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Professor P. Bishop MAC1105 (MDN) 1.3 Complex Numbers; Quadratic Equations with a Negative Discriminant The complex number system enables us to take even roots of negative numbers by means of the imaginary unit i, which is equal to the square root of –1; that is i2 = -1 and i = 1 . By factoring –1 out of a negative expression, it becomes positive and an even root can be taken: -b = i b . Standard form for complex expression is a + bi, where a is the real part and bi is the imaginary part. All properties of exponents hold when the base is i, thus i1 = i, i2 = -1, i3 = i2(i) = -1i = -i, i4 = i2(i2) = -1(-1) = 1. In general, for in, divide n by 4: if the remainder is 0, in= 1; if the remainder is 1, in = i, if the remainder is 2, in= -1; if the remainder is 3, in= -i. The product of a complex number (a + bi) and its conjugate (a – bi) is a nonnegative real number (a2 + b2). Write in a + bi form: 1. 25 2. a. 50 b. 50 Evaluate (simplify to 1, -1, i, -i): 4. i21 5. i39 6. i52 Write in a + bi form: 8. (-1 + 6i) + (5 – 4i) 9 10. 2 10 15. 4i10 + 6i5 7. i42 (-6 + 4i) - (2 – i) + (7 – 3i) 11. -5i7(2 + 3i) 13. (4 + 3 i)3 12. (3 – 2 i)(2 – 5i) 14. + 3i4 + 5i3 3. 3 16 2 16. (2 + 6i)(2 – 6i) 8 2 17. 3 5i 2i 18. 5 6i 6 3i Solve in the complex number system: 19. x2 + 49 = 0 20. 5x2 + 4x + 8 = 0 21. x3 – 64 = 0 22. x4 + 3x2 – 4 = 0 23. x3 + 4x2 – 4x – 16 = 0 TRY THESE: 24. Write in a + bi form: 2i7(5 + 3i3)(6 – 2i9) 25. Solve in the complex number system: 2x4 – 112 = 25z2 Professor P. Bishop MAC1105 (MDN) 5.4 COMPLEX ZEROS; FUNDAMENTAL THEOREM OF ALGEBRA Fundamental Theorem of Algebra: Every complex polynomial function of degree n ≥ 1 or higher has at least one complex zero. Theorem: Every complex polynomial of degree n ≥ or higher can be factored into n linear factors (not necessarily distinct). Conjugate Pairs Theorem: Given a polynomial with real coefficients, if r = a + bi is a zero, then r a bi is also a zero of the polynomial. Information is given about a polynomial f(x) whose coefficients are real numbers. Find the remaining zeros: 1. Degree 4; zeros: 3, 5, 4 + i 2. Degree 7; zeros: 2, 3i, 5 – i, 3 + i Form a polynomial f(x) with real coefficients having the given degree and zeros: 3. Degree 4; zeros: 4 – 3i, -3 multiplicity 2 4. Degree 5; zeros: 3 multiplicity 3, 2 – i Use the given zero to find the remaining zeros: 5. f(x) = x4 – 7x3 + 14x2 – 38x + 60; zero: 1 + 3i Find the complex zeros of the polynomial function. Write f in factored form: 6. f(x) = 2x4 + x3 – 35x2 – 113x + 65 Max. Zeros: Potential Zeros: Real Zeros: Positive Zeros: Negative Zeros: Try These (5.4) Pg. 386: #24. #32.