Math 1230, Notes 8. Sep. 23, Math 1230, Notes 8 Sep. 23, / 28

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1 Math 1230, Notes 8 Sep. 23, 2014 Math 1230, Notes 8 Sep. 23, / 28

2 algebra and complex numbers Math 1230, Notes 8 Sep. 23, / 28

3 algebra and complex numbers Math 1230, Notes 8 Sep. 23, / 28

4 Classification of numbers: Math 1230, Notes 8 Sep. 23, / 28

5 Classification of numbers: positive integers (Greece and before) Math 1230, Notes 8 Sep. 23, / 28

6 Classification of numbers: positive integers (Greece and before) positive rational numbers (Greece and before) Math 1230, Notes 8 Sep. 23, / 28

7 Classification of numbers: positive integers (Greece and before) positive rational numbers (Greece and before) positive irrational numbers (Greece) Math 1230, Notes 8 Sep. 23, / 28

8 Classification of numbers: positive integers (Greece and before) positive rational numbers (Greece and before) positive irrational numbers (Greece) zero and negative numbers (India, 7th century - Brahmagupta) Math 1230, Notes 8 Sep. 23, / 28

9 Classification of numbers: positive integers (Greece and before) positive rational numbers (Greece and before) positive irrational numbers (Greece) zero and negative numbers (India, 7th century - Brahmagupta) algebraic numbers (Muslim world; al-khwarizimi, 830 AD; Europe, 18th century) Math 1230, Notes 8 Sep. 23, / 28

10 real numbers, (Dedekind and Cantor, 19th century) Math 1230, Notes 8 Sep. 23, / 28

11 real numbers, (Dedekind and Cantor, 19th century) complex numbers, (Cardano and Bombelli, 16th century) Math 1230, Notes 8 Sep. 23, / 28

12 Definition: An algebraic number is a solution to an algebraic equation of the form x n + a n 1 x n a 1 x + a 0 = 0 where a 0,..., a n 1 are rational real numbers. Math 1230, Notes 8 Sep. 23, / 28

13 Definition: An algebraic number is a solution to an algebraic equation of the form x n + a n 1 x n a 1 x + a 0 = 0 where a 0,..., a n 1 are rational real numbers. (al-khwarizimi only considered n = 2, known already to the Babylonians, but he introduced algebraic notation, a huge advance.) Math 1230, Notes 8 Sep. 23, / 28

14 Definition: An algebraic number is a solution to an algebraic equation of the form x n + a n 1 x n a 1 x + a 0 = 0 where a 0,..., a n 1 are rational real numbers. (al-khwarizimi only considered n = 2, known already to the Babylonians, but he introduced algebraic notation, a huge advance.) In Greece, and even to al-khwarizimi, had no solution. x + 6 = 0 Math 1230, Notes 8 Sep. 23, / 28

15 Definition: An algebraic number is a solution to an algebraic equation of the form x n + a n 1 x n a 1 x + a 0 = 0 where a 0,..., a n 1 are rational real numbers. (al-khwarizimi only considered n = 2, known already to the Babylonians, but he introduced algebraic notation, a huge advance.) In Greece, and even to al-khwarizimi, had no solution. Babylonians knew how to solve x + 6 = 0 ax 2 + bx + c = 0 in practice (for explicit examples with positive solutions) but had no formula. Math 1230, Notes 8 Sep. 23, / 28

16 Definition: An algebraic number is a solution to an algebraic equation of the form x n + a n 1 x n a 1 x + a 0 = 0 where a 0,..., a n 1 are rational real numbers. (al-khwarizimi only considered n = 2, known already to the Babylonians, but he introduced algebraic notation, a huge advance.) In Greece, and even to al-khwarizimi, had no solution. Babylonians knew how to solve x + 6 = 0 ax 2 + bx + c = 0 in practice (for explicit examples with positive solutions) but had no formula. As we saw in Euclid, neither did the Greeks, who considered different cases geometrically. Math 1230, Notes 8 Sep. 23, / 28

17 Brahmagupta (628) : To the absolute number (the c ) multiplied by four times the (coefficient of the) square, add the square of the (coefficient of the) middle term, the square root of the same, less the (coefficient of the) middle term, being divided by twice the (coefficient of the) square is the value. Math 1230, Notes 8 Sep. 23, / 28

18 Brahmagupta (628) : To the absolute number (the c ) multiplied by four times the (coefficient of the) square, add the square of the (coefficient of the) middle term, the square root of the same, less the (coefficient of the) middle term, being divided by twice the (coefficient of the) square is the value. Brahmagupta allowed zero and negative numbers as possible solutions. Math 1230, Notes 8 Sep. 23, / 28

19 Brahmagupta (628) : To the absolute number (the c ) multiplied by four times the (coefficient of the) square, add the square of the (coefficient of the) middle term, the square root of the same, less the (coefficient of the) middle term, being divided by twice the (coefficient of the) square is the value. Brahmagupta allowed zero and negative numbers as possible solutions. al-khwarizimi did not, but he expressed Brahmagupta s words as an algebraic formula: x = b ± b 2 4ac. 2a Math 1230, Notes 8 Sep. 23, / 28

20 Up to about 1572, x = 0 had no solution. Until 1545 no one worried about this; it was just a fact. Math 1230, Notes 8 Sep. 23, / 28

21 Up to about 1572, x = 0 had no solution. Until 1545 no one worried about this; it was just a fact. Cardano (1545) tried to solve x 3 + ax 2 + bx + c = 0. Math 1230, Notes 8 Sep. 23, / 28

22 Up to about 1572, x = 0 had no solution. Until 1545 no one worried about this; it was just a fact. Cardano (1545) tried to solve x 3 + ax 2 + bx + c = 0. Let y = x + a 3 and we find that y 3 = py + q (1) for some numbers p and q which depend on a, b, c. Cardano knew how to solve quadratic equations, and thought of a clever way to obtain a quadratic equation from (1). Let y = u + v. Then 3uv = p u 3 + v 3 = q. (2) Math 1230, Notes 8 Sep. 23, / 28

23 Eliminate v and we get ( p ) 3 u 3 + = q. 3u Math 1230, Notes 8 Sep. 23, / 28

24 Eliminate v and we get ( p ) 3 u 3 + = q. 3u Multiply by u 3 to get ( u 3 ) 2 qu 3 + ( p 3 )3 = 0. Math 1230, Notes 8 Sep. 23, / 28

25 Eliminate v and we get ( p ) 3 u 3 + = q. 3u Multiply by u 3 to get ( u 3 ) 2 qu 3 + ( p 3 )3 = 0. Use the quadratic equation to get a solution for u 3, use (2) to get v 3, and we find that y = u + v = 3 q (q ) 2 ( p ) q (q ) 2 ( p ) Math 1230, Notes 8 Sep. 23, / 28

26 Eliminate v and we get ( p ) 3 u 3 + = q. 3u Multiply by u 3 to get ( u 3 ) 2 qu 3 + ( p 3 )3 = 0. Use the quadratic equation to get a solution for u 3, use (2) to get v 3, and we find that y = u + v = 3 q (q ) 2 ( p ) q (q ) 2 ( p ) This might yield a real number. If it doesn t then apparently there is no solution. So what? We already know there are algebraic equations with no solution. Math 1230, Notes 8 Sep. 23, / 28

27 Cardano considered the particular case x 3 = 15x + 4, and found that x = Math 1230, Notes 8 Sep. 23, / 28

28 Cardano considered the particular case x 3 = 15x + 4, and found that x = So, no solution. Math 1230, Notes 8 Sep. 23, / 28

29 Cardano considered the particular case x 3 = 15x + 4, and found that x = So, no solution. Except, Cardano could see that x = 4 is a solution. Math 1230, Notes 8 Sep. 23, / 28

30 Bombelli, 1572: Math 1230, Notes 8 Sep. 23, / 28

31 Bombelli, 1572: Express 3 2 ± 121 in the form α ± β γ 2. Math 1230, Notes 8 Sep. 23, / 28

32 Bombelli, 1572: Express 3 2 ± 121 in the form α ± β γ ± 121 = 2 ± 1. Math 1230, Notes 8 Sep. 23, / 28

33 Bombelli, 1572: Express 3 2 ± 121 in the form α ± β γ 2. Proof: cube both sides. 3 2 ± 121 = 2 ± 1. Math 1230, Notes 8 Sep. 23, / 28

34 Bombelli, 1572: Express 3 2 ± 121 in the form α ± β γ 2. Proof: cube both sides. x = 3 2 ± 121 = 2 ± 1. ( 2 + ) ( ) 1 = 4. Math 1230, Notes 8 Sep. 23, / 28

35 Bombelli, 1572: Express 3 2 ± 121 in the form α ± β γ 2. Proof: cube both sides. x = 3 2 ± 121 = 2 ± 1. ( 2 + ) ( ) 1 = 4. The solution was real, and he found it using symbols that seemed not to represent anything real. Math 1230, Notes 8 Sep. 23, / 28

36 Is 1 a number? Timeline: Math 1230, Notes 8 Sep. 23, / 28

37 Is 1 a number? Timeline: natural numbers {1,2,3,...} prehistory Math 1230, Notes 8 Sep. 23, / 28

38 Is 1 a number? Timeline: natural numbers {1,2,3,...} prehistory rational numbers: before 2000 bc Math 1230, Notes 8 Sep. 23, / 28

39 Is 1 a number? Timeline: natural numbers {1,2,3,...} prehistory rational numbers: before 2000 bc irrational, numbers: Pythagoreans, 500 bc Math 1230, Notes 8 Sep. 23, / 28

40 Is 1 a number? Timeline: natural numbers {1,2,3,...} prehistory rational numbers: before 2000 bc irrational, numbers: Pythagoreans, 500 bc zero, negative rational numbers, 628 Math 1230, Notes 8 Sep. 23, / 28

41 Is 1 a number? Timeline: natural numbers {1,2,3,...} prehistory rational numbers: before 2000 bc irrational, numbers: Pythagoreans, 500 bc zero, negative rational numbers, 628 real numbers, 1870 Math 1230, Notes 8 Sep. 23, / 28

42 Depict real numbers on a line. Math 1230, Notes 8 Sep. 23, / 28

43 Depict real numbers on a line. A need was found for the symbol 1 in 1572, but is this a number? Math 1230, Notes 8 Sep. 23, / 28

44 Depict real numbers on a line. A need was found for the symbol 1 in 1572, but is this a number? Argand: 1808: Depict complex numbers in a plane. Math 1230, Notes 8 Sep. 23, / 28

45 Depict real numbers on a line. A need was found for the symbol 1 in 1572, but is this a number? Argand: 1808: Depict complex numbers in a plane. Definition: A complex number is an ordered pair (a, b) of real numbers with the following definitions of addition and multiplication: (a, b) + (c, d) = (a + b, c + d) (a, b) (c, d) = (ac bd, ad + bc) Math 1230, Notes 8 Sep. 23, / 28

46 Depict real numbers on a line. A need was found for the symbol 1 in 1572, but is this a number? Argand: 1808: Depict complex numbers in a plane. Definition: A complex number is an ordered pair (a, b) of real numbers with the following definitions of addition and multiplication: Equivalent to: (a, b) + (c, d) = (a + b, c + d) (a, b) (c, d) = (ac bd, ad + bc) (a + ib) (c + id) = ac bd + i (ad + bc). Math 1230, Notes 8 Sep. 23, / 28

47 Depict real numbers on a line. A need was found for the symbol 1 in 1572, but is this a number? Argand: 1808: Depict complex numbers in a plane. Definition: A complex number is an ordered pair (a, b) of real numbers with the following definitions of addition and multiplication: Equivalent to: (a, b) + (c, d) = (a + b, c + d) (a, b) (c, d) = (ac bd, ad + bc) (a + ib) (c + id) = ac bd + i (ad + bc). associative, commutative, distributive Math 1230, Notes 8 Sep. 23, / 28

48 Depict real numbers on a line. A need was found for the symbol 1 in 1572, but is this a number? Argand: 1808: Depict complex numbers in a plane. Definition: A complex number is an ordered pair (a, b) of real numbers with the following definitions of addition and multiplication: Equivalent to: (a, b) + (c, d) = (a + b, c + d) (a, b) (c, d) = (ac bd, ad + bc) (a + ib) (c + id) = ac bd + i (ad + bc). associative, commutative, distributive Real numbers form a line. Complex numbers form a plane, which includes the real line of numbers (r, 0). Big advantage of complex numbers: They are algebraically closed: Math 1230, Notes 8 Sep. 23, / 28

49 Theorem: If a 0, a 2,..., a n 1 are complex numbers, then the equation z n + a n 1 z n 1 + a n 2 z n a 1 z + a 0 = 0 (3) has at least one solution z, which is a complex number. Math 1230, Notes 8 Sep. 23, / 28

50 Theorem: If a 0, a 2,..., a n 1 are complex numbers, then the equation z n + a n 1 z n 1 + a n 2 z n a 1 z + a 0 = 0 (3) has at least one solution z, which is a complex number. (Complex includes real) Math 1230, Notes 8 Sep. 23, / 28

51 Theorem: If a 0, a 2,..., a n 1 are complex numbers, then the equation z n + a n 1 z n 1 + a n 2 z n a 1 z + a 0 = 0 (3) has at least one solution z, which is a complex number. (Complex includes real) Result false if natural, rational, or real replace complex. Math 1230, Notes 8 Sep. 23, / 28

52 Theorem: If a 0, a 2,..., a n 1 are complex numbers, then the equation z n + a n 1 z n 1 + a n 2 z n a 1 z + a 0 = 0 (3) has at least one solution z, which is a complex number. (Complex includes real) Result false if natural, rational, or real replace complex. Various early attempts at proofs: Math 1230, Notes 8 Sep. 23, / 28

53 Theorem: If a 0, a 2,..., a n 1 are complex numbers, then the equation z n + a n 1 z n 1 + a n 2 z n a 1 z + a 0 = 0 (3) has at least one solution z, which is a complex number. (Complex includes real) Result false if natural, rational, or real replace complex. Various early attempts at proofs: d Alembert (1746) Math 1230, Notes 8 Sep. 23, / 28

54 Theorem: If a 0, a 2,..., a n 1 are complex numbers, then the equation z n + a n 1 z n 1 + a n 2 z n a 1 z + a 0 = 0 (3) has at least one solution z, which is a complex number. (Complex includes real) Result false if natural, rational, or real replace complex. Various early attempts at proofs: d Alembert (1746) Euler (1749) Math 1230, Notes 8 Sep. 23, / 28

55 Theorem: If a 0, a 2,..., a n 1 are complex numbers, then the equation z n + a n 1 z n 1 + a n 2 z n a 1 z + a 0 = 0 (3) has at least one solution z, which is a complex number. (Complex includes real) Result false if natural, rational, or real replace complex. Various early attempts at proofs: d Alembert (1746) Euler (1749) Lagrange (1772) Math 1230, Notes 8 Sep. 23, / 28

56 Theorem: If a 0, a 2,..., a n 1 are complex numbers, then the equation z n + a n 1 z n 1 + a n 2 z n a 1 z + a 0 = 0 (3) has at least one solution z, which is a complex number. (Complex includes real) Result false if natural, rational, or real replace complex. Various early attempts at proofs: d Alembert (1746) Euler (1749) Lagrange (1772) Laplace (1795) Math 1230, Notes 8 Sep. 23, / 28

57 Gauss (1799, in his Ph. D. thesis): Math 1230, Notes 8 Sep. 23, / 28

58 Gauss (1799, in his Ph. D. thesis): Finds defects in all the previous proofs Math 1230, Notes 8 Sep. 23, / 28

59 Gauss (1799, in his Ph. D. thesis): Finds defects in all the previous proofs Gives a new proof, which is by implication the first correct proof. Math 1230, Notes 8 Sep. 23, / 28

60 Gauss (1799, in his Ph. D. thesis): Finds defects in all the previous proofs Gives a new proof, which is by implication the first correct proof. However he does include a remark concerning a crucial step in the proof: Math 1230, Notes 8 Sep. 23, / 28

61 Gauss (1799, in his Ph. D. thesis): Finds defects in all the previous proofs Gives a new proof, which is by implication the first correct proof. However he does include a remark concerning a crucial step in the proof: As far as I know, nobody has raised any doubts about this. However should someone demand it then I will undertake to give a proof that is not subject to any doubt, on some other occasion. Math 1230, Notes 8 Sep. 23, / 28

62 Gauss (1799, in his Ph. D. thesis): Finds defects in all the previous proofs Gives a new proof, which is by implication the first correct proof. However he does include a remark concerning a crucial step in the proof: As far as I know, nobody has raised any doubts about this. However should someone demand it then I will undertake to give a proof that is not subject to any doubt, on some other occasion. The other occasion did not occur for fifty years. rewrote his 1799 proof, the same gap remained When finally he It was not fixed until Math 1230, Notes 8 Sep. 23, / 28

63 Outline of the proof of Gauss Math 1230, Notes 8 Sep. 23, / 28

64 Outline of the proof of Gauss Both z and p (z) are complex numbers, for any complex number z (including those of the from (a, 0) ), Suppose that z = x + iy. Then p (z) = f (x, y) + ig (x, y). Math 1230, Notes 8 Sep. 23, / 28

65 Outline of the proof of Gauss Both z and p (z) are complex numbers, for any complex number z (including those of the from (a, 0) ), Suppose that z = x + iy. Then p (z) = f (x, y) + ig (x, y). We show that (2) has a root by looking for (x, y) such that f (x, y) = g (x, y) = 0. Math 1230, Notes 8 Sep. 23, / 28

66 Outline of the proof of Gauss Both z and p (z) are complex numbers, for any complex number z (including those of the from (a, 0) ), Suppose that z = x + iy. Then p (z) = f (x, y) + ig (x, y). We show that (2) has a root by looking for (x, y) such that f (x, y) = g (x, y) = 0. Each of the equations f (x, y) = 0 and g (x, y) = 0 defines a curve in the (x, y) plane. To find a solution of (2) we look for a place where these curves intersect. Math 1230, Notes 8 Sep. 23, / 28

67 Example: p (z) = z 2. Since (x + iy) 2 = x 2 y 2 + 2ixy, the curves f = 0 and g = 0 are Math 1230, Notes 8 Sep. 23, / 28

68 Example: p (z) = z 2. Since (x + iy) 2 = x 2 y 2 + 2ixy, the curves f = 0 and g = 0 are x 2 y 2 = 0 2xy = 0. Math 1230, Notes 8 Sep. 23, / 28

69 Example: p (z) = z 2. Since (x + iy) 2 = x 2 y 2 + 2ixy, the curves f = 0 and g = 0 are x 2 y 2 = 0 2xy = 0. The set of points (x, y) satisfying x 2 + y 2 = 0 is the union of the lines y = x and y = x. Math 1230, Notes 8 Sep. 23, / 28

70 Example: p (z) = z 2. Since (x + iy) 2 = x 2 y 2 + 2ixy, the curves f = 0 and g = 0 are x 2 y 2 = 0 2xy = 0. The set of points (x, y) satisfying x 2 + y 2 = 0 is the union of the lines y = x and y = x. The set of points (x, y) satisfying 2xy = 0 is also the union of two lines, the x and y axes. The point (0, 0) is the only point where the two types of lines intersect, so z = 0 is the only solution of the equation. Math 1230, Notes 8 Sep. 23, / 28

71 Math 1230, Notes 8 Sep. 23, / 28

72 Here is another example: Find the solutions of z 2 4i = 0. Again letting z = x + iy, we get x 2 + 2ixy y 2 4i = 0. Math 1230, Notes 8 Sep. 23, / 28

73 Here is another example: Find the solutions of z 2 4i = 0. Again letting z = x + iy, we get x 2 + 2ixy y 2 4i = 0. We again get two equations: x 2 y 2 = 0 2xy 4 = 0. Math 1230, Notes 8 Sep. 23, / 28

74 The graph of the first equation is again the lines y = x and y = x. The graph of the second equation is the hyperbola y = 4 2x. Math 1230, Notes 8 Sep. 23, / 28

75 Math 1230, Notes 8 Sep. 23, / 28

76 Observe: two solutions Math 1230, Notes 8 Sep. 23, / 28

77 Observe: two solutions Math 1230, Notes 8 Sep. 23, / 28

78 Observe: two solutions Math 1230, Notes 8 Sep. 23, / 28

79 On a large circle, say x 2 + y 2 = 0, the red and blue curves of the last two examples cross this circle at almost the same places. (Recall that the blue curves are exactly the same in the two examples.) Math 1230, Notes 8 Sep. 23, / 28

80 Finally we do a more complicated example: z 5 z 1 = 0 z = x + iy x 5 + 5ix 4 y 10x 3 y 2 10ix 2 y 3 + 5xy 4 x + iy 5 iy 1 = 0. Math 1230, Notes 8 Sep. 23, / 28

81 Finally we do a more complicated example: z 5 z 1 = 0 z = x + iy x 5 + 5ix 4 y 10x 3 y 2 10ix 2 y 3 + 5xy 4 x + iy 5 iy 1 = 0. f (x, y) = x 5 x 10x 3 y 2 + 5xy 4 1 = 0 g (x, y) = y 5 y + 5x 4 y 10x 2 y 3 = 0 Math 1230, Notes 8 Sep. 23, / 28

82 Finally we do a more complicated example: z 5 z 1 = 0 z = x + iy x 5 + 5ix 4 y 10x 3 y 2 10ix 2 y 3 + 5xy 4 x + iy 5 iy 1 = 0. f (x, y) = x 5 x 10x 3 y 2 + 5xy 4 1 = 0 g (x, y) = y 5 y + 5x 4 y 10x 2 y 3 = 0 Do these two curves intersect somewhese in the (x, y) plane? Math 1230, Notes 8 Sep. 23, / 28

83 Math 1230, Notes 8 Sep. 23, / 28

84 Math 1230, Notes 8 Sep. 23, / 28

85 Math 1230, Notes 8 Sep. 23, / 28

86 An easier and more complete proof: (See Stillwell, pg. 268 of 2nd edition) Math 1230, Notes 8 Sep. 23, / 28

87 An easier and more complete proof: (See Stillwell, pg. 268 of 2nd edition) Basic idea: d Alembert (1746) Math 1230, Notes 8 Sep. 23, / 28

88 An easier and more complete proof: (See Stillwell, pg. 268 of 2nd edition) Basic idea: d Alembert (1746) Important Lemma made clearer: Argand, 1806 Math 1230, Notes 8 Sep. 23, / 28

89 An easier and more complete proof: (See Stillwell, pg. 268 of 2nd edition) Basic idea: d Alembert (1746) Important Lemma made clearer: Argand, 1806 Final detail for a rigorous proof: Weierstrass, Math 1230, Notes 8 Sep. 23, / 28

90 Recall: z = x + iy is a the point (x, y) in R 2. Math 1230, Notes 8 Sep. 23, / 28

91 Recall: z = x + iy is a the point (x, y) in R 2. Addition of complex numbers is equivalent to vector addition in the plane: (x + iy) + (u + iv) = (x + u) + i (y + v) = (x + u, y + v). Math 1230, Notes 8 Sep. 23, / 28

92 Recall: z = x + iy is a the point (x, y) in R 2. Addition of complex numbers is equivalent to vector addition in the plane: (x + iy) + (u + iv) = (x + u) + i (y + v) = (x + u, y + v). Definition: If z = x + iy is a complex number, then z = x 2 + y 2. Math 1230, Notes 8 Sep. 23, / 28

93 Recall: z = x + iy is a the point (x, y) in R 2. Addition of complex numbers is equivalent to vector addition in the plane: (x + iy) + (u + iv) = (x + u) + i (y + v) = (x + u, y + v). Definition: If z = x + iy is a complex number, then z = x 2 + y 2. d Alembert: Math 1230, Notes 8 Sep. 23, / 28

94 Recall: z = x + iy is a the point (x, y) in R 2. Addition of complex numbers is equivalent to vector addition in the plane: (x + iy) + (u + iv) = (x + u) + i (y + v) = (x + u, y + v). Definition: If z = x + iy is a complex number, then z = x 2 + y 2. d Alembert: Lemma 1. If for some z 0, p (z 0 ) = 0, then any neighborhood z 0 in the complex plane contains a point z 1 with p (z 1 ) < p (z 0 ). Math 1230, Notes 8 Sep. 23, / 28

95 Recall: z = x + iy is a the point (x, y) in R 2. Addition of complex numbers is equivalent to vector addition in the plane: (x + iy) + (u + iv) = (x + u) + i (y + v) = (x + u, y + v). Definition: If z = x + iy is a complex number, then z = x 2 + y 2. d Alembert: Lemma 1. If for some z 0, p (z 0 ) = 0, then any neighborhood z 0 in the complex plane contains a point z 1 with p (z 1 ) < p (z 0 ). Argand s proof of Lemma 1 requires the material in the first day of math Math 1230, Notes 8 Sep. 23, / 28

96 Lemma 2. There is an R > 0 such that if z R, then p (z) > p (0). Math 1230, Notes 8 Sep. 23, / 28

97 Lemma 2. There is an R > 0 such that if z R, then p (z) > p (0). The proof of Lemma 2 depends on showing that for large z, the term z n is larger than all the other terms in p (z) combined. This again requires a day s worth of complex variable theory, if that. Math 1230, Notes 8 Sep. 23, / 28

98 Lemma 2. There is an R > 0 such that if z R, then p (z) > p (0). The proof of Lemma 2 depends on showing that for large z, the term z n is larger than all the other terms in p (z) combined. This again requires a day s worth of complex variable theory, if that. Proof of theorem: The function p (z) can be considered a continuous function from R 2 to [0, ). Let D denote the unit disk z R in the complex plane. Math 1230, Notes 8 Sep. 23, / 28

99 Lemma 2. There is an R > 0 such that if z R, then p (z) > p (0). The proof of Lemma 2 depends on showing that for large z, the term z n is larger than all the other terms in p (z) combined. This again requires a day s worth of complex variable theory, if that. Proof of theorem: The function p (z) can be considered a continuous function from R 2 to [0, ). Let D denote the unit disk z R in the complex plane. We now use the theorem of Weierstrass that a continuous function on a bounded set (such as z R ) takes on its minimum value somewhere in this disk. Lemma 2 implies that this minimum value is not on the boundary z = R. Hence it is inside this circle. Math 1230, Notes 8 Sep. 23, / 28

100 Lemma 2. There is an R > 0 such that if z R, then p (z) > p (0). The proof of Lemma 2 depends on showing that for large z, the term z n is larger than all the other terms in p (z) combined. This again requires a day s worth of complex variable theory, if that. Proof of theorem: The function p (z) can be considered a continuous function from R 2 to [0, ). Let D denote the unit disk z R in the complex plane. We now use the theorem of Weierstrass that a continuous function on a bounded set (such as z R ) takes on its minimum value somewhere in this disk. Lemma 2 implies that this minimum value is not on the boundary z = R. Hence it is inside this circle. Let z 0 be a point in D where f (z) is a minimum. Suppose f (z 0 ) = 0. Then by Lemma 1 there is a point in D where f (z) < f (z 0 ). But this contradicts the definition of z 0. Hence, f (z 0 ) = 0. Math 1230, Notes 8 Sep. 23, / 28

101 Subsequent developments Math 1230, Notes 8 Sep. 23, / 28

102 Subsequent developments 1806: Argand develops the model of complex numbers as points in a plane. Math 1230, Notes 8 Sep. 23, / 28

103 Subsequent developments 1806: Argand develops the model of complex numbers as points in a plane Gauss gives a second, different proof. Math 1230, Notes 8 Sep. 23, / 28

104 Subsequent developments 1806: Argand develops the model of complex numbers as points in a plane Gauss gives a second, different proof Gauss gives a third proof. Math 1230, Notes 8 Sep. 23, / 28

105 Subsequent developments 1806: Argand develops the model of complex numbers as points in a plane Gauss gives a second, different proof Gauss gives a third proof Bolzano gives a proof which recognizes the importance of continuity in the argment. But continuity was not well understood until the work of Dedekind, so no completely correct proof was possible until then. Math 1230, Notes 8 Sep. 23, / 28

106 Subsequent developments 1806: Argand develops the model of complex numbers as points in a plane Gauss gives a second, different proof Gauss gives a third proof Bolzano gives a proof which recognizes the importance of continuity in the argment. But continuity was not well understood until the work of Dedekind, so no completely correct proof was possible until then Gauss rewrites his proof from The gap still remains. Math 1230, Notes 8 Sep. 23, / 28

107 Subsequent developments 1806: Argand develops the model of complex numbers as points in a plane Gauss gives a second, different proof Gauss gives a third proof Bolzano gives a proof which recognizes the importance of continuity in the argment. But continuity was not well understood until the work of Dedekind, so no completely correct proof was possible until then Gauss rewrites his proof from The gap still remains. Gauss kept trying to get the geometry out of his proof. But now days the theorem is viewed as primarily topological, rather than algebraic. Math 1230, Notes 8 Sep. 23, / 28

108 Subsequent developments 1806: Argand develops the model of complex numbers as points in a plane Gauss gives a second, different proof Gauss gives a third proof Bolzano gives a proof which recognizes the importance of continuity in the argment. But continuity was not well understood until the work of Dedekind, so no completely correct proof was possible until then Gauss rewrites his proof from The gap still remains. Gauss kept trying to get the geometry out of his proof. But now days the theorem is viewed as primarily topological, rather than algebraic. Once continuity was understood, most of the previous proofs could easily be fixed to be correct. An exception was Gauss s first proof. Math 1230, Notes 8 Sep. 23, / 28

109 Subsequent developments 1806: Argand develops the model of complex numbers as points in a plane Gauss gives a second, different proof Gauss gives a third proof Bolzano gives a proof which recognizes the importance of continuity in the argment. But continuity was not well understood until the work of Dedekind, so no completely correct proof was possible until then Gauss rewrites his proof from The gap still remains. Gauss kept trying to get the geometry out of his proof. But now days the theorem is viewed as primarily topological, rather than algebraic. Once continuity was understood, most of the previous proofs could easily be fixed to be correct. An exception was Gauss s first proof Ostrowsky fixes Gauss s first proof. Math 1230, Notes 8 Sep. 23, / 28

110 Often the proof of the fundamental theorem of algebra is based on theorems in the theory of complex analysis. These theorems are basically topological. One is the important Jordan Curve Theorem: Suppose that γ : [0, 1] R 2 is a continuous function with the properties that (i) γ (0) = γ (1), (ii) γ (s) = γ (t) unless s, t {0, 1}. Then there are connected disjoint sets A, B, and C = γ ([0, 1]) such that A B C = R 2. Further, we can assume that A is bounded and B is unbounded. Math 1230, Notes 8 Sep. 23, / 28

111 Often the proof of the fundamental theorem of algebra is based on theorems in the theory of complex analysis. These theorems are basically topological. One is the important Jordan Curve Theorem: Suppose that γ : [0, 1] R 2 is a continuous function with the properties that (i) γ (0) = γ (1), (ii) γ (s) = γ (t) unless s, t {0, 1}. Then there are connected disjoint sets A, B, and C = γ ([0, 1]) such that A B C = R 2. Further, we can assume that A is bounded and B is unbounded. Informal version: Every continuous simple closed curve has an inside and an outside, disjoint from each other. Math 1230, Notes 8 Sep. 23, / 28