 12 = 0

21



Part III Lesson 7 It is puzzling that the cubic formula produces what seems to be a meaningless expression as the root of a cubic equation even in cases where we know the equation does have a root. For example, we have seen in Lesson 6 that the equation x 3 − 15 x − 4 = 0 has exactly three roots, x =−2 ± 3 and x = 4 , and yet the cubic formula does not seem to produce any of them; instead

it gives x= 3 2 + 11 −1 + 3 2 − 11 −1 as a root. Q: How is this possible? A: One of the observations we made in Lesson 6 may provide a clue as to how to pursue an investigation of this puzzle. We saw that the cubic formula sometimes produces complicated expressions (Surprise 2), which when manipulated they turn into simple roots. So, let’s try to simplify the expression x= this might lead us.

3

2 + 11 −1 + 3 2 − 11 −1 and see where

To do so, we will treat—temporary—the expression number,

−1 must obey the rule:

(

−1

)

2

−1 as if it were a number. As a

= −1 . We begin by trying to get rid of the cubic

root in the expressions, 3 2 + 11 −1 and 3 2 − 11 −1 , in a similar manner to that we did in the previous lesson (see Problem 1 in the HP on Lesson 6 and the lesson’s discussions related to it). We set: 2 + 11 −1 = (u + r −1)3 . Expanding, we get:

(u + r

)

3

−1 =u 3 + 3u 2 r −1 + 3r 2

(

u 3 + 3u 2 r −1 + 3 ( −1) ur 2 + r 3 u 3 + 3u 2 r −1 − 3ur 2 − r 3

We have therefore:

(

)

)

2

−1 u + r 3

(

−1

)( 2

(

)

)

−1 =

−1 =( u 3 − 3ur 2 ) + ( 3u 2 r − r 3 ) −1

2 + 11 −1 = ( u 3 − 3ur 2 ) + ( 3u 2 r − r 3 ) −1

Equating separately the rational and irrational expressions, we get: (1) (2)

3

−1 =

= 2 u 3 − 3ur 2

11 = −1

22

( 3u r − r ) 2

3

−1



Recall our goal is merely to find u and r so that 2 + 11 −1 = (u + r −1)3 . So let’s try to find convenient r and u . Clearly r ≠ 0 . (Why?) Let’s try r = 1 . Then equations (1) and (2) become

= 2 u 3 − 3u

(3) (4)

11 −= 1

( 3u

2

− 1) −1

From equation (3), we get:

3u 2 − 1 = 11 This implies u = 2 or u = −2 . Since only u = 2 satisfies equation (1), we conclude: 2 + 11 −1 = (2 + −1)3 . Checking: (2 + −1)3 = 8 + 12 −1 − 6 − −1 = 2 + 11 −1 .

In a similar way, we find: 2 − 11 −1 = (2 − −1)3 .

Hence: x=

3

4. 2 + 11 −1 + 3 2 − 11 −1 = (2 − −1) + (2 + −1) =

We see here a remarkable phenomenon: by treating x=

3

−1 as a number, the expression

2 + 11 −1 + 3 2 − 11 −1 produced by the cubic formula turned out to be the number 4 .

This suggests that −1 and the expressions 2 + 11 −1 and 2 − 11 −1 may be meaningful after all. Q: What might this meaning be? A: This question is the topic of the next lesson. We will investigate the meaning of expressions of the form a + bi , where a and b are real numbers and i= such that i 2 = −1 .

−1 is a symbol

It took a long time in the history of mathematics to find a meaning for these expressions and to fully appreciate their contribution to mathematics and science. The process began in the 16th Century, when these expressions first appeared in the solution of cubic equations, and continued until the 19th Century. No wonder the mathematicians of that time called these expressions complex numbers. To begin with, it was hard to conceive of the expression z= a + bi , which consists of different parts, as a number. They called a the real part of z , denoting it by Re( z ) ; and b the imaginary part of z , denoting it by Im( z ) . These names and

23



symbols remained in use to this day, and they are the names and symbols we will use from now on. Homework Problems on Lesson 7 1. 2 is a root of the equation, x3 − 6 x + 4 =. 0 Prove that the root obtained by the cubic formula to this equation is indeed 2.

24



Lesson 8 We state again: Definition: A complex number is an expression of the form a + bi , where i 2 = −1 and a and b are real numbers. a is called the real part of z and is denoted by Re( z ) . b is called Q:

A:

the imaginary part of z and is denoted by Im( z ) . Are complex numbers “genuine”? That is to say, do they fulfill our expectations about numbers? The numbers we know (integers, fractions, and irrationals, which collectively are called real numbers) represent quantities, such as length, width, area, weight, temperature, speed, work, etc., and they can be located on the number line. The real numbers can be added, subtracted, multiplied, and divided, and these operations obey certain rules and have quantitative and geometric meaning. And, most important, with the real numbers we can solve problems, mathematical problems as well as application problems. Do the complex numbers have these qualities? The answer to this question will unfold as you progress in your study of complex numbers. Here you will learn some basic concepts necessary for a college-level course in complex numbers (called Complex Analysis).

In the rest of this unit, we divide our investigation into the meaning of complex numbers in two parts. The first part deals with the algebraic meaning of complex numbers (Lessons 8 and 9). The second part deals with the geometric meaning of complex numbers (Lessons 10, 11, and 12). We begin our algebraic investigation by agreeing that 0 + 1i = i , so as to make i a complex number. And we denote by  the set consisting of all complex numbers. Now we are ready to define addition, subtraction, multiplication, and division on complex numbers. The operations are simply those we normally perform on algebraic expressions. Definition: Let z1= a + bi and z2 = c + di be two complex numbers. Then Addition: z1 + z2 = (a + bi ) + (c + di ) = (a + c) + (b + d )i

Multiplication: z1 ⋅ z2 = (a + bi ) ⋅ (c + di ) = (ac − bd ) + (ad + bc)i

(We usually omit the multiplication sign and write z1 z2 instead of z1 ⋅ z2 .) Subtraction: z1 − z2 = (a + bi ) − (c + di ) = (a − c) + (b − d )i

Division requires a bit more work. The idea here is similar to the idea of rationalizing expressions with radicals in their denominator. 25



Division: z1 a + bi (a + bi )(c − di ) ac + bd bc − ad = = = + i z2 c + di (c + di )(c − di ) c 2 + d 2 c 2 + d 2

( c2 + d 2 ≠ 0 )

There is a simple way to remember the definition of division. Here is how: For a complex number z= a + bi , denote z= a − bi . Also, denote= z

a 2 + b 2 ( z is called the

conjugate of z and z is called the absolute value of z . Later, in Lesson 11, we will learn the geometric meanings of these new terms.) With these notations, the division of complex numbers z zz z1 a + bi (a + bi )(c − di ) z1 z2 z1 z2 . becomes 1 = 1 2 , because = = = 2= 2 z2 c + di (c + di )(c − di ) c + d z2 z2 z2 Q:

A:

Q: A:

Each of these operations is defined only on a pair of complex numbers. However, we want to be able to add, multiply, subtract and divide between real numbers and complex numbers. Can we do so? Yes, we can, because any real number can be viewed as complex number, simply by agreeing that expression u + 0i is u . Hence we can say that  , the set of real numbers, is a subset of  . That is to say, every real number is a complex number. With this view of the real numbers as complex numbers, do the four operations we have just defined agree with the real-number operations? Yes. Take two real numbers a and c and view them as complex numbers a + 0i and ac . Indeed: c + 0i , respectively. We expect that (a + 0i )(c + 0i ) =

(a + 0i )(c + 0i ) = (ac − 00) + (a 0 + 0c)i = ac + 0i = ac Likewise, when we calculate the division

By the definition of multiplication By the properties of the real numbers By agreement

a + 0i ( c ≠ 0 ) using the definition of division c + 0i

a , as expected: c a + 0i ac + 00 0c − a 0 i = 2 + c + 0i c + 02 c 2 + 02 a = + 0i c a = c

of complex numbers we get



26

By the definition of division By the properties of the real numbers By agreement


Q: A:


What about equality between complex numbers—how do we decide when two complex numbers are equal? As one would expect, two complex numbers are equal when their real parts are equal and their imaginary parts are equal, as stated in the following definition:

Definition: Let z1= a + bi and z2 = c + di be two complex numbers. Then z1 = z2 if and only if a = c and b = d . Q: A:

Do the complex-number operations satisfy all the rules of the operations on the real numbers? Yes, as you will show in the HP on this lesson.

Summary 1. We have created a new type of numbers, called complex numbers. We denoted the set of these numbers by  . 2. Complex numbers are expressions of the form a + bi , where a and b are real numbers. 3. A complex number can be a real number because we can think of a real number a as a + 0i . Hence  ( the set of the real numbers) is a subset of  . 4. We defined four arithmetic operations (addition, subtraction, multiplication, and division) on the complex numbers, and insured that they satisfy a list of rules. 5. We observed that each of the operations we defined on complex numbers agrees with its counterpart operation on the real numbers. This means when taking two real numbers u and v and viewing them as complex numbers z= u + 0i and w= v + 0i , respectively, each of the operations on z and w gives the same result as the corresponding operation on u and v . 6. This last observation and the fact that  is a subset of  are expressed by the phrase,  is an extension of  . Since complex numbers encompass the real numbers, it is important to keep in mind that when we say “ z is a complex number,” we mean that z may or may not be a real number. If we want to indicate that z is complex and not real, we have to say so, or simply write Im( z ) ≠ 0 . Homework Problems on Lesson 8 1. Let z1 , z2 , z3 ,  be an arithmetic sequence of complex numbers. Must the sequences Re( z1 ) , Re( z2 ) , Re( z3 ) ,  and Im( z1 ) , Im( z2 ) , Im( z3 ) ,  be arithmetic sequences too?

2. If the sequences Re( z1 ) , Re( z2 ) , Re( z3 ) ,  and Im( z1 ) , Im( z2 ) , Im( z3 ) ,  are arithmetic sequences, must the z1 , z2 , z3 ,  be an arithmetic sequence? 27



3. Let z1 , z2 , z3 ,  be a geometric sequence of complex numbers, and let q be its ratio. State a condition on q so that each of the sequences Re( z1 ) , Re( z2 ) , Re( z3 ) ,  and Im( z1 ) , Im( z2 ) , Im( z3 ) ,  is geometric.

4. Let z1 , z2 , z3 ,  be a sequence of complex numbers, where each of the sequences Re( z1 ) , Re( z2 ) , Re( z3 ) ,  and Im( z1 ) , Im( z2 ) , Im( z3 ) ,  is geometric. Must z1 , z2 , z3 ,  be

geometric? 5. Show that i105 + i 23 + i 20 − i 34 = 2 , and then compute i 2000 + i1999 + i 82 + i 47 . 6. Show that the solutions to the following equations are complex numbers. a. zi + 2 = 3 − 4i i b. (2 − i ) z + 5 = c. 3 z − 5i =2 − 4iz d. (5 − 2i ) z + 7 − i =0 7. Let z= a + bi . Prove z = 0 if and only if z = 0 . 8. Let z be a complex number different from zero. 1 a. Show that is also a complex number. z 1 1 1 is called the reciprocal of z because z ⋅ = ⋅ z = 1 . Verify this fact. b. z z z 1 c. Find a formula for . That is, if z= a + bi is a non-zero complex number and z w= x + yi is its reciprocal, express x and y in terms of a and b . d. Prove that there is only one reciprocal for any non-zero complex member. 9. Solve the equation, zz1 = zz2 , where z1= 5 − 3i and z2 =−1 + 2i . 10. To solve the equation in Problem 9, you likely obtained the equation z ( z1 − z2 ) = 0 and concluded that z = 0 . Your conclusion is based on the following theorem. Prove it. Theorem: The product of two complex numbers is zero if and only if at least one of the numbers is zero. This theorem is often called the Cancellation Theorem. Can you guess why? 11. Prove that if z is a complex number so is z n for any integer n . 12. Prove that the complex-number operations satisfy all the rules of the operations on the real numbers. Specifically, show that for any complex numbers z1 , z2 , z3 the following rules hold: a. z1 + z2 = z2 + z1 b. z1 + ( z2 + z3 ) = ( z1 + z2 ) + z3 c.

z + 0 = z + 0 = z ( 0 is the complex number 0 + 0i )

28



d.

z + (− z ) = (− z ) + z = 0 (if z= x + yi , by − z we mean the complex number − x + (− y )i )

e.

z1 z2 = z2 z1

f. 1= z z= 1 z −1 −1 g. zz = z= z 1 h. z1 ( z2 + z3 ) = z1 z2 + z1 z3 Team Project In the next 6 problems you will prove or be helped to understand the proofs of important facts about roots of polynomials. It is strongly recommended that you work on these problems collaboratively with one or two classmates. 13. a. Divide the polynomial P( z )= 5 z 4 − 4 z 3 − 6 z 2 + z − 1 by z − 2 . b. Divide the polynomial P( z ) = (3 − 4i ) z 3 − (5 + i ) z 2 + z − 2 by z − (2 + 3i ) 14. Can you explain why the algorithm you have just used works? 15. The answer to this question is in the proof of the following theorem: Division Theorem: Let P( z ) be a polynomial with complex coefficients, and let s be any complex number. It is always possible to construct a polynomial Q( z ) so that

P( z ) = ( z − s )Q( z ) + r , where r is a complex number. As you read the proof below, reflect on the process you used to solve Problem 13. This will help you understand the proof. The proof describes the exact steps for constructing Q( z ) . Proof: Let P( z )= an z n + an −1 z n −1 +  + a1 z1 + a0 , where a1 , a2 , , an are complex numbers. Define the polynomial P1 ( z )= P ( z ) − an z n −1 ( z − s )= an z n + an −1 z n −1 +  + a1 z1 + a0 − an z n + san z n −1 = (an −1 + san ) z n −1 +  + a1 z1 + a0 .

P1 ( z ) is a polynomial of degree n − 1 , one degree less than the degree of the

polynomial P( z ) . (0) For simplicity, denote a= an −1 + san . Then n −1

( z ) an(0)−1 z n −1 +  + a1 z1 + a0 . P1=

(P: What is P1 ( z ) in the divisions you carried out in Problem 13?) In a similar manner, define a polynomial P2 ( z ) P2 ( z )= P1 ( z ) − an(0)−1 z n − 2 ( z − s )= (an(0)−1 + san − 2 ) z n − 2 +  + a1 z1 + a0 .

29



P2 ( z ) is a polynomial of degree n − 2 , two degrees less than the degree of the

polynomial P( z ) . (1) Again, for simplicity, denote a= . Then an(0) n −1 −1 + san − 2

(P: What is P2 ( z ) in the divisions you carried out in Problem 13?) P2= ( z ) an(1)−1 z n − 2 +  + a1 z1 + a0 .

If we continue to define in this way the polynomials, P3 ( z ) , P4 ( z ) , …, Pn ( z ) , the degree of Pn ( z ) must be 0 ; that is, Pn ( z ) is just a complex number, say, r . Then Pn ( z ) = r

(P: What is r in the divisions you carried out in Problem 13?) We have, thus: P( z ) = P1 ( z ) + an z n −1 ( z − s ) P1 ( z ) = P2 ( z ) + an(1)−1 z n − 2 ( z − s ) n −3 P2 ( z ) = P3 ( z ) + an(2) ( z − s) −2 z

 Pn −1 ( z ) =Pn ( z ) + an( n−−21) ( z − s )

Pn ( z ) = r

Adding all these polynomials we get n −3 P ( z ) = ( z − s )(an z n −1 + an(1)−1 z n − 2 + an(2) +  + an( n−−21) ) + r −2 z n −3 Let Q( z= ) an z n −1 + an(1)−1 z n − 2 + an(2) +  + an( n−−21) . Then we have −2 z

P( z ) = ( z − s )Q( z ) + r ,

16.

as required.  Reproduce the proof of the Division Theorem independently. Use your own notation. Now repeat the proof’s steps for the division of P( z ) = −5iz 4 + iz 3 + 4 z 2 − 2 z + 2 − i by z − 2i

17. a. Check with a few examples the following conjecture: Let P( z ) = 0 be a polynomial equation with complex coefficients. Then If a complex number s is a root of P( z ) = 0 , then z − s divides P( z ) with zero remainder. And conversely, • If z − s divides P( z ) with zero remainder, then s is a root of P( z ) = 0 . b. Prove this conjecture. Now that you have proved this conjecture, it is a theorem. It is called Factor Theorem. 18. Conclude from the Division Theorem and the Factor theorem the following theorem: •

30



a. If for some positive integer k , ( z − s ) k divides a polynomial P( z ) with no remainder, then s appears k times as a root of the equation P( z ) = 0 . And, conversely, b. If s appears k times as a root of the equation P( z ) = 0 , then ( z − s ) k divides P( z ) with no remainder.

31



Lesson 9 Q:

A:

For a quadratic equation, Az 2 + Bz + C = 0 , with real number coefficients, it is easy to show that its roots are complex numbers (P: show this). What about a quadratic equation with complex number coefficients; must its roots also be complex numbers? Yes, the roots of any quadratic equation with complex coefficients are complex numbers.

Proof Let Az 2 + Bz + C = 0 be a quadratic equation ( A ≠ 0) with complex coefficients. Using the same algebraic manipulation as in the case of real coefficients, we get z =

B 2 − 4 AC −B ± 2A 2A

Clearly, B 2 − 4 AC is a complex number (P: why?). To complete the proof, we only need to show that the square root of any complex number is a complex number (P: explain). Let α + β i be a complex number, where β ≠ 0 . We are to find real numbers a and b so that a + bi =

α + βi

This equation is equivalent to the following system (P: explain) a 2 − b 2 = α  2ab = β The solutions of this system are: 1

 −α + α 2 + β 2  2  −α + α 2 + β 2  , b = ± a = ±    2 2    where the signs must be chosen to satisfy 2ab = β (P: confirm). We conclude what was to be proved (P: how?).  Q:

A:

1

2   , 

We have just shown that the square root of complex number is a complex numbers. What about the cubic roots of a complex number, and, in general, the n th root of a complex number; are they too complex numbers? A complete answer to this question will be given in Lesson 11. For now, we answer this question for three special cases.

ICP 8: Solve the equation, z 2 = i . After you are done, compare your solution to the one below. 32



Solution P: Explain each step as needed.

z2 = i z= ± i

Let a + bi =i . From the previous proof, 2 2 and b = 2 2

a=

or 2 2 and b = − 2 2

a= −

Hence, the solutions to z 2 = i are: 2 2 2 2 i. + i, z = − − 2 2 2 2

= z

 ICP 9: Solve the equation z 4 = −2 . After you are done compare your solution to the solution below Solution P: Explain each step as needed. z 4 = −2

z 2 =± −2 =±i 2 1

1

±2 4 i −i z = ±2 4 i or z = 1 1  2 −  −1  2  z= i  = ±2 4  + ±2 4 + 2 4i, 2     2

1 1  2 −  −1  2  z= ±2 4 i  − i  = ±2 4 − 2 4i 2     2 Hence, the equation z 4 = −2 has four roots: −

1 4

−

1 4

−

1 4

−

1 4

−

1 4

−

1 4

−

1 4

−

1 4

= z 2 +2 i, z = −2 − 2 i ,= z 2 −2 i, z = −2 + 2 i  ICP 10: Solve the equation, z 3 = 8 . After you are done compare your solution to the solution below. Solution P: Explain each step as needed. Let z= a + bi be a solution to the equation z 3 = 8 . Then (a + bi )3 = 8 33



a 3 − 3b 2 a + (3a 2b − b3 )i = 8 a 3 − 3b 2 a = 8  2 3 0 3a b − b = This system has three solutions: a = 2 and b = 0 a = −1 and b = 3 a = −1 and b = − 3 Accordingly the solution to the equation, z 3 = 8 , are: z = 2 , z =−1 + i 3 , z =−1 − i 3  Homework Problems on Lesson 9 1. Find all the complex roots of the following equation, and express them in the form a + bi , where a and b are real numbers. a.

z 2 = 5 − 12i

2 b. (i + 6) z =282 − 212i c. z 2 − 8 z + 80 = 0

d. 4 z 2 − 24iz − 35 = 0 e. z 2 + (3i − 2) z − 6i = 0 f.

z 2 − 2 z (1 + i ) =3 − 6i

g. (1 + 2i ) z 2 − 2 z (3 + i ) = 25 2. The following equations have four complex roots. Find them. a. z 4 = 1 b. z 4 = −16 3. The following equations have three complex roots. Find them.

z 3 = −8 b. z 3 − i =0 a.

z 3 = −216 d. z 3 = 1 − i e. z 3 = 1 + i c.

34



Lesson 10 Q:

A:

We have just shown that the roots of any quadratic equation are members of  . We have also shown that the roots of some simple cubic equations are members of  . Are the roots of any cubic equation members of  ? If the answer is negative, then our extension of  into  is incomplete, because for those cubic equations whose roots are neither real nor complex, we would need to extend  further in order to accommodate the new roots. We can ask the same question about any polynomial equation: Are the roots of any polynomial equation members of  , or that we will have to keep extending  with every discovery of a new type or roots? Remarkably, the answer to this question is in the affirmative; there is no need to further extend  , as is stated in the following theorem: Theorem: Any polynomial equation, cn z n + cn −1 z n −1 +  + c1 z1 + c0 = 0 , where c0 , c1 ,  , cn are complex numbers, has a solution, and all its solutions are members of  . We proved this theorem only for the case of quadratic equations. As you saw, the proof was a bit intricate, and so were the solutions for simple cubic equations. Imagine we had to approach the problem in a similar manner for the general polynomial equation. This is unlikely to succeed, and indeed attempts to do so even for special cases, such as the general 5th degree polynomial equation, failed. Not until the late 1800 were mathematicians able to provide a complete proof for this theorem. By now, many proofs have been found, but all use ideas outside of the scope of this unit. This theorem is remarkable because it frees us from the need to further extend  in order to solve polynomial equations of different degrees. Historically, it was this theorem that gave the complex numbers a credible status. It gave the mathematicians the certainty they sought—that the only roots any polynomial equation has are complex numbers! It is no wonder, therefore, that mathematicians dubbed this theorem, The Fundamental Theorem of Algebra (FTA).

Q:

A:

An immediate consequence of the FTA is an answer to the following question: The FTA guarantees that any polynomial equation has a complex root and all its roots are complex numbers, but can we tell by the degree of the polynomial how many roots the equation has? From experience we conjecture that a polynomial equation of degree n has n roots, This conjecture is true, as it is stated in the following theorem: Theorem: (a) Any polynomial equation of degree n has at most n distinct roots. And

35



(b) If we count all the roots of the equation, including those that appear more than once, then the equation has exactly n roots. Proof of (a) Let P ( z ) = 0 be a polynomial equation of degree n with complex coefficients. By the FTA, all the roots of the equation are complex numbers. Let s1 , s2 , , sk be all the distinct roots of the equation. We are to show that k ≤ n . By the Factor Theorem (Problem 17 in the HP on Lesson 8), P( z ) is divisible by z − s1 , z − s2 , , z − sk .

Hence P( z ) = ( z − s1 )( z − s2 ) ( z − sk )Q( z ) , where Q( z ) is a polynomial n − k degree (P: explain). Clearly, n − k ≥ 0 , which implies k ≤ n .  Proof of (b) Let, again, s1 , s2 , , sk be all the distinct roots of P ( z ) = 0 , where P( z ) is a polynomial of degree n . We are to show that if s1 appears l1 times, s2 appears l2 times,  , sk appears lk times, then l1 + l2 +  + lk = n. By the Factor Theorem, P( z ) can be expressed as P( z ) = ( z − s1 )l1 ( z − s2 )l2  ( z − sk )lk q

where q is a complex number (P: explain). The product ( z − s1 )l ( z − s2 )l  ( z − sk )l q is a polynomial of degree l1 + l2 +  + lk (P: explain) Since the degree of P( z ) is n , l1 + l2 +  + lk = n. 1

2

k

 Homework Problems on Lesson 10 Team Project We have just seen that the FTA, together with the Division Theorem and the Factor Theorem, entails important facts about polynomials and polynomial equations. In this and the next few problems, you will prove more facts on the basis of these theorems. It is strongly recommended that you work on these problems collaboratively with one or two classmates.

36



1. In Secondary School you learned that any whole number p can be factored into a product of primes: p = s1q1 s2 q2  s j q j . For example, 95256 = 23 ⋅ 57 ⋅ 7 2 . Here the prime factors of 95256 are 2 , 5 , and 7 , where 2 repeats 3 times, 3 repeats 5 times, and 7 repeats twice. Linear polynomials of the form s ( z )= z − a , where a is a complex number, are like prime numbers. Any polynomial can be factored into linear polynomials, as stated by the following theorem: Theorem: Any polynomial P( z ) of degree n can be factored into P( z ) = a ( z − z1 ) k1 ( z − z2 ) k2  ( z − z j ) j , where z1 , z2 , , z j are complex numbers, k

k1 + k2 +  + k j = n , and a is the coefficient of the highest term in the polynomial.

Prove this theorem. 2. Examine the following conjecture with a few examples of polynomial equations of degree 2 and 3 . Conjecture: For any polynomial equation with real coefficients, if a + bi is a root of the equation, then a − bi is also a root of the equation. For z= a + bi , z= a − bi is called the conjugate of z . 3. The goal of this and next problem is to prove the conjecture in Problem 2. Prove the following properties: a. The relation z = z holds if and only if z is a real number. b. For any complex number z the relation z = z holds. c. For any complex number z the number z ⋅ z is a nonnegative real number. d. z1 + z2 = z1 + z2 (the conjugate of a sum is the sum of conjugates). e.

z1 ⋅ z2 = z1 ⋅ z2 (the conjugate of a product is the product of conjugates).

f. For any nonzero complex number z the relation z −1 = ( z ) −1 holds.  z1  z1 g. = , z2 ≠ 0 (the conjugate of a quotient is the quotient of conjugates).    z2  z2 z+z z−z h. The formulas Re( z ) = and Im( z ) = are valid for all complex numbers. 2 2i 4. Graph the solution sets of each of the equations x 2 + 4 + i y − 4 =10

i.

z+

ii.

1 2 (Hint: substitute z= x + yi . Factor the equation in x and y obtained from the = z

substitution into the equation ( ( x 2 + y 2 − 2 y − 1)( x 2 + y 2 + 2 y − 1) = 0 .) 4. Solve

z z + = 1. z z 37



5. Let P( z )= an z n + an −1 z n −1 +  + a1 z1 + a0 be a polynomial with real coefficients. a. Show that an z n + an −1 z n −1 +  + a1 z1 + a0= an z n + an −1 z n −1 +  + a1 z 1 + a0 b. Conclude that if z is a root of the equation, P( z ) = 0 , then z is also a root of the same equation. 6. Let P( z ) = 0 be a polynomial equation of degree n with real coefficients. Prove that if n is odd, then the equation must have at least one real root.

38



Lesson 11 One of the properties of the real numbers is that they are ordered. That is to say, for any two real numbers a and b , only one of the following is true: a = b , a < b , a > b . This allows us to locate the real numbers on the number line orderly: If two real numbers a and b correspond to two points A and B on the number-line, respectively, then b > a if and only if B is to the right of A . Q: A:

Can complex numbers be ordered? Complex numbers cannot be ordered. Here is why: Let z= a + bi be any non-zero complex number where b ≠ 0 (that is, z is not a real number). Now consider the order relation between z and 0 . There can be two possibilities z < 0 or z > 0 . If z =a + bi < 0 , then, a < −bi . Hence, depending on the signs of a and b , either i < 0 or i > 0 (P: explain). This is not possible because if i < 0 then −1 > 0 , and if i > 0 then −1 > 0 . In both cases we get a contradiction. If z =a + bi > 0 , a similar reasoning leads to the same contradiction.

Q:

Each point on the number line corresponds to one and only one real number; and conversely, each real number corresponds to one and only one point on the number line. Therefore, there is no room for complex numbers that are not real on the number line. What geometric meaning, then, do complex numbers have? We observe that each complex number a + bi can be assigned to the ordered pair (a, b) . This observation allows us to locate each complex number at a point on the coordinate plane, as illustrated in Figure 2.

A:

4

4 + 3i 2

O

-

5 2

5

-i -2

Figure 2

39



The assignment a + bi → (a, b) allows us to identify unambiguously complex numbers with points in the coordinate plane, and points in the coordinate plane with complex numbers. Q:

Q:

The last sentence needs clarification. What is it meant by the sentence “The assignment a + bi → (a, b) allows us to identify unambiguously complex numbers with points in the coordinate plane, and points in the coordinate plane with complex numbers”? By this sentence we mean that the assignment a + bi → (a, b) have three properties: (1) Each complex number is located at exactly one point. That is, it never happens that (a1 , b1 ) ≠ (a2 , b2 ) and yet a1 + b1i = a2 + b2i . [Recall this property makes the assignment a + bi → (a, b) a function.] (2) Each point houses at most one complex number. That is, it is not possible that (a1 , b1 ) = (a2 , b2 ) and yet a1 + b1i ≠ a2 + b2i . [This property makes the assignment a + bi → (a, b) a one-to-one function.] (3) Each point is housed by a complex number. That is, for any ordered pair of real number (a, b) , there is a complex number z= x + yi so that x = a and

y = b . [This property makes the assignment a + bi → (a, b) an onto function.] Thanks to these properties, we can speak of a complex number as a point in the coordinate plane, and of a point in the coordinate plane as a complex number. In this sense, we can talk of the distance between two complex numbers, z= a1 + b1i and z= a1 + b1i , to mean the 1 2 distance between the points whose coordinates are (a1 , b1 ) and (a2 , b2 ) . As perhaps you have seen in your physics classes, an ordered pair (a, b) of real numbers can be thought of as a directed-line segment, or vector, emanating from the origin and ending at the point with the coordinates (a, b) . Since a complex number z= a + bi is identified with the ordered pair (a, b) , we can think of z as a vector as well. As with real numbers, we denote the distance of a complex number z from the origin by z . As with the real numbers, we call z , the absolute value of z . Figure 3 explains why if z= a + bi , then= z

a 2 + b 2 (P: how?).

40



z = a 2 + b2

4

a

2

z

(a, b)

b

O

5

Figure 3

The distance between z1 and z2 is equal to the distance of z1 − z2 from the origin (P:

( a1 − a2 ) + ( b1 − b2 )

why?). Hence, for z= a1 + b1i and z= a1 + b1i , z1 − z2 = 1 2 Q:

Q:

2

2

(P: explain).

We have just given a geometric meaning to complex numbers. (A complex number can be thought of as a point in the coordinate plane or as a vector emanating from the origin.) What does this entail for the geometric meaning of the four operations on complex numbers? Let z1= a + bi and z2 = c + di be two complex numbers. Then z1 is at A(a, b) and z2 is at B (c, d ) as shown in Figure 4. Geometric meaning of addition: Since w = z1 + z2 = (a + bi ) + (c + di ) = (a + c) + (b + d )i , w is at C (a + c, b + d ) . We want to know where C is located relative to A and B . By observation, we conjecture that if O is the origin, then the quadrilateral OACB is a parallelogram (Figure 3). Prove this conjecture.

C B (a +c)+ i(b + d) c + id

a + ib

O

Figure 4 Geometric meaning of multiplication: 41

A



Since w =z1 z2 =(a + bi )(c + di ) =(ac − bd ) + (ad + bc)i , w is at D(ac − bd , ad + bc) . We want to know where C is located relative to A and B . This time a figure representing A , B , C , and O on the coordinate system may not be as helpful to conjecture an answer to this question as in the case of addition. Let us calculate w , the distance of the point D from the origin: w = z1 z2 = (ac − bd ) 2 + (ad + bc) 2 = 2

2

(a 2 c 2 − 2abcd + b 2 d 2 ) + (a 2 d 2 + 2abcd + b 2 c 2 ) = a 2 c 2 + b 2 d 2 + a 2 d 2 + b 2 c 2= a 2 (c 2 + d 2 ) + b 2 (c 2 + d 2 )=

(a 2 + b 2 )(c 2 + d 2 ) = ( z1 z2

)

2

.

Hence = w

z= 1 z2

z1 z2 .

In other words, the distance of the product w = z1 z2 from the origin is equal to the product of the distances of z1 and z2 from the origin. Hence, the complex number w = z1 z2 is on the circle with radius r = z1 z2 and center at the origin (Figure 5).

r = z1 z2

C r

O

Figure 5

To find the location of w on this circle, we need to determine the angle θ that the segment OC creates with the positive direction of the x − axis . Let θ1 and θ 2 be, respectively, the angles that the segments OA and OB create with the positive direction of the x − axis . We will express θ in terms of θ1 and θ 2 . Let r1 = z1 and r2 = z2 . Then a1 = r1 cos θ1

b1 = r1 sin θ1

a2 = r2 cos θ 2

b2 = r2 sin θ 2 .

42



Hence, z1 =a1 + ib1 =r1 ( cos θ1 + i sin θ1 ) z2 =a2 + ib2 =r2 ( cos θ 2 + i sin θ 2 ) r1r2 ( cos θ1 + i sin θ1 )( cos θ 2 + i sin θ 2 ) w= z1 z2 =+ ( a1 ib1 )( a2 + ib2 ) =

= r1r2 ( ( cos θ1 cos θ 2 − sin θ1 sin θ 2 ) + i ( sin θ1 cos θ 2 + sin θ 2 cos θ1 ) ) = r1r2 ( cos (θ1 + θ 2 ) + i sin (θ1 + θ 2 ) )

We see that the angle between the vector OC and the positive direction of the x − axis is the sum of the angles θ1 and θ 2 created by the vectors OA and OB , respectively, with the positive direction of the x − axis . (Figure 6)

C 8

6

4

B

2

θ1+ θ2 θ2

A

θ1 O

5

Figure 6

Summary: 1. Forms of complex numbers: A complex number z can be written in two forms: and z r (cos θ + i sin θ ) , where r is the radius vector of z , and θ is the z= a + bi = angle of z . 2. Geometric meaning of complex numbers: When a complex number is written in the form z= a + bi , it can be thought of as the point (a, b) in the coordinate plane, or as the vector OA , where O is the origin and A is the point (a, b) .

43



z r (cos θ + i sin θ ) , it = When a complex number is written in its trigonometric form can be thought of as a vector of length r emanating from the origin and creating an angle θ with the positive direction of the x -axis. 3. Geometric meaning of addition: Let z1 , z2 be any two complex numbers. Then the points, z1 , z2 z1 + z2 , 0 form a parallelogram. 4. Geometric meaning of multiplication: Let z1 , z2 be any two complex numbers, and let w = z1 z2 . Let θ1 , θ 2 , θ be the angles of z1 , z2 , and w . Then = w

z= 1 z2

z1 z2 , and θ= θ1 + θ 2 . In other words, the radius vector of the product

w = z1 z2 is the product of the radius vectors of z1 and z2 ; and the angle of the

product w = z1 z2 is the sum of the angles of z1 and z2 . Homework Problems on Lesson 11 1. a. = Let z r (cos θ + i sin θ ) . Prove = z 2 r 2 (cos 2θ + i sin= 2θ ) , z 3 r 3 (cos 3θ + i sin 3θ ) , = z 4 r 4 (cos 4θ + i sin 4θ ) .

b. Explain why the formula z n r n (cos nθ + i sin nθ ) is true for any positive integer n . = This formula is known as the de Moivre's formula. The formula is saying that z 2 is obtained by rotating z by the angle θ and magnifying its length by the factor r .

= z r (cos θ + i sin θ ) , z 3 is obtained by rotating z by the angle 2θ and magnifying its length by the factor r 2 , etc. c. It follows from (b) that it is natural to define z 0 = 1 . (P.: Why?) z r (cos θ + i sin= θ ) , z −1 r −1 (cos θ − i sin θ ) , = d. Prove that for a non-zero complex number z −2 r −2 (cos 2θ − i sin= 2θ ) , z −3 r −3 (cos 3θ − i sin 3θ ) . =

e. Conclude that de Moivre's formula, = z n r n (cos nθ + i sin nθ ) , is true for any integer n . 2. = Let z1 r1 (cos θ1 + i sin θ1 ) = and z2 r2 (cos θ 2 + i sin θ 2 ) . Prove that z1 r1 = ( cos (θ1 − θ 2 ) + i sin (θ1 − θ 2 ) ) . Interpret this result geometrically. z2 r2

3. = Let z r (cos θ + i sin θ ) . 1 1 a. Show that = (cos θ − i sin θ ) . Interpret this result geometrically. z r 1 b. Conclude = z −n (cos(nθ ) − i sin(nθ )) for any integer n . rn 4. Prove

a. sin 3θ = 3cos 2 θ sin θ − sin 3 θ = 3sin θ − 4sin 3 θ b. sin 5t = 16sin 5 t − 20sin 3 t + 5sin t 44



c. cos 5t = 16 cos5 t − 20 cos3 t + 5cos t Hint: Use the Binomial Theorem: (a + b) n = ∑ ( nk ) a k bn−k . n

k =0

45



Lesson 12 = z r (cos θ + i sin θ ) , emerged as we The trigonometric form of a complex number, probed into the geometric meaning of the operation of multiplication on complex numbers. As you will now see, this form is very useful to finding the roots of z n = a . We will also discover a beautiful geometric property of these roots. ICP 11: Understand the following solution to the equation, z 3 = 1 . Solution P: Explain each step, as needed.

= Set z r (cos θ + i sin θ ) . Clearly the solution must be located on the circle with radius 1 and center at the origin. Therefore, r = 1 . By the de Moivre's formula,

z 3 cos 3θ + i sin 3θ = Therefore, z 3 = 1 if and only if cos 3θ + i sin 3θ = 1. Since cos 0 + i sin 0 = 1, (1) cos 3θ + i sin 3θ = cos 0 + i sin 0 . Hence, = = cos= 3θ cos = 0 1 and sin 3θ sin 3θ 0 Solving for θ , we get 2π k , k = 0, ± 1, ± 2, θ= 3 2π k 2π k , where k is an Equation (1) seems to have infinitely many roots: z cos = + i sin 3 3 integer, but by a theorem we proved in Lesson 10, there should be exactly three! This can easily be explained by the periodicity of the sine and cosine functions. Any integer k can be one of three forms: k = 3m , = k 3m + 1 , and = k 3m + 2 (P: why?) For k = 3m , z =cos 2π m + i sin 2π m =1 + i 0 =1 For = k 3m + 1 , 2π (3m + 1) 2π (3m + 1) + i sin = z =cos 3 3 2π 2π 1 3  2π   2π  cos + i sin = − +i cos  + 2π m  + i sin  + 2π m  = 3 3 2 2  3   3 

46



For = k 3m + 2 , 2π (3m + 2) 2π (3m + 2)  4π   4π  + i sin = cos  z =cos + 2π m  + i sin  + 2π m  = 3 3  3   3  4π 4π 1 3 . + i sin = − −i 3 3 2 2 Thus, our initial equation, z 3 = 1 , has exactly three roots as expected. They are: cos

1 3 1 3 , and − − i . z = 1 , z =− + i 2 2 2 2

Now it is easy to see that 2π 4π and z1 0 z3 = 3 3 Hence, the three roots z1 , z2 , and z3 are the vertices of the equilateral triangle inscribed z1 0 z2 =

in the unit circle (Figure 7). z2

z1 O

1

z3

Figure 7

 ICP 12: Explain without doing any calculation why for the roots we have just found z2 2 = z3 z32 = z2 , and z33 = 1 .

ICP 13: Solve the equation z n = 1 , where n is a positive integer. Compare your solution to the following solution Solution P: Explain each step, as needed.

47



Set z r (cos θ + i sin θ ) . = By the de Moivre's formula, z n = r n (cos nθ + i sin nθ ) = 1

Therefore r = 1 , cos nθ + i sin nθ = 1 We get cos nθ = 1 , sin nθ = 0 Therefore 2π k , k = 0, ± 1, ± 2, n Like before our equation seems to have infinitely many roots: 2π k 2π k , z cos = + i sin n n where k is an integer, we know it should have no more than n . As before, this is easily explained by the periodicity of the sine and cosine functions. Any integer k can be one of n forms: k = nm , = k nm + 1 , and = k nm + 2 ,  ,

θ=

k = nm + (n − 1)

Each of these k s provides one root. k nm + j , where 0 ≤ j ≤ n − 1 , For = 2π (nm + j ) 2π (nm + j )  2π j   2π j  + i sin =cos  + 2π m  + i sin  + 2π m  = n n  n   n  2π j 2π j cos + i sin n n . = So there are exactly n distinct roots corresponding to j 0,1, 2, , n − 1 . z =cos

These roots form the vertices of the regular polygon with n sides inscribed in the unit circle. One of these vertices is 1 since for j = 0 , z = 1 .

 Recall in Lesson 6 we saw that the cubic formula did not seem to produce all the solutions (Surprise 1). Now we know from the FTA that it does have three solutions (with possible repetitions). Let’s now learn how to find them. ICP 14: Find all the solutions to the cubic equation, x 3 − 12 x + 16 = 0. Solution Using the cubic formula we get

48



16 16  16   −12   16   −12  x = − +   +  + 3 − −   +  = 2 2  2  3   2  3  2

3

2

3

3

3

−8 +

(8) + ( −4 ) 2

3

(8) + ( −4 )

+ 3 −8 −

2

3

=

3

−8 + 3 −8

Let z= 3 −8 . Then z 3 = −8 Therefore, 3

 z    =1  −2  . The solutions to this equation are z 1 3 z 1 3 z , =− + i = − −i = 1, −2 2 2 −2 2 2 −2 We have three values for z=

3

−8 :

z1 = −2 , z2 = 1 − i 3 , or z3 = 1 + i 3 .

This gives us six possible values for x = 3 −8 + 3 −8 (P: how?) There are several ways to determine which of these values are roots of our equation. Way 1: We simply substitute the six values of x in the equation and eliminate those that do not satisfy it. Way 2: We begin as in Way 1, except that we stop once we find the first root. For example, when we substitute z2 and z3 we get x=

3

(

) (

)

−8 + 3 −8 = z2 + z3 = 1 − i 3 + 1 + i 3 = 2 . We stop here since 2 is a root

to our equation. Now we simply divide the polynomial x3 − 12 x + 16 by x − 2 , and solve the quadratic equation that results from the division. Way 3: We go back to how we developed the cubic formula. We reduce the equation x 3 − 12 x + 16 = 0 into the system uv = 4 (1) ,  3 3 −16 u + v = where x= u + v . Any u and v for which x= u + v is a solution to our equation must satisfy system (1). Therefore, it remains to check which pair out of z1 , z2 , and z3 satisfy system (1). There are six possible values for u and v : u= v= z1 49



u= v= z2 u= v= z3 u = z1 and v = z2 ; u = z1 and v = z3 u = z2 and v = z3 Proceeding this way, we find that the equation has two roots, 2 and −4 , where 2 repeats twice. x =z1 + z1 =−2 − 2 =−4 x = z2 + z3 = 1 − i 3 + 1 + i 3 = 2 x = z3 + z2 = 1 + i 3 + 1 − i 3 = 2 You will witness the advantage of the third way in the solution of some of the equations in Problem 3 in the next HP.  Homework Problems on Lesson 12 1. Find all the solutions to the following equations:

z 4 + 16 = 0 b. z 3 − 27 = 0 3 c. z + 8 = 0 4 2 d. z + z + 1 =0

a.

2.

Solve the equations: a. z 2 + z + 1 =0 b. z 3 = 1 c.

z2 + 2 z = 4 2

d. z + z =3 + 4i e. iz 2 + (1 + 2i ) z + 1 =0 3.

f. z 4 + 6(1 + i ) z 2 + 5 + 6i = 0 Find all the roots of the following equations and compare them to the roots provided in parenthesis: 1 a. 4 x3 − 5 x − 1 =0 ( x = 1, x = − ) 2 35 200 2 1 ( x = −2 , x = −1 ) b. x 3 + 6 x 2 + x + = 0 3 27 3 3 5 75 125 c. x3 − x + ( x = −5 , x = ) = 0 2 4 4 50



d. x3 + 45 x − 98 = 0

( x = 2 , x =−1 ± 4 3i )

e.

x3 + 3x 2 + 27 x + 81 = 0

( x = −3 , x = ±3 3i )

f.

x3 − 3 3 2 x + 2 = 0

3 ( x == 4, x

6

 6 2 2  −  ) 2 2  

4. How many solutions does the equation z n −1 = iz have? Find all the solutions. 5. Let B, C be complex numbers with C ≠ 0 . The solutions p, q of the quadratic equation

z 2 + Bz + C = 0 are on the same circle with center at the origin. Prove that 6.

7.

p is real. q

1 . Show that the 2 product of the distance of z from i and −i can never be greater than 1. Let z= a + bi be a complex number for which the point (a, b) is on a circle with radius r A point z is in the interior of a circle whose center is −1 and radius

and centered at the origin, and let w= x + yi be its reciprocal. Where is the point ( x, y ) relative to the point (a, b) on the coordinate system? 8. 9.

x + yi . Prove that the point ( Re z , Im z ) is on the unit circle. x − yi Prove the converse of the statement in the previous problem. That is, prove that if z is a

Let z =

complex number for which ( Re z , Im z ) is on the unit circle, then there exist real numbers x and y so that z =

x + yi . x − yi

10.

Let f ( z ) = z 2 − z + 1 . Find maximum and minimum of f over the unit circle.

11.

Vieta’s Theorem: Let z1 and z2 be the solutions to the quadratic equation Az 2 + Bz + C = 0 . Then z1 + z2 = − z1 z2 =

B A

C . A

12.

Prove that for any two complex numbers, z1 and z2 , z1 z2 + z2 z1 is real.

13.

Let z1 , z2 be complex numbers for which Im( z1 ) ≠ 0 and Im( z2 ) ≠ 0 . State necessary and sufficient condition for

14.

z1 to be real. z2

Let Az 2 + Bz + C = 0 be a quadratic equation where A = B= C . Prove: a. If one of the roots of the equation is on the unit circle, then the other root and the sum of the two roots are also on the unit circle. 51



b. If one of the roots of the equation is on the unit circle, then B is the geometric mean of A and C . (i.e., B 2 = AC ) 15. Can the sum of a complex number different from zero and its reciprocal be real? If yes, find all such numbers. 16. a, b, c are three complex numbers on a circle centered at the origin so that b is geometric mean of a, c and c is geometric mean of a, b . Prove that: a. a is geometric mean of b, c. b. The triangle whose vertices are a, b, c is equilateral.

52