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Chapter 3 Section 3.4 The Fundamental Theorem of Algebra The Fundamental Theorem Complex Numbers and the Imaginary i Definition: The number x such that x2 = –1 is defined to be i Note: Since x = –1 then i = –1 and –i = – –1 iy Standard Form a + bi , b = 0 a + bi Real Numbers Complex Numbers a + bi , b ≠ 0 Imaginary Numbers ● a + bi x Sometimes b ≠ 0 and a = 0 The Complex Plane Note: a + bi is also written a + ib for real a and b … especially when b is a radical or other functional form 4/28/2010 Section 4.4 v5.0 2 The Fundamental Theorem Radical Expressions and Arithmetic The expression –a –a = –1 a = i a The sum and difference of complex numbers (a + bi) ± (c + di) = (a ± c) ± (b ± d)i Examples: (3 + 4i) – (2 – 5i) = (3 – 2) + (4 + 5)i = 1 + 9i (7 – 3i) + (2 – 5i) = (7 + 2) – (3 + 5)i = 9 – 8i The product of complex numbers (a + bi)(c + di) = ac + bdi2 + (ad + bc)i = (ac – bd) + (ad + bc)i Example: (3 + 4i)(2 – 5i) = 6 – 20i2 + (8i – 15i) = 26 – 7i 4/28/2010 Section 4.4 v5.0 3 The Fundamental Theorem Complex Conjugates Definition: a + bi and a – bi are a complex conjugate pair Example: 7 + 3i and 7 – 3i are complex conjugates Fact: The product of complex conjugates is always real (a + bi ) (a – bi) = a2 + b2 Example: (7 + 3i) (7 – 3i) = 72 + 32 = 49 + 9 = 58 The quotient of complex numbers (ac + bd) + (bc – ad)i a + bi a + bi c – di = = c + di c2 + d2 c + di c – di ac + bd ac + bd (bc – ad)i bc – ad = 2 = 2 i + + 2 c + d2 c + d2 c2 + d2 c + d2 ( 4/28/2010 Section 4.4 v5.0 ) ( ) 4 The Fundamental Theorem Quotient Examples 1. (15 + 65i) (1 – 2i) 15 + 130 + (65 – 30)i 15 + 65i = (1 + 2i)(1 – 2i) = 12 + 22 1 + 2i = 29 + 7i 2. 3. 3 3 ( –i ) = = –3i i ( i )( –i ) (–2 + i) (–2i) –2 + i (–2 + i) (1 – i )2 = = (1 + i )2 (1 + i )2 (1 – i )2 ((1 + i )(1 – i ))2 2 + 4i = ( 2 )2 = 4/28/2010 1 +i 2 Section 4.4 v5.0 5 The Fundamental Theorem Quadratic Formula Examples 1. Solve: x2 – 2x + 1 = 0 x= –b ± b2 – 4ac 2a = Solution set: { 1 } +2 ± (–2)2 – 4(1)(1) 2(1) = 2± 0 2 =1 Question: Is there and easier way ? 2. Solve: x2 – 4x – 5 = 0 x= –b ± b2 – 4ac 2a = +4 ± (–4)2 – 4(1)(–5) 2(1) = 4 ± 36 = 2 –1 5 Solution set: { –1, 5 } 3. Solve: x2 – 2x + 5 = 0 x= –b ± b2 – 4ac 2a = +2 ± (–2)2 – 4(1)(5) 2(1) = 2 ± –16 = 1± 2i 2 Solution set: { 1 – 2i , 1 + 2i } 4/28/2010 Section 4.4 v5.0 6 The Fundamental Theorem Complex Solutions of Quadratic Equations Complex solutions occur in pairs Each has a complex conjugate solution Depends on discriminant: x = –b ± b2 – 4ac 2a WHY ? The Quadratic Formula b2 – 4ac = 0 One real solution b2 – 4ac > 0 Two real solutions b2 – 4ac < 0 Two complex (non-real) solutions Note the radical 4/28/2010 Section 4.4 v5.0 7 The Fundamental Theorem The Fundamental Theorem of Algebra A polynomial of degree n ≥ 1 has at least one complex zero Consequence: Every polynomial has a complete factorization WHY ? For polynomial f(x) there is some zero k1 so that f(x) = (x – k1)Q1(x) Then Q1(x) = (x – k2)Q2(x) so that f(x) = (x – k1)(x – k2)Q2(x) Applying the Fundamental Theorem n times f(x) = (x – k1)(x – k2)...(x – kn)Cn ... where Cn is a non-zero constant 4/28/2010 Section 4.4 v5.0 8 The Fundamental Theorem The Number of Zeros Theorem A polynomial of degree n has at most n distinct zeros Follows from the Fundamental Theorem factorization f(x) = (x – k1)(x – k2)...(x – kn)Cn If all the ki are distinct there are n distinct zeros ... otherwise one or more zeros are repeated 4/28/2010 Examples 1. f(x) = 5x(x + 3)(x – 1)(x – 4) 2. f(x) = 4x2(x – 2)3(x – 7) How many distinct zeros ? How many distinct zeros ? Total zeros ? Section 4.4 v5.0 9 The Fundamental Theorem The Conjugate Zeros Theorem If f(x) is a polynomial with only real coefficients then its complex zeros occur in conjugate pairs Examples 1. f(x) = x4 + 5x2 – 36 Let u = x2 = u2 + 5u – 36 = (u – 4)(u + 9) = (x2 – 4)(x2 + 9) = (x + 2)(x – 2)(x + 3i)(x – 3i) Zeros ? 2. f(x) = x3 + x = x(x2 + 1) = x(x + i)(x – i) 4/28/2010 Zeros ? Section 4.4 v5.0 10 The Fundamental Theorem Solving Equations Using Zeros Factor polynomial Set completely factored form equal to 0 Apply zero product property Examples 1. f(x) = x4 + 5x2 – 36 = (x + 2)(x – 2)(x + 3i)(x – 3i) = 0 (x + 2) = 0 OR (x – 2) OR (x + 3i) = 0 OR (x – 3i) = 0 Solution set is { –2, 2, –3i, 3i } 2. f(x) = x3 + x = x(x + i)(x – i) = 0 x=0 OR x+i=0 OR x–i=0 Solution set is { 0, – i , i } 4/28/2010 Section 4.4 v5.0 11 Think about it ! 4/28/2010 Section 4.4 v5.0 12