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 p 2 Part 2, Polynomials We introduce a number i = −1 defined to satisfy the equation x = −1: 2 2 Lecture 2.4a, Complex Numbers (Of course if i = −1 then x = −i also satisfies x = −1:) The complex numbers are defined as all numbers of the form a + bi Dr. Ken W. Smith Write Sam Houston State University C := fa + bi : a; b 2 Rg: 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 The nicest version of the Fundamental Theorem of Algebra says that every Any \number" can be written as a complex number in the form a + bi: polynomial of degree n has exactly n zeroes. The number 3i = 0 + 3i has real part 0 and is said to be \purely But this is not quite true. Or is it? imaginary"; 2 2 2 the number 5 = 5 + 0i has imaginary part 0 and is \real". Consider the functions f(x) = x − 1; g(x) = x and h(x) = x + 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: z¯ = a − bi: Thus the conjugate of 2 + i is 2 + i = 2 − i; p p p the conjugate of 3 − πi is 3 − πi = 3 + πi. The conjugate of i is ¯i = −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) f(x) = x2 − 1 = (x − 1)(x + 1) 2 has two zeroes. Move the green parabola up one unit: g(x) = x . What g(x) = x2 = (x − 0)(x − 0) happened to our two zeroes? They merged into the single x-intercept at 2 the origin. We claim that g(x) = x2 still has two zeroes, if we are willing What if we move the parabola up one more step and graph h(x) = x + 1? to count multiplicities. This makes some sense because we can write Now, suddenly, there are no solutions. The graph never touches the x-axis. 2 Suddenly we have lost our pair of solutions! Can this be fixed? Smith (SHSU) g(x) = Elementaryx = (x Functions− 0)(x − 0) 2013 5 / 35 Smith (SHSU) Elementary Functions 2013 6 / 35 Motivation for the complex numbers 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 f(x) = (x − 1)(x + 1) x2 + 1 = 0 2 still hasSmith two (SHSU) zeroes, i and −i. TheElementary quadratic Functions x + 1 now factors as2013 7 / 35 Smith (SHSU) Elementary Functions 2013 8 / 35 x2 + 1 = (x − i)(x + i): Motivation for the complex numbers Motivation for the complex numbers f(x) = (x − 0)(x + 0) f(x) = (x − i)(x + i) 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 Modern cell phone signals rely on sophisticated signal analysis; we would to allow for a new number, one whose square was −1, quite a lot of not have cell phones without the mathematics of complex numbers. mathematics got simpler! More analysis of electrical wiring and electrical signaling uses complex (They particularly noticed that they could solve quadratic and cubic numbers. equations!) Complex numbers appear throughout all of mathematics and greatly This \imaginary" number was therefore very useful. simplify many mathematical problems! Over time, the term \imaginary" has stuck, even though scientists and In the next presentation we will look at complex numbers in quadratic engineers now use complex numbers all the time. 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 p x = −b± b2−4ac Dr. Ken W. Smith 2a The expression b2 − 4ac under the radical sign is called the discriminant Sam Houston State University of the quadratic equation and is often abbreviated by ∆: 2013 If ∆ = b2 − 4ac is positive then the square root of ∆ is a real number and so the quadratic equation has two real solutions: p p −b+ ∆ −b− ∆ x = 2a and x = 2a : Smith (SHSU) Elementary Functions 2013 13 / 35 Smith (SHSU) Elementary Functions 2013 14 / 35 Complex numbers in Quadratic Equations Complex numbers in Quadratic Equations The quadratic equation ax2 + bx + c = 0 The quadratic equation has solutions p ax2 + bx + c = 0 −b± b2−4ac x = 2a p has solutions 2 p If ∆ = b − 4ac is negative then ∆ is imaginary and so our solutions are −b± b2−4ac complex numbers which are not real. x = 2a p p To be explicit, if ∆ is negative then −∆ is positive and so ∆ = −∆ i: If ∆ = b2 − 4ac is zero then there is only one solution since The solutions to the quadratic formula are then p p p p x = −b± ∆ = − b± 0 = − b : −b+ −∆ i −b− −∆ i 2a 2a 2a x = 2a and x = 2a This single solution occurs with multiplicity two. p In this case, the plus/minus sign (±) in front of ∆ assures us that we will get two complex numbers as solutions. 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 Some worked examples. x2 + x + 1 = 0 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 are p p p p p p p p p −1± 12−4(1)(1) 1± −3 1± 3 i 1 3 −1± −3 −1± 3 −1 −1± 3 i 1 3 2 = 2 = 2 ± 2 i: 3 = 2 = 2 = 2 = − 2 ± 2 i: Since the two solutions to the equation x2 − x + 1 = 0 are the complex Thus the two solutions to the equation x2 + x + 1 = 0 are the complex numbers conjugate pairs p p p p 1 + 3 i and 1 − 3 i: 1 3 1 3 2 2 2 2 − 2 + 2 i and − 2 − 2 i: then the polynomial x2 − x + 1 factors as Since these are the two zeroes of the polynomial x2 + x + 1 then we can p p factor 1 3 1 3 p p (x − ( 2 + 2 i))(x − ( 2 − 2 i)) 2 1 3 1 3 x + x + 1 = (x − (− 2 + 2 i))(x − (− 2 − 2 i)) p p 1 3 1 3 p p = (x − − i)(x − + i) 1 3 1 3 2 2 2 2 = (x + 2 − 2 i)(x + 2 + 2 i) Smith (SHSU) Elementary Functions 2013 17 / 35 Smith (SHSU) Elementary Functions 2013 18 / 35 Complex numbers in Quadratic Equations 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 p p Solve the quadratic equation 5 31 5 31 x = − 4 + 4 i and x = − 4 − 4 i 2x2 + 5x + 7 = 0 and since c is a zero of a polynomial if and only if x − c is a factor, then p p Solution. According to the quadratic formula, (x − (− 5 + 31 i))(x − (− 5 − 31 i)) p 4 4 4 4 2 p p p p p −5± 5 −4(2)(7) −5± −31 −5± 31 −1 −5± 31 i 5 31 2 x = 4 = 4 = 4 = 4 = − 4 ± 4 i: must be a factor of 2x + 5x + 7: But if we check the leading coefficient of the polynomial in the expression above, we see that we need to multiply Our two solutions are the conjugate pairs by 2 to complete the factorization. So 2x2 + 5x + 7 factors as p p 5 31 5 31 p p x = − 4 + 4 i and x = − 4 − 4 i: 5 31 5 31 2(x − (− 4 + 4 i))(x − (− 4 − 4 i)) p p 5 31 5 31 = 2(x + 4 − 4 i)(x + 4 + 4 i) Smith (SHSU) Elementary Functions 2013 19 / 35 Smith (SHSU) Elementary Functions 2013 20 / 35 Complex numbers in Quadratic Equations Elementary Functions In the next presentation, we explore the algebra and geometry of complex Part 2, Polynomials numbers. 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 Smith (SHSU) Elementary Functions 2013 22 / 35 Geometric interpretation of complex numbers 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.