Complex Numbers and Polynomials PDF

CHAPTER I - F I E L D S A N D P O LY N O M I A L S 14 Active Reading Task 1 1. List all elements in Z8. 2. Compute · , +. 3. Is it true that for all x 2 Z8 there exists y 2 Z8 so that x + y = ? Our goal now is to understand what algebraic properties Zn satisfies. In particular, we w...

CHAPTER I - F I E L D S A N D P O LY N O M I A L S 14 Active Reading Task 1 1. List all elements in Z8. 2. Compute · , +. 3. Is it true that for all x 2 Z8 there exists y 2 Z8 so that x + y = ? Our goal now is to understand what algebraic properties Zn satisfies. In particular, we will investigate if Zn is a field in class and in HW. As a final note, we we will generally drop the cumbersome [x] notation, and moving forward we will © ™ denote Zn = 0, 1, · · · , n ° 1 , except in cases where confusion may arise. If you are ever unsure, just ask. C OMPLEX N UMBERS The quadratic formula for the roots of the equation ax 2 + bx + c = 0 is the starting point for this section, so let’s begin by reminding ourselves what it says. Theorem 3: Quadratic Formula Let a, b, c 2 R. The quadratic equation ax 2 + bx + c = 0 (where a 6= 0) has: p °b ± b 2 ° 4ac 1. Solutions x = if b 2 ° 4ac ∏ 0. 2a 2. No solutions if b 2 ° 4ac < 0. Using the quadratic formula we can see that in order for every quadratic equation over R to have a root, we need to enlarge our number system R to a system where we can take the square root of a negative number. It turns out that simply dealing with the quadratic equations, we will create a number system where every polynomial equation has a root (not just the quadratic ones). This number system is called the “Complex Numbers". Basics Definition 3 p Let i = °1. I.e. i is a number with the property that i 2 = °1. © ™ Let C = a + bi | a, b 2 R. We call C the set of complex numbers and we define addition and  CHAPTER I - F I E L D S A N D P O LY N O M I A L S 15 multiplication +, · in the obvious ways: (x + yi ) + (a + bi ) = (x + a) + (y + b)i (x + yi ) · (a + bi ) = ax + bxi + a yi + b yi 2 = (ax ° b y) + (a y + bx)i Note that for multiplication, we simply expand brackets and collect like terms as expected, but use i 2 = °1 when mutliplying i y with Example 4 i b. In particular, this means that there is no need to memorize the formulae in the definition. Compute (1 + 3i ) + (2 ° 5i ) and (3 + 2i )(1 ° i ). Solution: For adding we simply collect “like terms": (1 + 3i ) + (2 ° 5i ) = (1 + 2) + (3 + (°5))i = 3 ° 2i. For multiplication, we “expand and collect like terms" making use of i 2 = °1: (3 + 2i )(1 ° i ) = 3 ° 3i + 2i ° 2i 2 = 3 ° 3i + 2i ° 2(°1) = (3 + 2) + (°3 + 2)i = 5°i Active Reading Task 2 Compute (2 + i ) + ((1 + i )(2 + 4i )). The following theorem confirms for us that C is a field. Theorem 4: Basic Properties of Arithmetic for Complex Numbers The set of complex numbers C, with addition and multiplication defined above, is a field. Proof. We set the special elements 0, 1 to be 0 = 0 + 0i and 1 = 1 + 0i. We will leave axioms 1,3,4 as exercises, and prove 2 and 5 here. For 2, let z 1 = x 1 + y 1 i , z 2 = x 2 + y 2 i , z 3 = x 3 + y 3 i. The we have:  CHAPTER I - F I E L D S A N D P O LY N O M I A L S 16 ° ¢ (z 1 + z 2 ) + z 3 = (x 1 + y 1 i ) + (x 2 + y 2 i ) + (x 3 + y 3 i ) = ((x 1 + x 2 ) + (y 1 + y 2 )i ) + (x 3 + y 3 i ) by definition of addition of complex numbers = ((x 1 + x 2 ) + x 3 ) + ((y 1 + y 2 ) + y 3 )i by definition of addition of complex numbers = (x 1 + (x 2 + x 3 )) + (y 1 + (y 2 + y 3 ))i since addition of real numbers is associative = (x 1 + y 1 i ) + (x 2 + x 3 ) + (y 2 + y 3 )i by definition of addition of complex numbers ° ¢ = (x 1 + y 1 i ) + (x 2 + y 2 i ) + (x 3 + y 3 i ) by definition of addition of complex numbers = z 1 + (z 2 + z 3 ) by definition of addition of complex numbers. A similar computation, using the associativity of multiplication of real numbers can be done to prove that complex multiplication is associative. For 5, we need to show that C has additive and multiplicative inverses. Let z = x + yi 2 C. Set w = (°x)+(°y)i. Then z + w = (x + yi )+((°x)+(°y)i ) = (x +(°x))+(y +(°y))i = 0 + 0i = 0. So additive inverses exist. 1 Now suppose that z = x + yi 6= 0. Then at least one of x, y 6= 0 and hence x 2 + y 2 6= 0. Set w = x 2 +y 2 (x ° *You may ask “where does this yi ).* Since x 2 + y 2 6= 0 this complex number makes sense. choice of multiplicative inverse come from?" which is a very valid question. There are a couple of We now check that it is the inverse: ways to arrive here. One is to write 1 w = a + bi , compute zw, and set it zw = (x + i y)( (x ° yi )) equal to 1 = 1+0i. This leads to a set x2 + y 2 of equations that a, b must satisfy, 1 which leads you to the correct = (x + yi )(x ° yi ) x2 + y 2 definition for the inverse. The 1 second requires the “conjugate" = 2 (x 2 ° x yi + y xi + y 2 ) and appears in Theorem 5. x + y2 1 = 2 (x 2 + y 2 ) x + y2 = 1. Active Reading Task 3 (2 + 3i ) Find (1 ° i )°1 , and then compute. 1°i Geometric Interpretation of Complex Numbers Real numbers are visualized using the number line. To define a complex number requires the inde- pendent choice of two real numbers, and so the set of all possible complex numbers z = x + i y can be thought of as the same thing as the set of all pairs of real numbers (x, y). The latter is the x y-plane R2.  CHAPTER I - F I E L D S A N D P O LY N O M I A L S 17 We can visualize the complex numbers in the x y-plane by identifying z = x + yi with the point (x, y) 2 R2. y z = x+iy x Definition 4 Given a complex number z = x + yi , we define its conjugate by: z = x ° yi. We define the length (or modulus) of a complex number by: q |z| = x2 + y 2. Note that in the x y-plane, we obtain z, the conjugate of z, by reflecting z in the x-axis, and the length of a complex number is just the usual distance from z to the origin in the x y-plane. y z = x + yi |z| x °y z = x ° yi Active Reading Task 4 For z = 1 + 3i , compute z and |z|. Theorem 5: Properties of Conjugate and Length For any z, w 2 C we have: 1. z ± w = z ± w.  CHAPTER I - F I E L D S A N D P O LY N O M I A L S 18 2. zw = z w. ° ¢ 3. wz = wz (provided w 6= 0). 4. z = z. 5. zz = |z|2. z 6. z °1 = |z|2 (provided z 6= 0). 7. |zw| = |z||w|. Ø Ø |z| 8. Ø wz Ø = |w| (provided w 6= 0). 9. |z + w| ∑ |z| + |w| (“Triangle Inequality for Complex Numbers"). Example 5 Let’s verify properties 5,6 in Theorem 5. Solution: For property 5, we must compute the product zz and compare with |z|2. For z = x + i y, we have |z|2 = x 2 + y 2. Then we get: zz = (x + i y)(x ° i y) = x2 ° i x y + i y x ° y 2i 2 = x2 + y 2 = |z|2. For property 6*: if z = x + i y 6= 0 we have *It is possible to give an alternate proof of (6) using (5). Think about how you can do this. z x +i y x °i y x y = = = ° i = z °1. |z|2 x 2 + y 2 x 2 + y 2 x 2 + y 2 x 2 + y 2 Active Reading Task 5 Verify property 1, by setting z = x + i y, w = a + bi and computing the left and right side of the equation z + w = z + w. Repeat for the equation z ° w = z ° w. Polar Form One way to think of how to travel from the origin to a complex number z = x + i y, or its point (x, y) in the x y-plane, is using the x y-coordinates of that point: the first coordinate tells us how far to travel  CHAPTER I - F I E L D S A N D P O LY N O M I A L S 19 in the x-direction (left/right), and the second coordinate tells us how far to travel in the y-direction (up/down). However there are other ways to arrive at the same point, and one that has many useful applications involves using what is called the “polar form". In this approach, we arrive at (x, y) by specifying the direction that we should travel (specified as an angle from the positive x-axis) and how far we should travel in that direction. This is illustrated in the diagram below. y = r sin µ z = x +i y |z| µ x = r cos µ If z = x + yi , then using a bit of trigonometry and geometry, we see that x = r cos µ, y = r sin µ, where r = |z|, and µ is the angle between z and the positive x axis (measured counterclockwise, in radians). Definition 5 p For z = x + yi , we define its polar form as z = r e i µ , where r = |z| = x 2 + y 2 and µ is the angle between z and the positive x axis (measured counterclockwise, in radians). The angle µ is called the argument of z, and r is called the length (or modulus) of z. Example 6 p Express z = 1 ° 3i in polar form. Solution: p We first plot z = 1 ° 3i. µ x =1 2 p p y =° 3 z = 1 ° 3i  CHAPTER I - F I E L D S A N D P O LY N O M I A L S 20 q p p Note that r = |z| = 12 + ( 3)2 = 4 = 2. So µ is a fourth quadrant angle sastisfying: 1 1 = 2 cos µ, or cos µ = 2 p p ° 3 ° 3 = 2 sin µ, or sin µ = 2. This tells us that µ = 5º 3. (You can also deduce the “reference angle" from this information and the diagram above to find µ.) p 5º So 1 ° 3i = 2e 3 i. Active Reading Task 6 3º Express the complex number z = 2e i 2 in the form z = x + yi and plot it in the x y-plane. Express w = 1 + i in polar form w = r e i µ. Begin by plotting w in the x y-plane. You might ask yourself why do we use exponential notation for the polar form of a complex number. The following theorem helps provide an explanation. Theorem 6 Let z = r e i µ , w = Re i ¡. zw = r Re i (µ+¡) z n = r n e i nµ This tells us that the angles behave as exponents when multiplying complex numbers in polar form. This helps explain the use of exponential notation.* *There is more to the connection between polar form and exponen- tial notation than what we see in this theorem. You can look at the This theorem is quite useful for computing powers of complex numbers as well as solving power Exercise 20 for more details. equations or computing nth roots of complex numbers. Active Reading Task 7 p i 2º For z = 2e 3 , find z 13. Express your final answer in the form a + bi.  CHAPTER I - F I E L D S A N D P O LY N O M I A L S 21 P OLYNOMIALS Polynomials play a fundamental role in many parts of mathematics, including in linear algebra. We will see that in a lot of ways, polynomials are similar to integers: we can add and multiply them, there is division with remainder, which leads notions of divisibility and divisors. We begin our study by reminding ourselves of the definition. Let F be a field (fixed throughout, but unspecified). Definition 6 A polynomial p with coefficients from F is an expression p(x) = c 0 + c 1 x + c 2 x 2 + · · · + c n x n where c i 2 F. We call the field elements c 0 ,... , c n the “coefficients" of p. The largest exponent n so that c n 6= 0 is called the degree of p, and we typically write deg p = n. Constant polynomials are of degree 0. The set of all polynomials over F is denoted by P(F). The set of all polynomials of degree less than or equal to n is denoted by Pn (F). Example 7 Let F = Q. Then p(x) = 3x 5 + 7x 2 + x ° 45 is a polynomial of degree 5, while q(x) = 2x 3 + 5x 2 + 23x ° 1 is a polynomial of degree 3. The expression g (x) = 2x 3 + 4x + 5 + x52 is not a polynomial due to the presence of the term 5 x2. Since we know how to add and multiply the coefficients of polynomials (they are field elements) we can use that to add, subtract and multiply polynomials using the usual rules for “opening brackets" and “collecting like terms". For example, if F = R we have: (x 2 + x + 3)(2x 3 + 3x 2 + 5x + 1) + (2x 4 ° 3x 2 + 4x + 4) = = (2x 5 + 3x 4 + 5x 3 + x 2 + 2x 4 + 3x 3 + 5x 2 + x + 6x 3 + 9x 2 + 15x + 3) + (2x 4 ° 3x 2 + 4x + 4) = (2x 5 + 5x 4 + 14x 3 + 15x 2 + 16x + 3) + (2x 4 ° 3x 2 + 4x + 4) = 2x 5 + 7x 4 + 14x 3 + 12x 2 + 20x + 7. One very useful fact about polynomials is that we can perform “long division with remainder" like we can with integers. This will allow us to define divisibility for polynomials, like we do for integers.  CHAPTER I - F I E L D S A N D P O LY N O M I A L S 22 Theorem 7 Let F be a field, and f , g 2 P(F) be non-zero polynomials. Then there exist unique polynomials q, r 2 P(F) so that: 1. f (x) = q(x)g (x) + r (x) 2. deg(r ) < deg g if deg(g ) 6= 0 3. r = 0 if deg(g ) = 0. Proof. If deg(g ) = 0, then g is a constant and there is nothing to prove. Let n = deg f and m = deg g. Write f (x) = A n x n + A n°1 x n°1 + · · · + A 0 and g (x) = B m x m + B m°1 x m°1 + · · · + B0. If n < m, we set q(x) = 0 and r (x) = f (x). If n = m we have f (x) = A n x n + A n°1 x n°1 + · · · + A 0 and g (x) = B n x n + B n°1 x n°1 + · · · + B 0 where A n , B n 6= 0. Then f (x) = A n B n°1 g (x) + (A n°1 x n°1 + · · · + A 0 ) ° (A n B n°1 )(B n°1 x n°1 + · · · + B 0 ). So we set q(x) = A n B n°1 and r (x) = (A n°1 x n°1 + · · · + A 0 ) ° (A n B n°1 )(B n°1 x n°1 + · · · + B 0 ). Suppose that 0 ∑ m < n. For m = 0, since g is constant and g 6= 0, we have g (x) = B for some B 6= 0 2 F. Then set q(x) = B °1 f (x) and r (x) = 0. Then we have that q(x)g (x) + r (x) = B °1 f (x)B + 0 = f (x) as required. For m > 0, we use induction on n. For n = 1 either m > n, or m = n = 1: each case has been dealt with above. So the result holds for n = 1. Now suppose that the result holds n = k. For n = k + 1 we need to use the polynomials f , g to con- struct a polynomial of degree k to which we can apply the induction hypothesis. Since deg f > deg g we must multiply g (x) by an appropriate power to obtain a polynomial of degree equal to f. This leads us to consider multiples of the form Ax n°m g (x). We need to choose A carefully so that F (x) = f (x) ° Ax n°m g (x) is guaranteed to be of degree less than n = deg f. In other words we must choose A so that the coefficient of x n in Ax n°m g (x) is A n. By expanding Ax n°m g (x) we see that we should set A = B m °1 An. °1 Consider the polynomial F (x) = f (x) ° B m A n x n°m g (x). We have that deg F < n so by induction hypothesis we have that F (x) = q 1 (x)g (x) + r (x)  CHAPTER I - F I E L D S A N D P O LY N O M I A L S 23 where deg r < deg g. °1 Note that we can write f (x) = F (x) + B m A n x n°m g (x). Using our expression for F we get: °1 f (x) = F (x) + B m A n x n°m g (x) °1 = q 1 (x)g (x) + r (x) + B m A n x n°m g (x) °1 = (q 1 (x) + B m A n x n°m )g (x) + r (x) °1 So we set q(x) = q 1 (x) + B m A n x n°m and r (x) = r (x). To see that such an expression is unique suppose that f (x) = q 1 g (x) + r 1 (x) and f (x) = q 2 g (x) + r 2 (x) where deg r 1 , deg r 2 < deg g. Then we must have: 0 = f (x) ° f (x) = (q 1 g (x) + r 1 (x)) ° (q 2 g (x) + r 2 (x)) = (q 1 (x) ° q 2 (x))g (x) ° (r 2 (x) ° r 1 (x)) (q 1 (x) ° q 2 (x))g (x) = (r 2 (x) ° r 1 (x)) But since deg r 2 °r 1 < deg g the only way this is possible is if q 1 ° q 2 = 0 and hence q 1 = q 2. (Otherwise deg(q 1 ° q 2 ) · deg g > deg r 2 ° r 1 , and we cannot have these polynomials equal to each other.) This then implies that r 2 ° r 1 = 0, and hence r 2 = r 1. Example 8 Let f (x) = x 3 + 3x ° 5, g (x) = x 2 + 2x + 4 2 P(R). Then we can obtain q, r using long division of polynomials as illustrated below: ∂ x °2 x 2 + 2x + 4 x 3 +0x 2 +3x °5 x 3 +2x 2 +4x °2x 2 °x °5 °2x 2 °4x °8 3x +3 So we conclude that x 3 + 3x ° 5 = (x ° 2)(x 2 + 2x + 4) + (3x + 3) Active Reading Task 8 Let F = R. For each pair, use the division algorithm to write f (x) = q(x)g (x) + r (x). 1. f (x) = x 2 + x + 1, g (x) = x + 2. 2. f (x) = x 4 + 3x 2 + 2, g (x) = x.  CHAPTER I - F I E L D S A N D P O LY N O M I A L S 24 3. f (x) = x 4 + 3x 2 + 2, g (x) = x 2 + 1. 4. f (x) = x, g (x) = x + 1. Like we do with integers, we can define a notion of “divisibility" for polynomials. Definition 7 Let F be a field and f , g 2 P(F). We say that g divides f if f (x) = q(x)g (x) for some polynomial q 2 P(F). In the integers, the “prime numbers" play an important role: all other numbers factor into prod- ucts of the primes. There are polynomials which play a similar role: we call them the “irreducible" polynomials. Definition 8 We say that a non-constant polynomial* p 2 P(F) is “irreducible" if we cannot express p as a *That is we only apply this defi- nition to polynomials p so that product of polynomials of smaller degree. deg p ∏ 1. I.e. p is irreducible if we cannot write p(x) = g (x)q(x) for any polynomials g , q 2 P(F) with the property that both deg g , deg q < deg p. Example 9 p Let F = Q. The polynomial f (x) = x 2 ° 2 is irrreducible. (Otherwise 2 would be rational, which it is not.) p p Let F = R. The polynomial f (x) = x 2 ° 2 is not irreducible: f (x) = (x ° 2)(x + 2). Let F = C*. The polynomial f (x) = 3x + (2 + 5i ) is irreducible. To see this, since deg f = 1 we would *Does the choice of field really matter in this argument? Try to need f (x) = g (x)q(x) with deg g , deg q = 0. This would imply that g , q are both constant generalize this to a statement which polynomials. Since the product of two constant polynomials is again a constant polynomial, it is holds for any field F. not possible to find such g , q. Active Reading Task 9 Find an element of p 2 P(Q) satisfying the following conditions: 1. deg p = 3 2. p is irreducible. Does such a polynomial exist if we replace Q with R? Explain.  CHAPTER I - F I E L D S A N D P O LY N O M I A L S 25 The following theorem is used (often implicitly) when we solve polynomial equations by factoring. It is a very useful theorem. Theorem 8 Let F be a field, p 2 P(F) and deg p ∏ 1. Then a 2 F is a root of p if and only if x ° a divides p. Proof. First, suppose that a is a root of p. By Theorem 7 we can write p(x) = (x ° a)q(x) + r (x) where deg r < deg(x ° a) = 1. So we have that deg r = 0, and hence r (x) = r is a constant polynomial. Since a is a root of p we must have: p(a) = (a ° a)q(a) + r = 0+r =r = 0. So r (x) = r = 0, and p(x) = (x ° a)q(x). I.e. x ° a divides p. Now suppose that x ° a divides p. Then we can write p(x) = (x ° a)q(x) and we have that p(a) = (a ° a)q(a) = 0 · q(a) = 0 and hence a is a root of p. The following Theorems is one of the key results about complex numbers. We will not prove it in this *You can find a proof here which book*, but you are free to make use of it when needed. is the same proof as is found in Friedberg, Insell, Spence’s “Linear Algebra" in Appendix D. Theorem 9: Fundamental Theorem of Algebra Every non-constant polynomial has a root over C. In fact, every non-constant polynomial factors completely into a product of linear terms over C. A DDITIONAL E XERCISES Exercise 1 1. Compute the following in Z9. (a) 3 · 5 (b) 4 + 7  CHAPTER I - F I E L D S A N D P O LY N O M I A L S 26 (c) (4 · 5) + 3 2. Compute the following in Z12. (a) 5 + 7 (b) (3 · 5) + (5 · 7) + 1. (c) 34 + (8 · 7) + 11. 3. In each case, solve the equation over Z5 or show that no solutions exist. (a) 2x + 3 = 0 (b) x 2 + 3 = 0 (c) x 2 + x + 1 = 0 (d) x 2 + 1 = 0 Exercise 2 Let z = 1 ° i , w = 3 + 4i. 1. Compute z + w, w ° i z, zw. w 2. Compute z. 3. |z| and w. 4. Solve for v if zv = i w + i (1 ° i ). 5. Sketch z and w in the x y-plane. 6. Find the polar form of z. 7. Compute z 43. Exercise 3 In each case, perform long division with remainder. 1. f (x) = x 2 + x + 1, g (x) = x + 1; where F = Q. 2. f (x) = x 3 + 3x 2 + 2x + 1, g (x) = x 2 + 3x + 1; where F = Q. 3. f (x) = x 3 + 3x 2 + 2x + 1, g (x) = x 2 + 3x + 1; where F = Z5. 4. f (x) = x 2 + 1, g (x) = x 2 + x + 2; where F = R. 5. f (x) = x + 1, g (x) = x 2 ; where F is any field.  CHAPTER I - F I E L D S A N D P O LY N O M I A L S 27 Exercise 4 Determine if the given polynomial is irreducible over the given field. As always, justify your answer. 1. p(x) = x 2 + 1 over F = Q. 2. p(x) = x 2 + 1 over F = Z2. 3. p(x) = x 2 + 1 over F = Z3. 4. p(x) = x 2 + 1 over F = Z5. 5. p(x) = x 4 + 3x 2 + 2 over F = Q. 6. p(x) = x 4 + 3x 2 + 2 over F = Q. Exercise 5 Let F be a field. Prove that any polynomial of degree 1 is irreducible. Exercise 6 In Definition 1 we referred to addition and multiplication as “operations on the set F". In this exercise we will make this a bit more formal. An operation on F is a function F £ F ! F. So a field is then a set F with two special elements 0, 1 2 F and two functions A : F £ F ! F and M : F £ F ! F, where A denotes addition and M denotes multiplication. 1. Rewrite the field axioms as conditions on the functions A, M. 2. What, if any, symmetry do the functions A, M possess? 3. In the case of F = Q write out these functions explicitly. Exercise 7 1. Compute the addition and multiplication tables of Zn for n = 2, 3, 4,... , 12, 13. 2. For which of the values above is Zn not a field? How can you tell? 3. Conjecture which values of n have the property that Zn is a field. 4. Prove your conjecture. (Hint: what field axioms do all Zn satisfy? What is left to show?)  CHAPTER I - F I E L D S A N D P O LY N O M I A L S 28 Exercise 8 Prove the remaining parts of Theorem 1. Exercise 9 Let n > 1. Prove that if a 2 Zn is invertible if and only if gcd(a, n) = 1. As a hint: recall that if d = gcd(x, y) then there exist s, t 2 Z so that sx + t y = d. Exercise 10 Let F be a finite field. Prove that the addition and multiplication tables for F have the following properties: 1. They are symmetric. 2. The first row and column of the multiplication table consists of 0’s. 3. Each row and each column of the addition table contains every element of F exactly once. 4. Each non-zero row and non-zero column of the multiplication table contains every element of F exactly once. © ™ 5. Use these properties to deduce whether the following tables for the set 0, 1, a, b, c describe a field. + 0 1 a b c · 0 1 a b c 0 0 1 a b c 0 0 0 0 0 0 1 1 c c a 0 1 0 1 c b a a a c b c b a 0 c b c 1 b b a 0 c a b 0 b a c 1 c c 0 b a 1 c 0 a 1 1 a Exercise 11 © ™ Let F4 = 0, 1, x, y be a field with four elements and that 1 + 1 = 0.* *This is true in F4 without adding it as an assumption, but we are 1. Prove that x + x = 0 and y + y = 0. adding it to simplify the problem. 2. Prove that it’s impossible that x + y = x. 3. Prove that it’s impossible that x + y = y. 4. Prove that it’s impossible that x + y = 0. 5. Deduce that x + y = 1. 6. Prove that it’s impossible that x · y = 0.  CHAPTER I - F I E L D S A N D P O LY N O M I A L S 29 7. Prove that it’s impossible that x · y = x. 8. Prove that it’s impossible that x · y = y. 9. Deduce the value of x · y. 10. Complete the addition and multiplication tables for this field. Exercise 12 In this exercise we will investigate the solutions to the equation z n = 1 over the field F = C. 1. Solve the equation z n = 1 for n = 3, 4, 5, 6 using polar form. For each value of n sketch the solutions. 2. Prove that every solution to z n = 1 lies on the unit circle. (Hint: what can you say about |z| if z is a solution to the equation z n = 1?) 3. Solve the equation z n = 1 for arbitrary n 2 N. Express your solutions in polar form. 4. How many solutions does the equation z n = 1 have? 5. What geometric shape do the solutions form? Exercise 13 The exponential function on complex numbers is defined as follows: For z = x + i y 2 C we define e z = e x+i y = e x e i y. º º For example: e 2+ 4 i = e 2 e 4 i = e 2 (cos( º4 ) + i sin( º4 )). 1. Prove the famous formula: e i º + 1 = 0. 2. Prove that e 0 = 1. 3. Let z = x + i y, w = s + i t 2 C. Prove that if e z = e w , then x = s. 4. Find all z so that e z = 1. (Hint: The y component of z becomes the “angle" for the cos, sin functions in e i y = cos(y) + i sin(y).) 5. Given w = x + i y 6= 0, find all z 2 C so that e z = w. That is, find all “logarithms" of w. © ™ 6. Is there a well-defined function log : C \ 0 ! C, which is inverse to the exponential function? Explain. © ™ 7. If there is a well-defined logarithm log : C \ 0 ! C, define it. If no, “restrict" the domain of the exponential function, so that you can define its inverse.  CHAPTER I - F I E L D S A N D P O LY N O M I A L S 30 Exercise 14 Find all complex solutions to the equation z 6 = °64. Express your answers in the form z = x + i y. Sketch your solutions in the x y-plane. Exercise 15 © p(x) ™ Let R(F) = f (x) = q(x) | p, q 2 P(F) be the set of rational functions over F. p(x) r (x) For f (x) = q(x) , g (x) = s(x) 2 R(F), define +, · as follows: p(x)s(x) + q(x)r (x) ( f + g )(x) = q(x)s(x) p(x)r (x) ( f · g )(x) =. q(x)s(x) Prove that R(F) is a field. (Hint: This is very similar to proving that Q is a field.) Exercise 16 Let F be a field. We define the characteristic of F to be the smallest value of n so that 1 | +1+ {z· · · + 1} = 0. We denote this number by ch(F). If no such n exists, we say that ch(F) = 0. n times For example: Z2 has characteristic 2, because 1 + 1 = 0, but 1 6= 0, so 2 is the smallest n so that 1 | +1+{z· · · + 1} = 0. n times 1. Prove that ch(Q) = 0. 2. Prove that ch(Zp ) = p. 3. Prove that if ch(F) 6= 0, then ch(F) = p must be prime. (Hint: Suppose that ch(F) = a · b is composite. Consider the product (1 | +1+ {z· · · + 1}) · (1 | +1+ {z· · · + 1}) to arrive at a contradiction.) a times b times 4. What is ch(F4 ), where F4 is the field with 4 elements from Exercise 11. Exercise 17 Let F = Z3. Find all irreducible polynomials of degrees 1, 2, 3. Exercise 18 Complete the proof of the remaining parts of Theorem 5  CHAPTER I - F I E L D S A N D P O LY N O M I A L S 31 Exercise 19 Complete the proof of Theorem 4, by verifying that the remaining field axioms are satisfied in C. Exercise 20 Recall the following power series expansions: X1 1 et = xn n=0 n ! X 1 (°1)n 2n+1 sin(t ) = x n=0 (2n + 1)! X1 (°1)n cos(t ) = x 2n. n=0 (2n)! It is possible to show that the representation by power series for e t works for complex values of t , not just real ones! This will be assumed here, but is proved in a course on complex functions (e.g. a “complex analysis", or “complex variables course".) 1. Write t = x + i y and recall that we define e x+i y = e x e i y. Compute e i y as a power series, using the power series representation above, by substituting t = i y. 2. Write out by hand, the first ten terms of this series. 3. Simplify the powers of i k that appear in (b). What do you notice? 4. Find a general formula for i k depending on k. 5. Prove that you can split the sum e i y = cos(y) + i sin(y) by regrouping the even and odd terms of the power series e i y.