Complex numbers The complex number system is an extension of the - PowerPoint PPT Presentation

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 x 2 = − 1 . Part 2, Polynomials (Of course if i 2 = − 1 then x = − i also satisfies x 2 = − 1 . ) 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 ℜ = a and imaginary part ℑ = 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 3 i = 0 + 3 i has real part 0 and is said to be “purely But this is not quite true. Or is it? imaginary”; Consider the functions f ( x ) = x 2 − 1 , g ( x ) = x 2 and h ( x ) = x 2 + 1 . We the number 5 = 5 + 0 i has imaginary part 0 and is “real”. 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 ) = x 2 − 1 = ( x − 1)( x + 1) f ( x ) = x 2 − 1 = ( x − 1)( x + 1) has two zeroes. Move the green parabola up one unit: g ( x ) = x 2 . What g ( x ) = x 2 = ( x − 0)( x − 0) happened to our two zeroes? They merged into the single x -intercept at What if we move the parabola up one more step and graph h ( x ) = x 2 + 1 ? the origin. We claim that g ( x ) = x 2 still has two zeroes, if we are willing Now, suddenly, there are no solutions. The graph never touches the x -axis. to count multiplicities. This makes some sense because we can write Suddenly we have lost our pair of solutions! Can this be fixed? g ( x ) = x 2 = ( x − 0)( x − 0) Smith (SHSU) Elementary Functions 2013 5 / 35 Smith (SHSU) Elementary Functions 2013 6 / 35 Motivation for the complex numbers Motivation for the complex numbers Algebraically, h ( x ) = x 2 + 1 does not have any real zeroes because that requires x 2 + 1 = 0 = ⇒ x 2 = − 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) x 2 + 1 = 0 still has two zeroes, i and − i . The quadratic x 2 + 1 now factors as Smith (SHSU) Elementary Functions 2013 7 / 35 Smith (SHSU) Elementary Functions 2013 8 / 35

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) i 2 = − 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 ax 2 + bx + c = 0 Elementary Functions Part 2, Polynomials Lecture 2.4b, Complex Numbers in Quadratic Equations By completing the square we can solve for x and find that √ b 2 − 4 ac x = − b ± 2 a Dr. Ken W. Smith The expression b 2 − 4 ac under the radical sign is called the discriminant Sam Houston State University of the quadratic equation and is often abbreviated by ∆ . 2013 If ∆ = b 2 − 4 ac is positive then the square root of ∆ is a real number and so the quadratic equation has two real solutions: √ √ x = − b + ∆ ∆ and x = − b − . 2 a 2 a 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 ax 2 + bx + c = 0 The quadratic equation has solutions √ ax 2 + bx + c = 0 b 2 − 4 ac x = − b ± 2 a √ has solutions If ∆ = b 2 − 4 ac is negative then ∆ is imaginary and so our solutions are √ b 2 − 4 ac x = − b ± complex numbers which are not real. 2 a √ √ To be explicit, if ∆ is negative then − ∆ is positive and so ∆ = − ∆ i. If ∆ = b 2 − 4 ac is zero then there is only one solution since The solutions to the quadratic formula are then √ √ x = − b + √− ∆ i and x = − b −√− ∆ i ∆ 0 x = − b ± = − b ± = − b 2 a . 2 a 2 a 2 a 2 a √ 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. Solve the quadratic equation x 2 − x + 1 = 0 . Also, factor x 2 − x + 1 . x 2 + x + 1 = 0 Solution. By the quadratic formula the solutions to x 2 − x + 1 = 0 are are − 1 ± √ 1 ±√− 3 √ √ = − 1 ±√− 3 √ 3 √− 1 √ √ = 1 ± 3 i = 1 3 2 ± 2 i. 1 2 − 4(1)(1) = − 1 ± = − 1 ± 3 i = − 1 3 2 ± 2 i. 2 2 3 2 2 2 Since the two solutions to the equation x 2 − x + 1 = 0 are the complex Thus the two solutions to the equation x 2 + x + 1 = 0 are the complex numbers conjugate pairs √ √ 1 2 i and 1 3 3 2 + 2 − 2 i. √ √ − 1 2 i and − 1 3 3 2 + 2 − 2 i. then the polynomial x 2 − x + 1 factors as Since these are the two zeroes of the polynomial x 2 + x + 1 then we can √ √ factor ( x − ( 1 2 i ))( x − ( 1 3 3 2 + 2 − 2 i )) √ √ x 2 + x + 1 = ( x − ( − 1 2 i ))( x − ( − 1 3 3 2 + 2 − 2 i )) √ √ 3 3 = ( x − 1 2 i )( x − 1 2 − 2 + 2 i ) √ √ = ( x + 1 2 i )( x + 1 3 3 2 − 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 2 x 2 + 5 x + 7 to factor 2 x 2 + 5 x + 7 . Solution. Since the two solutions to the equation 2 x 2 + 5 x + 7 = 0 are √ √ Solve the quadratic equation x = − 5 4 i and x = − 5 31 31 4 + 4 − 4 i 2 x 2 + 5 x + 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 4 i ))( x − ( − 5 31 31 4 + 4 − 4 i )) x = − 5 ± √ √ = − 5 ±√− 31 31 √− 1 √ √ 5 2 − 4(2)(7) must be a factor of 2 x 2 + 5 x + 7 . But if we check the leading coefficient = − 5 ± = − 5 ± 31 i = − 5 31 4 ± 4 i. 4 4 4 4 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 2 x 2 + 5 x + 7 factors as √ √ √ √ x = − 5 4 i and x = − 5 31 31 4 + 4 − 4 i. 31 31 2( x − ( − 5 4 i ))( x − ( − 5 4 + 4 − 4 i )) √ √ = 2( x + 5 4 i )( x + 5 31 31 4 − 4 + 4 i ) Smith (SHSU) Elementary Functions 2013 19 / 35 Smith (SHSU) Elementary Functions 2013 20 / 35