4.4 The Fundamental Theorem of Algebra What is -1 ? BUT, whatever it is Define a new number: i1 = i2 = i3 = i. i2 = i4 = i2. i2 = i5 = etc. i -1 -i 1 i not a real number its square should be -1 i -1 (hence i2 = -1) i70 = i68i2 = (i4)17(-1) = -1 By adding i to the real number system, we create a larger number system: the complex number system is the set of all numbers of the form a + bi, where a and b are real numbers C = { a + bi, a and b real} Examples: 3 + 2i 5i (= 0 + 5i) imaginary) 6 (= 6 + 0i) 0 a (real part) 3 0 b (imaginary part) 2 5 (pure 6 0 0 (real) 0 Every real number is also a complex number! (but not viceversa): 12 = 12 + 0i 4.4-1 Operations on complex numbers Easy! think of i as a "variable" use all the laws we already know to do the computation reduce any powers of i that you get write the final answer in standard form: a + bi (1 + 2i) + (2 - 3i) (1 + 2i)(2 + 3i) =3-i = (by FOIL) 2 + 7i + 6i2 = -4 + 7i Division trick: to do 1 + 2i 2 - 3i , do the following: (1 + 2i) . (2 + 3i) (2 - 3i) (2 + 3i) (multiply numerator and denominator by complex conjugate of denominator) = - 4 + 7i 2 2 - 9i 2 = - 4 + 7i = -4/13 + (7/13)i 13 Notes: conjugate of 20i is -20i the square root of a negative can be written using i: - 7 = -1 7 = i 7 always turn square root of a negative into imaginary before you do any computing: ( - 3 )2 = (i 3 )2 = i2( 3 )2 = -3 ( - 3 )2 = - 3 - 3 = 9 = 3 YES NO 4.4-2 FUNDAMENTAL THEOREM OF ALGEBRA All zeroes of all polynomials with real or imaginary coefficients are complex numbers or, to put it another way LINEAR FACTORIZATION THEOREM Any polynomial f(x) = anxn + . . . a1x + a0 (degree n) can be factored (by repeated division) into n linear factors: f(x) = an(x - c1)(x - c2) . . . (x - cn) where the ci’s are complex numbers. Constructing a polynomial having given zeros Easy! If given zeros are 3, -2, -i and i, and the leading coefficient is -2, the polynomial f(x) = -2(x - 3)(x +2)(x - i)(x + i) will clearly do the trick! If you have to multiply it out, multiply these two factors first: (x - i)(x + i) = x2 - i2 = x2 + 1 getting rid of the presence of i in subsequent multiplications 4.4-3 Conjugate Zeros Theorem If a polynomial with real coefficients has a zero a + bi, it also has a zero a - bi, and vice-versa. Imaginary zeros for such polynomials always occur in complex conjugate pairs. reason? think about x2 - 2x + 2 zeros are the roots of x2 - 2x + 2 = 0 2 4 quadratic formula gives us: = 1i 2 it's the in the quadratic formula that gives us the complex conjugate pairs. Exercise: Construct a cubic polynomial f with real coefficients two of whose zeros are 3 and 2 + i. Since it has real coefficients, it must also have 2 - i as a zero. F(x) = (x - 3)(x - (2 + i))(x - (2 - i)) This is the solution. If you are asked to multiply it out, do this: (x - 3)(x - 2 - i)(x - 2 + i) = (x - 3) ((x - 2) - i)((x - 2) + i) = (x - 3)( (x - 2)2 – i2 ) [(a – b)(a + b) = a2 – b2 ] = (x – 3)((x – 2)2 + 1) Now you can finish it without having to deal with i’s. 4.4-4