3.7 Complex Zeros; Fundamental Theorem of Algebra


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1 SECTION.7 Complex Zeros; Fundamental Theorem of Algebra 2.7 Complex Zeros; Fundamental Theorem of Algebra PREPARING FOR THIS SECTION Before getting started, review the following: Complex Numbers (Appendix, Section A.6, pp ) Now work the Are You Prepared? problems on page 27. Quadratic Equations with a Negative Discriminant (Appendix, Section A.6, pp ) OBJECTIVES Use the Conjugate Pairs Theorem 2 Find a Polynomial Function with Specified Zeros Find the Complex Zeros of a Polynomial In Section.6 we found the real zeros of a polynomial function. In this section we will find the complex zeros of a polynomial function. Finding the complex zeros of a function requires finding all zeros of the form a + bi. These zeros will be real if b = 0. A variable in the complex number system is referred to as a complex variable. A complex polynomial function f of degree n is a function of the form fx2 = a n x n + a n  x n  + Á + a x + a 0 () where a n, a n , Á, a, a 0 are complex numbers, a n Z 0, n is a nonnegative integer, and x is a complex variable. As before, a n is called the leading coefficient of f. A complex number r is called a (complex) zero of f if fr2 = 0. We have learned that some quadratic equations have no real solutions, but that in the complex number system every quadratic equation has a solution, either real or complex. The next result, proved by Karl Friedrich Gauss ( ) when he was 22 years old, * gives an extension to complex polynomials. In fact, this result is so important and useful that it has become known as the Fundamental Theorem of Algebra. Fundamental Theorem of Algebra Every complex polynomial function fx2 of degree n Ú has at least one complex zero. We shall not prove this result, as the proof is beyond the scope of this book. However, using the Fundamental Theorem of Algebra and the Factor Theorem, we can prove the following result: Theorem Every complex polynomial function fx2 of degree n Ú can be factored into n linear factors (not necessarily distinct) of the form fx2 = a n x  r 2x  r 2 2 # Á # x  rn 2 (2) where a n, r, r 2, Á, r n are complex numbers. That is, every complex polynomial function of degree n Ú has exactly n (not necessarily distinct) zeros. * In all, Gauss gave four different proofs of this theorem, the first one in 799 being the subject of his doctoral dissertation.
2 24 CHAPTER Polynomial and Rational Functions Proof Let fx2 = a n x n + a n  x n  + Á + a x + a 0 By the Fundamental Theorem of Algebra, f has at least one zero, say r. Then, by the Factor Theorem, x  r is a factor, and fx2 = x  r 2q x2 where q x2 is a complex polynomial of degree n  whose leading coefficient is a n. Again by the Fundamental Theorem of Algebra, the complex polynomial q x2 has at least one zero, say r 2. By the Factor Theorem, q x2 has the factor x  r 2, so q x2 = x  r 2 2q 2 x2 where q 2 x2 is a complex polynomial of degree n  2 whose leading coefficient is a n. Consequently, fx2 = x  r 2x  r 2 2q 2 x2 Repeating this argument n times, we finally arrive at fx2 = x  r 2x  r 2 2 # Á # x  rn 2q n x2 where q n x2 is a complex polynomial of degree n  n = 0 whose leading coefficient is a Thus, q n x2 = a n x 0 n. = a n, and so fx2 = a n x  r 2x  r 2 2 # Á # x  rn 2 We conclude that every complex polynomial function fx2 of degree n Ú has exactly n (not necessarily distinct) zeros. Conjugate Pairs Theorem Use the Conjugate Pairs Theorem We can use the Fundamental Theorem of Algebra to obtain valuable information about the complex zeros of polynomials whose coefficients are real numbers. Let fx2 be a polynomial whose coefficients are real numbers. If r = a + bi is a zero of f, then the complex conjugate r = a  bi is also a zero of f. In other words, for polynomials whose coefficients are real numbers, the zeros occur in conjugate pairs. Proof Let fx2 = a n x n + a n  x n  + Á + a x + a 0 where a n, a n , Á, a, a 0 are real numbers and a n Z 0. If r = a + bi is a zero of f, then fr2 = fa + bi2 = 0, so We take the conjugate of both sides to get a n r n + a n  r n  + Á + a r + a 0 = 0 a n r n + a n  r n  + Á + a r + a 0 = 0 a n r n + a n  r n  + Á + a r + a 0 = 0 The conjugate of a sum equals the sum of the conjugates (see the Appendix, Section A.6). a n r2 n + a n  r2 n  + Á + a r + a 0 = 0 The conjugate of a product equals the product of the conjugates. a n r2 n + a n  r2 n  + Á + a r + a 0 = 0 The conjugate of a real number equals the real number. This last equation states that fr2 = 0; that is, r = a  bi is a zero of f.
3 SECTION.7 Complex Zeros; Fundamental Theorem of Algebra 25 The value of this result should be clear. If we know that, + 4i is a zero of a polynomial with real coefficients, then we know that  4i is also a zero. This result has an important corollary. COROLLARY A polynomial f of odd degree with real coefficients has at least one real zero. Proof Because complex zeros occur as conjugate pairs in a polynomial with real coefficients, there will always be an even number of zeros that are not real numbers. Consequently, since f is of odd degree, one of its zeros has to be a real number. For example, the polynomial fx2 = x 5  x 4 + 4x  5 has at least one zero that is a real number, since f is of degree 5 (odd) and has real coefficients. EXAMPLE Using the Conjugate Pairs Theorem A polynomial f of degree 5 whose coefficients are real numbers has the zeros, 5i, and + i. Find the remaining two zeros. Solution Since complex zeros appear as conjugate pairs, it follows that 5i, the conjugate of 5i, and  i, the conjugate of + i, are the two remaining zeros. NOW WORK PROBLEM 7. 2 Find a Polynomial Function with Specified Zeros EXAMPLE 2 Finding a Polynomial Function Whose Zeros Are Given Figure (a) Find a polynomial f of degree 4 whose coefficients are real numbers and that has the zeros,, and 4 + i. (b) Graph the polynomial found in part (a) to verify your result. Solution (a) Since 4 + i is a zero, by the Conjugate Pairs Theorem, 4  i must also be a zero of f. Because of the Factor Theorem, if fc2 = 0, then x  c is a factor of fx2. So we can now write f as fx2 = ax  2x  2x i24x i24 where a is any real number. If we let a =, we obtain fx2 = x  2x  2x i24x i24 = x 22x + 2x i2x i2x i24  i24 = x 22x + 2x 2 + 4x  ix + 4x + ix i  4i  i 2 2 = x 22x + 2x 2 + 8x + 72 = x 4 + 8x + 7x 22x  6x 24x + x 2 + 8x + 7 = x 4 + 6x + 2x 226x + 7 (b) A quick analysis of the polynomial f tells us what to expect: At most three turning points. For large ƒxƒ, the graph will behave like y = x 4. A repeated real zero at so that the graph will touch the xaxis at. The only xintercept is at. Figure 80 shows the complete graph. (Do you see why? The graph has exactly three turning points and the degree of the polynomial is 4.)
4 26 CHAPTER Polynomial and Rational Functions Exploration Graph the function found in Example 2 for a = 2 and a = . Does the value of a affect the zeros of f? How does the value of a affect the graph of f? Now we can prove the theorem we conjectured earlier in Section.6. Theorem Every polynomial function with real coefficients can be uniquely factored over the real numbers into a product of linear factors and/or irreducible quadratic factors. Proof Every complex polynomial f of degree n has exactly n zeros and can be factored into a product of n linear factors. If its coefficients are real, then those zeros that are complex numbers will always occur as conjugate pairs. As a result, if r = a + bi is a complex zero, then so is r = a  bi. Consequently, when the linear factors x  r and x  r of f are multiplied, we have x  r2x  r2 = x 2  r + r2x + rr = x 22ax + a 2 + b 2 This seconddegree polynomial has real coefficients and is irreducible (over the real numbers). Thus, the factors of f are either linear or irreducible quadratic factors. Find the Complex Zeros of a Polynomial EXAMPLE Finding the Complex Zeros of a Polynomial Find the complex zeros of the polynomial function fx2 = x 4 + 5x + 25x x  8 Solution STEP : The degree of f is 4. So f will have four complex zeros. STEP 2: The Rational Zeros Theorem provides information about the potential rational zeros of polynomials with integer coefficients. For this polynomial (which has integer coefficients), the potential rational zeros are ;, ; 2, ;, ;2, ;, ;6, ;9, ;8 Figure STEP : Figure 8 shows the graph of f. The graph has the characteristics that we expect of this polynomial of degree 4: It behaves like y = x 4 for large ƒxƒ and has yintercept 8. There are xintercepts near 2 and between 0 and. STEP 4: Because f22 = 0, we known that 2 is a zero and x + 2 is a factor of f. We can use long division or synthetic division to factor f: fx2 = x + 22x  x x  92 From the graph of f and the list of potential rational zeros, it appears that may also be a zero of Since fa we know that is a zero of f. f. b = 0, We use synthetic division on the depressed equation of f to factor
5 SECTION.7 Complex Zeros; Fundamental Theorem of Algebra 27 Using the bottom row of the synthetic division, we find fx2 = x + 22ax  bx = x + 22ax  bx The factor x does not have any real zeros; its complex zeros are ;i. The complex zeros of fx2 = x 4 + 5x + 25x x  8 are 2,, i, i. NOW WORK PROBLEM.
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