1. Chapter 3 Section 3.4 The Fundamental Theorem of Algebra

2. 4/28/2010 Section 4.4 v5.0 2 Complex Numbers and the Imaginary i Definition: The number x such that x 2 = –1 is defined to be i Note: Since Standard Form The Fundamental Theorem  –1then i = a + bi Complex Numbers a + bi, b = 0 Real Numbers a + bi, b ≠ 0 Imaginary Numbers x iyiy ● a + bia + bi Note: a + bi is also written a + ib for real a and b … especially when b is a radical or other functional form Sometimes b ≠ 0 and a = 0 The Complex Plane and  –1 – i = –  –1 x = 

3. 4/28/2010 Section 4.4 v5.0 3 Radical Expressions and Arithmetic The expression The sum and difference of complex numbers ( a + bi ) ± ( c + di ) = ( a ± c ) ± ( b ± d ) i Examples: (3 + 4 i ) – (2 – 5 i ) (7 – 3 i ) + (2 – 5 i ) The product of complex numbers ( a + bi )  ( c + di ) Example: (3 + 4 i )  (2 – 5 i ) The Fundamental Theorem  –a–a  –a–a  –1  a =  a i = = (3 – 2) + (4 + 5) i = 1 + 9 i = (7 + 2) – (3 + 5) i = 9 – 8 i = ac + bdi 2 + ( ad + bc ) i = ( ac – bd ) + ( ad + bc ) i = 6 – 20 i 2 + (8 i – 15 i ) = 26 – 7 i

4. 4/28/2010 Section 4.4 v5.0 4 Complex Conjugates Definition: a + bi and a – bi are a complex conjugate pair Example: 7 + 3 i and 7 – 3 i are complex conjugates Fact: The product of complex conjugates is always real ( a + bi )  ( a – bi ) = a 2 + b 2 Example: (7 + 3 i )  (7 – 3 i ) = 7 2 + 3 2 = 49 + 9 = 58 The quotient of complex numbers The Fundamental Theorem a + bi c + di c – di = a + bi c + di = ( bc – ad ) i + ( ac + bd ) c2 + d2c2 + d2 = + ac + bd c2 + d2c2 + d2 ( bc – ad ) i c2 + d2c2 + d2 = + i ac + bd c2 + d2c2 + d2 ( ) bc – ad c2 + d2c2 + d2 ( )

5. 4/28/2010 Section 4.4 v5.0 5 Quotient Examples 1. 2. 3. The Fundamental Theorem 15 + 65 i 1 + 2i1 + 2i (15 + 65 i ) (1 + 2 i ) (1 – 2 i ) = 15 + 130 + (65 – 30) i 12 + 2212 + 22 = = 29 + 7 i 3 i = ( i )( – i ) 3 ( – i ) = –3 i (1 + i ) 2 –2 + i = (1 – i ) 2 (1 + i ) 2 (1 – i ) 2 (–2 + i ) = (1 – i ) ) 2 ( (1 + i ) (–2 i ) (–2 + i ) = ( 2 ) 2 2 + 4 i = 1 2 i +

6. 4/28/2010 Section 4.4 v5.0 6 Quadratic Formula Examples 1. Solve: x 2 – 2x + 1 = 0 2. Solve: x 2 – 4x – 5 = 0 3. Solve: x 2 – 2x + 5 = 0 The Fundamental Theorem x = 2a2a b 2 – 4 ac  ± –b–b = 1 = 2(1) ( –2 ) 2 – 4(1)(1)  ± +2+2 = 2 0  ± 2 Solution set: { 1 } Question: Is there and easier way ? x = 2a2a b 2 – 4 ac  ± –b–b = 2(1) ( –4 ) 2 – 4(1)(–5)  ± +4+4 = 2 36  ± 4 Solution set: { –1, 5 } = 5 –1 x = 2a2a b 2 – 4 ac  ± –b–b ± ( –2 ) 2 – 4(1)(5) = 2(1)  +2+2 = 2 –16  ± 2 = 1± 2 i Solution set: { 1 – 2 i, 1 + 2 i }

7. 4/28/2010 Section 4.4 v5.0 7 Complex Solutions of Quadratic Equations Complex solutions occur in pairs Each has a complex conjugate solution Depends on discriminant: The Fundamental Theorem WHY ? x = 2a2a b 2 – 4 ac  ± –b–b The Quadratic Formula = b 2 – 4 ac 0 0 > 0 < One real solution Two real solutions Two complex (non-real) solutions Note the  radical

8. 4/28/2010 Section 4.4 v5.0 8 The Fundamental Theorem of Algebra A polynomial of degree n ≥ 1 has at least one complex zero Consequence: Every polynomial has a complete factorization For polynomial f(x) there is some zero k 1 so that f(x) = (x – k 1 )Q 1 (x) Then Q 1 (x) = (x – k 2 )Q 2 (x) so that f(x) = (x – k 1 )(x – k 2 )Q 2 (x) Applying the Fundamental Theorem n times f(x) = (x – k 1 )(x – k 2 )...(x – k n )C n The Fundamental Theorem WHY ?... where C n is a non-zero constant

9. 4/28/2010 Section 4.4 v5.0 9 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 – k 1 )(x – k 2 )...(x – k n )C n If all the k i are distinct there are n distinct zeros... otherwise one or more zeros are repeated Examples 1. f(x) = 5x(x + 3)(x – 1)(x – 4) 2. f(x) = 4x 2 (x – 2) 3 (x – 7) The Fundamental Theorem How many distinct zeros ? Total zeros ?

10. 4/28/2010 Section 4.4 v5.0 10 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) = x 4 + 5x 2 – 36 = u 2 + 5u – 36 = (u – 4)(u + 9) = (x 2 – 4)(x 2 + 9) 2. f(x) = x 3 + x = x(x 2 + 1) = x(x + i )(x – i ) The Fundamental Theorem Let u = x 2 = (x + 2)(x – 2)(x + 3 i )(x – 3 i )Zeros ?

11. 4/28/2010 Section 4.4 v5.0 11 Solving Equations Using Zeros Factor polynomial Set completely factored form equal to 0 Apply zero product property Examples 1. f(x) = x 4 + 5x 2 – 36 (x + 2) = 0 OR (x – 2) OR (x + 3 i ) = 0 OR (x – 3 i ) = 0 Solution set is { –2, 2, –3 i, 3i } 2. f(x) = x 3 + x = x(x + i )(x – i ) x = 0 OR x + i = 0 OR x – i = 0 Solution set is { 0, – i, i } The Fundamental Theorem = (x + 2)(x – 2)(x + 3 i )(x – 3 i ) = 0

12. 4/28/2010 Section 4.4 v5.0 12 Think about it !