Honors Advanced Mathematics November 4, /2.6 summary and extra problems page 1 Recap: complex numbers

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1 November 4, 013.5/.6 summary and extra problems page 1 Recap: complex numbers Number system The complex number system consists of a + bi where a and b are real numbers, with various arithmetic operations. The real numbers are a subset of the complex numbers: a = a + 0i corresponding to (a, 0). So the real line is the horizontal axis of the complex plane. Operations To add or subtract complex numbers, just combine like terms. To multiply complex numbers, use the distributive property and the fact i = 1. To divide complex numbers, use this technique involving a conjugate: a bi c di c di c di A square root of a + bi is a complex number whose square equals a + bi. Every non-zero complex number has two square roots, which are opposites of each other. For example, ( 3i) = 5 1i and ( + 3i) = 5 1i. See.5 #60 for a problem where you re asked to find square roots. As of now, we don t have a definition for how to use a complex number as an exponent. That requires some trigonometry ideas that we haven t established yet. Other operations on a complex number z = a + bi: absolute value (or modulus) z = a b conjugate z = a bi

2 November 4, 013.5/.6 summary and extra problems page Summary: theorems about zeros of polynomials Name of theorem; location in Precalculus text Factor p. 199 Fundamental of Algebra p. 10 Complex Conjugates p. 11 Fundamental connection between zeros and x-intercepts Applies when the coefficients meet this requirement Pertains to what kind of zeros What the theorem says complex coeffs. complex zeros x = c is a zero (x c) is a factor. complex coeffs. complex zeros P(x) of degree n has exactly n zeros in the complex numbers (when counted using their multiplicities). real coeffs. complex zeros If a + bi is a zero, then a bi is also a zero. real coeffs. real zeros x = c is a zero the graph has an x-intercept at x = c. Remarks We proved this using the Remainder. An equivalent statement is that P(x) (x c) has remainder 0. Equivalently: every polynomial can be factored into n linear factors. The complex number system is called complete because it has this complete factorization property. To understand why the coefficients in this theorem must be real, multiply out a counterexample (ex. a + bi is a zero but a bi is not). Note that non-real complex zeros do not appear on graphs as x-intercepts. Intermediate Value (applied to finding zeros of polynomials) p. 190 Odd Degree p. 14 Irrational Conjugates not in textbook Rational Zeros p. 01 real coeffs. real zeros If P(c) and P(d) have opposite signs, then there must be a zero between c and d. real coeffs. real zeros Odd degree polynomials have at least one real zero. rational coeffs. real zeros If a + b is a zero (where b is irrational) then a b is also a zero. integer coeffs. rational zeros The only possible rational zeros are of the factor of erm form constant t. factor of leadingcoeff. The last two theorems aren t required knowledge for this course, but included in case you are interested. This is a consequence of the fact that polynomial graphs are continuous. The same fact applies for any continuous function. This theorem can be understood graphically by thinking about end behavior. This theorem can be thought of as the irrational analogue to the Complex Conjugates. Often used to prove that certain polynomials have no rational zeros, or to prove that a certain zero is irrational.

3 November 4, 013.5/.6 summary and extra problems page 3 Problems about finding solutions and factors 1. Consider the equation x a. How many solutions does this equation have in the real number system? b. According to the Fundamental of Algebra, how many solutions does this equation have in the complex number system? c. Find all solutions of the equation in the complex number system. a. Find a quadratic polynomial with real coefficients having ( + 5i) as a zero. Write your answer in standard form (that is, multiply the factors, don t leave in factored form). b. Find a cubic polynomial with real coefficients having 4 and (3 i) as zeros. Write your answer in standard form. Hint: Easiest if you multiply the factors from the non-real zeros first. 3. To answer these questions you will need to write polynomials that have complex numbers as coefficients. It is not possible to write answers having real coefficients. a. Find a linear polynomial with complex coefficients having ( + 5i) as a zero. b. Find a quadratic polynomial with complex coefficients having 4 and (3 i) as zeros. Write your answer in standard form. 4. You are given this information about polynomial P(x): P(x) is a degree 5 polynomial with real coefficients. The graph of P(x) for real numbers x is given on the grid. Its intercepts are at x = 3, x = 4, and y =. In the complex numbers, P(i+1) = 0. a. When P(x) is divided by (x 3), what is the remainder? b. Factor P(x) completely in the complex number system. c. Factor P(x) as completely as possible in the real number system. 5. Suppose f(x) = x 4 8x 3 + 7x 50x a. Verify that f(1 i) = 0. b. Factor f(x) completely in the complex number system. c. Factor f(x) as completely as possible in the real number system.

4 November 4, 013.5/.6 summary and extra problems page 4 Problems about incorrect uses of theorems Directions for problems 6 and 7: Each of the following arguments reaches an incorrect conclusion. Identify the erroneous step in each argument; identify which theorem was misused and explain why the use is incorrect. The theorem summary on page may be helpful. Consider the type of coefficients required by each theorem Argument: The function f(x) = has values f(3) = 1 and f(5) = 1. Because one of these values x 4 is negative and the other is positive, there must be a zero in between. That is, f(x) must have a zero in the interval 3 < x < Argument: Let P be a polynomial with complex coefficients, having + 3i as a zero. Since non-real zeros occur in conjugate pairs, P cannot be a polynomial of degree 1. Follow-up question: Find polynomials of degrees 1,, and 3, each having + 3i as a zero. Make them polynomials with real coefficients, if possible. Problems about square roots and complex zeros 8. a. How could you show that 79 is the square root of 641 without using a calculator? b. How could you show that 3 i is a square root of 8-6i? 9. Show that 1 i is a square root of i. 10. Find the square roots of 3+4i. That is, find a + bi having the property (a + bi) = ( 3 + 4i). 11. a. Find the square roots of ( i). That is, find a + bi having the property (a + bi) = ( i). b. Check your answer by typing ( i) on your calculator. 1. Consider this polynomial with complex coefficients: f(x) = x + 3i x + (1 i). (Note that several of the theorems on page apply only to polynomials with real coefficients, so do not apply to this polynomial.) a. Use the Quadratic Formula to find the zeros of f(x). Hint: The result of problem 9 will be needed. b. Write a complete factorization of f(x). Problems about GCF and LCM 13. f(x) = x 4 8x 3 + 7x 50x + 50 and g(x) = 3x 5 6x x 3 44x 86x Find GCF(f(x), g(x)) and LCM(f(x), g(x)) given the hint that f(1 i) = 0. Definition: A pair of numbers or a pair of polynomials is called relatively prime if the pair has a greatest common factor (GCF) of Suppose f x and gx are polynomials. The zeros of f x are 3, 1, and 5. The zeros of gx are 4, 1, and 1. Are f x and gx relatively prime? Explain. 3 m and n x x 5x 8x 6. Find all the zeros (real and non-real) for both functions given the hint that m x and n x are not relatively prime Suppose x x x 4x 4x 4

5 November 4, 013.5/.6 summary and extra problems page 5 ANSWER NOTES: Depending on the problem, what s shown here might be a complete solution, or just the final answer, or just a hint or a first step. Problems about finding solutions and factors 1. c. 3, i, 3 i. a. x + 4x + 9 b. Begin from (x 4)(x (3 i)) (x (3 + i)). The easiest start is to multiply the second and third factors. That gives (x 4)(x 6x + 13). Then use the distributive property to get x 3 10x + 37x a. x ( + 5i) b. (x 4)(x (3 i)), then either distribute or apply the formula from a, to get x + ( 7 + i)x + (1 8i). 4. a. remainder = P(3) = 5 b. 1 (x + 3) (x 4)(x (1 + i))(x (1 i)) c (x + 3) (x 4)(x x + ) 5. b. (x (1 i))(x (1 + i))(x (3 + i))(x (3 i)) c. (x 6x + 10)(x x + 5) Problems about incorrect use of theorems 6. Intermediate Value does not apply because f(x) is not continuous on the interval. 7. Complex Conjugates does not apply when the coefficients are complex. (x ( + 3i)); x 4x + 13; x 3 4x + 13x Problems about square roots and complex zeros 79 b. 3 i 8 6i 8. a i i 10. +i and i. See problem 11 notes for more hints.

6 November 4, 013.5/.6 summary and extra problems page a. You need to solve the system a b = 13, ab = 8. Solve the second equation for a, substitute into the first equation. Multiply by b to get b 4 13b 16 = 0. Let x = b and apply the quadratic formula to x 13x 16 to get x = But x = b cannot be negative, so discard the negative possibility Square root again to get b =. Perhaps the easiest way to get a is to repeat the process solving for the other variable. Get a = So a + bi = i or i. 1. a. Quadratic formula gives root. 3i 13 8i, then substitute the answers to 9a in place of the square Problems about GCF and LCM 13. GCF: x - x + 5; LCM: (x x + 5)(x 6x + 10)(3x 3 8x 5x + 6) 14. No - because (x - 1) is a common factor for both functions. 15. m(x): x 1 i, x i n(x): x = 3, x 1 i