Elementary Functions Complex Numbers Motivation for the Complex

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 √ 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 := {a + bi : a, b ∈ R}. 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; √ √ √ 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 √ 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: √ √ −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 √ ax2 + bx + c = 0 −b± b2−4ac x = 2a √ has solutions 2 √ If ∆ = b − 4ac is negative then ∆ is imaginary and so our solutions are −b± b2−4ac complex numbers which are not real. x = 2a √ √ 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 √ √ √ √ x = −b± ∆ = − b± 0 = − b . −b+ −∆ i −b− −∆ i 2a 2a 2a x = 2a and x = 2a This single solution occurs with multiplicity two. √ 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 √ √ √ √ √ √ √ √ √ −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 √ √ √ √ 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 √ √ factor 1 3 1 3 √ √ (x − ( 2 + 2 i))(x − ( 2 − 2 i)) 2 1 3 1 3 x + x + 1 = (x − (− 2 + 2 i))(x − (− 2 − 2 i)) √ √ 1 3 1 3 √ √ = (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 √ √ 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 √ √ Solution. According to the quadratic formula, (x − (− 5 + 31 i))(x − (− 5 − 31 i)) √ 4 4 4 4 2 √ √ √ √ √ −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 √ √ 5 31 5 31 √ √ x = − 4 + 4 i and x = − 4 − 4 i. 5 31 5 31 2(x − (− 4 + 4 i))(x − (− 4 − 4 i)) √ √ 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.

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 2013 24 / 35 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 Thus i = 0 + 1i is equated with the point (0, 1) and the number 1 = 1 + 0i point (a, b) in the Cartesian plane. For example, 6 + 5i can be graphed as is equated with the point√ (1, 0). The point (2, 1) represents√ the number the point (6, 5). 2 + i. The number ( 3 + i)/2 is equated with the point ( 3/2, 1/2). The ordinary Cartesian plane then becomes a plane of complex numbers. In the complex plane the x-axis is called the “real” axis and the y-axis is called the imaginary” axis. Smith (SHSU) Elementary Functions 2013 25 / 35 Smith (SHSU) Elementary Functions 2013 26 / 35 The algebra of complex numbers Complex numbers

We multiply complex numbers (a + bi)(c + di) just as we would the The complex numbers have a natural addition, subtraction and polynomials (a + bx)(c + dx) (except that we remember that i2 = −1.) multiplication. We add and subtract complex numbers just as we would polynomials, For example, since keeping up with the real and imaginary parts. For example, (3 + 4x)(7 + 11x) = 21 + 61x + 44x2. (3 + 4i) + (7 + 11i) = 10 + 15i then and (3 + 4i) − (7 + 11i) = −4 − 7i. (3 + 4i)(7 + 11i) = 21 + 61i + 44i2 = 21 + 61i − 44 = −23 + 61i.

Smith (SHSU) Elementary Functions 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” Write the complex fractions below into the “Cartesian” form z = a + bi a + bi so there is a little twist involved in doing division with complex where a, b ∈ R. numbers. 3+2i 1 . Note that if z = a + bi then zz¯ = (a + bi)(a − bi) = a2 + b2. So if we are 7−3i 5 1 2 dividing by z, we may view z as 3+i 2 1 z¯ a b 3 = = − i. 1+i z zz¯ a2 + b2 a2 + b2 1 4 i Solution. Computationally, this means that anytime we have a fraction involving a 3+2i (3+2i)(7+3i) 15+23i 15 23 15 1 = = = + i . (So the real part is and complex number z in the denominator, we multiply numerator and 7−3i (7−3i)(7+3i) 58 58 58 58 23 denominator by z¯ and simplify. the imaginary part is 58 .) 5 5 3−i 15−5i 3 1 2 = ( )( ) = = − i . For example, 3+i 3+i 3−i 10 2 2 2 2 1−i 2(1−i) 3 + 4i 3 + 4i 7 + 11i (3 + 4i)(7 + 11i) −23 + 61i −23 61 3 = ( )( ) = = 1 − i . = · = = = +i . 1+i 1+i 1−i 2 7 − 11i 7 − 11i 7 + 11i (7 − 11i)(7 + 11i) 170 170 170 1 1 −i −i 4 i = ( i )( −i ) = 1 = −i . (Or 0 − i .) This process, multiplying the numerator and denominator of a fraction by the conjugateSmith (SHSU) of the denominator,Elementaryis called Functionsrationalizing the 2013 29 / 35 Smith (SHSU) Elementary Functions 2013 30 / 35 Usingdenominator complex. numbers and the factor theorem Some worked examples.

We want to be ready to use complex numbers when factoring polynomials Factor the polynomial f(x) = x3 − 8 and then find all solutions to x3 = 8. or solving polynomial equations. For example, let’s factor the polynomial Solution. Since f(x) = x3 − 8 is zero when x3 = 8 then we know that 3 2 f(x) = x − 2x + 9x − 18. one zero is x = 2. Since f(2) = 0 we know that x − 2 is a factor of f(x). Since this is a cubic polynomial and we don’t want to use the cubic Dividing x3 − 8 by x − 2 gives us x3 − 8 = (x − 2)(x2 + 2x + 4). The formula (there is a cubic formula but it is quite messy!) then we need to solutions to x3 − 8 = 0 are the solutions to (x − 2)(x2 + 2x + 4) = 0. By find a zero. the quadratic formula, the solutions to x2 + 2x + 4 = 0 are We could try some numbers (see some techniques in the next lectures) √ √ x = 1 (−2 ± 2 3i) = −1 ± 3i. and discover by trial-and-error that f(2) = 0. Or we could graph this 2 This implies that (x2 + 2x + 4) factors as polynomial on a graphing calculator and see that x = 2 is a zero. √ √ √ √ x2+2x+4 = (x−(−1+ 3 i))(x−(−1− 3 i) = (x+1− 3 i)(x+1+ 3 i). Since f(2) = 0 (check this!) then x − 2 is a factor of x3 − 2x2 + 9x − 18. So the factoring of f(x) is Now divide x − 2 into f(x) = x3 − 2x2 + 9x − 18 by synthetic division: √ √ 3 1 − 2 9 − 18 x − 8 = (x − 2)(x + 1 − 3 i)(x + 1 + 3 i) .

2 2 0 18 The full set of solutions to the equation x3 = 8 is the set of solutions to 1 0 9 0 the equation x3 − 8 = 0. These are the zeroes of f(x): 2 2 √ √ We seeSmith that (SHSU)f(x) factors as (x Elementary− 2)(x Functions+ 9). Since x + 9 = 0 implies2013 that 31 / 35 Smith (SHSU) x = 2, x = −Elementary1 + 3 Functionsi, x = −1 − 3 i . 2013 32 / 35 x = ±3 i then x2 + 9 factors as (x − 3i)(x + 3i). So f(x) factors as x3 − 2x2 + 9x − 18 = (x − 2)(x − 3i)(x + 3i). Another example Factoring x6 − 1

Factor completely g(x) = x6 − 1. 6 3 3 2 2 Solution.The Fundamental Theorem of Algebra tells us the x6 − 1 has six g(x) = x − 1 = (x + 1)(x − 1) = (x + 1)(x − x + 1)(x − 1)(x + x + 1). zeroes and therefore factors into six linear pieces. 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 One way to begin factoring g(x) = x6 − 1 is to view this polynomial as a formula and found that difference of squares: √ √ 1 3 1 3 g(x) = x6 − 1 = (x3)2 − 12 = (x3 + 1)(x3 − 1). x2 + x + 1 = (x − − i)(x − + i) 2 2 2 2 3 To further factor x − 1, notice that x = 1 is surely a zero and after and √ √ synthetic division we see that 1 3 1 3 x2 − x + 1 = (x + − i)(x + + i) x3 − 1 = (x − 1)(x2 + x + 1). 2 2 2 2 So In a similar way we should notice that x = −1 is a zero of x3 + 1 and so g(x) = x6 − 1 = (x + 1)(x2 − x + 1)(x − 1)(x2 + x + 1) x3 + 1 = (x + 1)(x2 − x + 1). √ √ √ √ 1 3 1 3 1 3 1 3 So we now have = (x + 1)(x + 2 − 2 i)(x + 2 + 2 i)(x − 1)(x − 2 − 2 i)(x − 2 + 2 i) . g(x) = x6 − 1 = (x3 + 1)(x3 − 1) = (x + 1)(x2 − x + 1)(x − 1)(x2 + x + 1). Notice that in this last example, relying on complex numbers and the We stillSmith have (SHSU) to factor the quadratics....Elementary Functions 2013 33 / 35 formulaSmith for (SHSU) difference of squares,Elementarywe were Functions able to break the sixth2013 degree 34 / 35 Complex Numbers polynomial x6 − 1 down into six linear terms as promised by the Fundamental Theorem of Algebra.

In the next presentation, we return to studying the zeroes of polynomials.

(END)

Smith (SHSU) Elementary Functions 2013 35 / 35