Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
List of important publications in mathematics wikipedia , lookup
Elementary algebra wikipedia , lookup
Large numbers wikipedia , lookup
Real number wikipedia , lookup
System of polynomial equations wikipedia , lookup
Mathematics of radio engineering wikipedia , lookup
Complex numbers The complex number system is an extension of the real number system. It unifies the mathematical number system and explains many mathematical phenomena. Elementary Functions √ We introduce a number i = −1 defined to satisfy the equation x2 = −1. (Of course if i2 = −1 then x = −i also satisfies x2 = −1.) Part 2, Polynomials Lecture 2.4a, Complex Numbers The complex numbers are defined as all numbers of the form a + bi Write Dr. Ken W. Smith C := {a + bi : a, b ∈ R}. Sam Houston State University A complex number of the form 2013 z = a + bi is said to have real part < = a and imaginary part = = b. Smith (SHSU) Elementary Functions 2013 1 / 35 Smith (SHSU) Elementary Functions 2013 2 / 35 Motivation for the complex numbers Motivation for the complex numbers Any “number” can be written as a complex number in the form a + bi. The nicest version of the Fundamental Theorem of Algebra says that every polynomial of degree n has exactly n zeroes. The number 3i = 0 + 3i has real part 0 and is said to be “purely imaginary”; the number 5 = 5 + 0i has imaginary part 0 and is “real”. But this is not quite true. Or is it? Consider the functions f (x) = x2 − 1, g(x) = x2 and h(x) = x2 + 1. We graph these functions below. The real numbers are a subset of the complex numbers. The conjugate of a complex number z = a + bi is created by changing the sign on the imaginary part: z̄ = a − bi. Thus the conjugate of 2 + i is 2 + i = 2 − i; √ √ √ the conjugate of 3 − πi is 3 − πi = 3 + πi. The conjugate of i is ī = −i and the conjugate of the real number 5 is merely 5. Smith (SHSU) Elementary Functions 2013 3 / 35 Smith (SHSU) Elementary Functions 2013 4 / 35 Motivation for the complex numbers Motivation for the complex numbers It is obvious that the quadratic graphed in green f (x) = x2 − 1 = (x − 1)(x + 1) has two zeroes. Move the green parabola up one unit: g(x) = x2 . What happened to our two zeroes? They merged into the single x-intercept at the origin. We claim that g(x) = x2 still has two zeroes, if we are willing to count multiplicities. This makes some sense because we can write Smith (SHSU) g(x) = Elementary x2 = (xFunctions − 0)(x − 0) 2013 5 / 35 Motivation for the complex numbers f (x) = x2 − 1 = (x − 1)(x + 1) g(x) = x2 = (x − 0)(x − 0) What if we move the parabola up one more step and graph h(x) = x2 + 1? Now, suddenly, there are no solutions. The graph never touches the x-axis. Suddenly we have lost our pair of solutions! Can this be fixed? Smith (SHSU) Elementary Functions 2013 6 / 35 2013 8 / 35 Motivation for the complex numbers Algebraically, h(x) = x2 + 1 does not have any real zeroes because that requires x2 + 1 = 0 =⇒ x2 = −1. If we square any positive real number, the result is positive. So we cannot get −1. But if we use imaginary numbers then the equation x2 f (x) = (x − 1)(x + 1) +1=0 2 still has two zeroes, i and −i. The quadratic Smith (SHSU) Elementary Functions x + 1 now factors as 2013 7 / 35 Smith (SHSU) Elementary Functions Motivation for the complex numbers Motivation for the complex numbers f (x) = (x − i)(x + i) f (x) = (x − 0)(x + 0) Smith (SHSU) Elementary Functions 2013 9 / 35 Smith (SHSU) Elementary Functions 2013 10 / 35 Motivation for the complex numbers Motivation for the complex numbers In the late middle ages, mathematicians discovered that if one were willing to allow for a new number, one whose square was −1, quite a lot of mathematics got simpler! Modern cell phone signals rely on sophisticated signal analysis; we would not have cell phones without the mathematics of complex numbers. More analysis of electrical wiring and electrical signaling uses complex numbers. (They particularly noticed that they could solve quadratic and cubic equations!) This “imaginary” number was therefore very useful. Complex numbers appear throughout all of mathematics and greatly simplify many mathematical problems! Over time, the term “imaginary” has stuck, even though scientists and engineers now use complex numbers all the time. In the next presentation we will look at complex numbers in quadratic equations. It is now common agreement to write i as an entity that satisfies (END) i2 = −1. Smith (SHSU) Elementary Functions 2013 11 / 35 Smith (SHSU) Elementary Functions 2013 12 / 35 Complex numbers in Quadratic Equations Complex numbers appear naturally in quadratic equations. Suppose we wish to solve the quadratic equation Elementary Functions ax2 + bx + c = 0 Part 2, Polynomials Lecture 2.4b, Complex Numbers in Quadratic Equations By completing the square we can solve for x and find that x= Dr. Ken W. Smith The expression b2 − 4ac under the radical sign is called the discriminant of the quadratic equation and is often abbreviated by ∆. Sam Houston State University 2013 If ∆ = b2 − 4ac is positive then the square root of ∆ is a real number and so the quadratic equation has two real solutions: x= Smith (SHSU) √ −b± b2 −4ac 2a Elementary Functions 2013 13 / 35 Complex numbers in Quadratic Equations Smith (SHSU) √ −b+ ∆ 2a and x = √ −b− ∆ . 2a Elementary Functions 2013 14 / 35 Complex numbers in Quadratic Equations The quadratic equation ax2 + bx + c = 0 The quadratic equation has solutions ax2 + bx + c = 0 x= √ If ∆ = b2 − 4ac is negative then ∆ is imaginary and so our solutions are complex numbers which are not real. √ √ To be explicit, if ∆ is negative then −∆ is positive and so ∆ = −∆ i. has solutions x= √ −b± b2 −4ac 2a If ∆ = b2 − 4ac is zero then there is only one solution since √ x= −b± ∆ 2a √ −b± b2 −4ac 2a The solutions to the quadratic formula are then √ b = − b±2a 0 = − 2a . x= This single solution occurs with multiplicity two. √ −b+ −∆ i 2a and x = √ −b− −∆ i 2a In this case, the plus/minus sign (±) in front of will get two complex numbers as solutions. √ ∆ assures us that we These two complex solutions come in conjugate pairs. Smith (SHSU) Elementary Functions 2013 15 / 35 Smith (SHSU) Elementary Functions 2013 16 / 35 Complex numbers in Quadratic Equations Complex numbers in Quadratic Equations Example. The solutions to the quadratic equation x2 are √ −1± 12 −4(1)(1) 3 = √ −1± −3 2 +x+1=0 √ √ −1± 3 −1 2 = Thus the two solutions to the equation conjugate pairs − 12 √ 3 2 i + Since these are the two zeroes of factor = (x + Smith (SHSU) − x2 = √ −1± 3 i 2 √ 1± −3 2 √ ± 3 2 i. √ 3 2 i))(x + 1 2 1 2 + x + 1 then we can − (− 12 − √ 3 2 i + 3 2 i)) 1 2 = (x − 3 2 i) 2013 17 / 35 Complex numbers in Quadratic Equations 1 2 √ 3 2 i. ± √ 1 2 and √ 3 2 i))(x (x − ( 12 + √ Elementary Functions = 3 2 i. − then the polynomial x2 − x + 1 factors as √ + √ 1± 3 i 2 = Since the two solutions to the equation x2 − x + 1 = 0 are the complex numbers √ and − − 23 i. the polynomial x2 √ 3 2 i)(x = − 12 + x + 1 = 0 are the complex 1 2 x2 + x + 1 = (x − (− 12 + 1 2 Some worked examples. Solve the quadratic equation x2 − x + 1 = 0. Also, factor x2 − x + 1. Solution. By the quadratic formula the solutions to x2 − x + 1 = 0 are Smith (SHSU) − − ( 12 − √ 3 2 i)(x − 1 2 √ 3 2 i)) √ 3 2 i) + Elementary Functions 2013 18 / 35 Complex numbers in Quadratic Equations Use the roots of 2x2 + 5x + 7 to factor 2x2 + 5x + 7. Solution. Since the two solutions to the equation 2x2 + 5x + 7 = 0 are Solve the quadratic equation x = − 54 + 2x2 + 5x + 7 = 0 Solution. According to the quadratic formula, √ √ √ √ −5± 52 −4(2)(7) −5± −31 −5± 31 −1 x= = = = 4 4 4 √ 31 4 i and x = − 54 − √ −5± 31 i 4 = − 54 ± (x − (− 54 + √ 31 4 i. Elementary Functions √ 31 4 i and x = − 54 − √ 31 4 i. 2(x − (− 54 + 2013 19 / 35 √ 31 4 i))(x − (− 54 − √ 31 4 i)) must be a factor of 2x2 + 5x + 7. But if we check the leading coefficient of the polynomial in the expression above, we see that we need to multiply by 2 to complete the factorization. So 2x2 + 5x + 7 factors as = 2(x + Smith (SHSU) 31 4 i and since c is a zero of a polynomial if and only if x − c is a factor, then Our two solutions are the conjugate pairs x = − 54 + √ Smith (SHSU) 5 4 √ 31 4 i))(x − (− 54 − √ − 31 4 i)(x + Elementary Functions 5 4 √ 31 4 i)) √ + 31 4 i) 2013 20 / 35 Complex numbers in Quadratic Equations Elementary Functions In the next presentation, we explore the algebra and geometry of complex numbers. Part 2, Polynomials Lecture 2.4c, The Geometry and Algebra of Complex numbers (END) Dr. Ken W. Smith Sam Houston State University 2013 Smith (SHSU) Elementary Functions 2013 21 / 35 Geometric interpretation of complex numbers Smith (SHSU) Elementary Functions 2013 22 / 35 2013 24 / 35 Complex numbers Mathematicians began to recognize the value of complex numbers sometime back in the Renaissance period (fifteenth and sixteenth centuries) but it was not until there was a geometric interpretation of the complex numbers that people began to feel comfortable with them. We may view the complex numbers as lying in the Cartesian plane. Let the traditional x-axis represent the real numbers and the traditional y-axis represent the numbers of the form yi. We equate a complex number x + yi with the point (x, y). (So the imaginary numbers yi are “perpendicular” to the real numbers!) (See the complex plane is drawn on the next page.) The complex plane Smith (SHSU) Elementary Functions 2013 23 / 35 Smith (SHSU) Elementary Functions Complex numbers Complex numbers The Argand Diagram The process of changing a + bi into the point (a, b) can be traced to Argand around 1800 and is sometimes called the “Argand diagram”. In the Argand diagram, the complex number z = a + bi is equated with the point (a, b) in the Cartesian plane. For example, 6 + 5i can be graphed as the point (6, 5). The ordinary Cartesian plane then becomes a plane of complex numbers. Smith (SHSU) Elementary Functions 2013 Thus i = 0 + 1i is equated with the point (0, 1) and the number 1 = 1 + 0i is equated with the point √ (1, 0). The point (2, 1) represents√the number 2 + i. The number ( 3 + i)/2 is equated with the point ( 3/2, 1/2). In the complex plane the x-axis is called the “real” axis and the y-axis is called the imaginary” axis. 25 / 35 The algebra of complex numbers Smith (SHSU) Elementary Functions 2013 26 / 35 Complex numbers We multiply complex numbers (a + bi)(c + di) just as we would the polynomials (a + bx)(c + dx) (except that we remember that i2 = −1.) The complex numbers have a natural addition, subtraction and multiplication. We add and subtract complex numbers just as we would polynomials, keeping up with the real and imaginary parts. For example, For example, since (3 + 4x)(7 + 11x) = 21 + 61x + 44x2 . (3 + 4i) + (7 + 11i) = 10 + 15i then and (3 + 4i) − (7 + 11i) = −4 − 7i. Smith (SHSU) Elementary Functions (3 + 4i)(7 + 11i) = 21 + 61i + 44i2 = 21 + 61i − 44 = −23 + 61i. 2013 27 / 35 Smith (SHSU) Elementary Functions 2013 28 / 35 Division of complex numbers Some worked examples. We would like our complex numbers to be written in “Cartesian form” a + bi so there is a little twist involved in doing division with complex numbers. Note that if z = a + bi then z z̄ = (a + bi)(a − bi) = a2 + b2 . So if we are dividing by z, we may view z1 as Write the complex fractions below into the “Cartesian” form z = a + bi where a, b ∈ R. 1 2 z̄ a 1 b = = 2 − 2 i. 2 z z̄ a +b a + b2 z 3 4 3+2i 7−3i . 5 3+i 2 1+i 1 i Solution. Computationally, this means that anytime we have a fraction involving a complex number z in the denominator, we multiply numerator and denominator by z̄ and simplify. 1 We want to be ready to use complex numbers when factoring polynomials or solving polynomial equations. For example, let’s factor the polynomial f (x) = x3 − 2x2 + 9x − 18. Since this is a cubic polynomial and we don’t want to use the cubic formula (there is a cubic formula but it is quite messy!) then we need to find a zero. We could try some numbers (see some techniques in the next lectures) and discover by trial-and-error that f (2) = 0. Or we could graph this polynomial on a graphing calculator and see that x = 2 is a zero. Since f (2) = 0 (check this!) then x − 2 is a factor of x3 − 2x2 + 9x − 18. Now divide x − 2 into f (x) = x3 − 2x2 + 9x − 18 by synthetic division: 1 2 1 −2 9 − 18 2 0 18 0 9 0 We seeSmith that f (x) factors as (x Elementary − 2)(x2Functions + 9). Since x2 + 9 = 0 implies that (SHSU) 2013 31 / 35 = (3+2i)(7+3i) (7−3i)(7+3i) = the imaginary part is For example, 3 + 4i 7 + 11i (3 + 4i)(7 + 11i) −23 + 61i −23 61 3 + 4i = · = = = +i . 7 − 11i 7 + 11i (7 − 11i)(7 + 11i) 170 170 170 7 − 11i This process, multiplying the numerator and denominator of a fraction by Smith (SHSU) Elementary Functions 2013 29 / 35 the conjugate of the denominator, is called rationalizing the denominator. Using complex numbers and the factor theorem 3+2i 7−3i 2 3 4 5 3+i 5 3−i = ( 3+i )( 3−i )= 15+23i 58 23 58 .) 15−5i 10 = 2(1−i) 2 2 1−i 1+i = ( 1+i )( 1−i ) = 2 1 −i −i 1 i = ( i )( −i ) = 1 = −i . Smith (SHSU) = 3 2 15 58 + 23 58 i . (So the real part is 15 58 and − 12 i . = 1−i . (Or 0 − i .) Elementary Functions 2013 30 / 35 Some worked examples. Factor the polynomial f (x) = x3 − 8 and then find all solutions to x3 = 8. Solution. Since f (x) = x3 − 8 is zero when x3 = 8 then we know that one zero is x = 2. Since f (2) = 0 we know that x − 2 is a factor of f (x). Dividing x3 − 8 by x − 2 gives us x3 − 8 = (x − 2)(x2 + 2x + 4). The solutions to x3 − 8 = 0 are the solutions to (x − 2)(x2 + 2x + 4) = 0. By the quadratic formula, the solutions to x2 + 2x + 4 = 0 are √ √ 1 x = 2 (−2 ± 2 3i) = −1 ± 3i. This implies that (x2 + 2x + 4) factors as √ √ √ √ x2 +2x+4 = (x−(−1+ 3 i))(x−(−1− 3 i) = (x+1− 3 i)(x+1+ 3 i). So the factoring of f (x) is x3 − 8 = (x − 2)(x + 1 − √ √ 3 i)(x + 1 + 3 i) . The full set of solutions to the equation x3 = 8 is the set of solutions to the equation x3 − 8 = 0. These are the zeroes of f (x): √ √ Smith (SHSU) Elementary 2013 32 / 35 x = 2, x = −1 + 3Functions i, x = −1 − 3 i . Factoring x6 − 1 Another example Factor completely g(x) = x6 − 1. Solution.The Fundamental Theorem of Algebra tells us the x6 − 1 has six zeroes and therefore factors into six linear pieces. One way to begin factoring g(x) = x6 − 1 is to view this polynomial as a difference of squares: g(x) = x6 − 1 = (x3 )2 − 12 = (x3 + 1)(x3 − 1). To further factor x3 − 1, notice that x = 1 is surely a zero and after synthetic division we see that x3 − 1 = (x − 1)(x2 + x + 1). In a similar way we should notice that x = −1 is a zero of x3 + 1 and so g(x) = x6 − 1 = (x3 + 1)(x3 − 1) = (x + 1)(x2 − x + 1)(x − 1)(x2 + x + 1). We are not done; we need to factor the two quadratics x2 − x + 1 and x2 + x + 1. In an early example, we factored these using the quadratic formula and found that √ √ 1 3 1 3 2 x + x + 1 = (x − − i)(x − + i) 2 2 2 2 and √ √ 1 3 1 3 i)(x + + i) x2 − x + 1 = (x + − 2 2 2 2 So g(x) = x6 − 1 = (x + 1)(x2 − x + 1)(x − 1)(x2 + x + 1) x3 + 1 = (x + 1)(x2 − x + 1). = (x + 1)(x + So we now have 1 2 √ − 3 2 i)(x + 1 2 √ + 3 2 i)(x − 1)(x − 1 2 √ − 3 2 i)(x − 1 2 √ + 3 2 i) g(x) = x6 − 1 = (x3 + 1)(x3 − 1) = (x + 1)(x2 − x + 1)(x − 1)(x2 + x + 1). Smithhave (SHSU) Elementary Functions We still to factor the quadratics.... 2013 33 / 35 Complex Numbers In the next presentation, we return to studying the zeroes of polynomials. (END) Smith (SHSU) Elementary Functions 2013 35 / 35 Notice that in this last example, relying on complex numbers and the Smith (SHSU) Elementary Functions 2013 34 / 35 formula for difference of squares, we were able to break the sixth degree polynomial x6 − 1 down into six linear terms as promised by the Fundamental Theorem of Algebra. .