Complex roots of polynomials: factor theorem and conjugates

The remainder theorem

• Remainder: p(a)
• Factor: p(a) = 0

1. p(z) = z³ + kz² + 3z − 4
2. p(2) = 2³ + k(2²) + 3(2) − 4substitute z = 2
3. = 8 + 4k + 6 − 4
4. = 10 + 4k
5. 10 + 4k = 2the remainder is 2
6. 4k = −8
7. k = −2

Two unknown constants

• p(1) = 0: factor theorem
• z + 1: substitute z = −1
• Add to eliminate b

1. p(z) = z³ + az² + bz − 5
2. p(1) = 1 + a + b − 5 = 0factor theorem
3. a + b = 4
4. p(−1) = −1 + a − b − 5 = −8z + 1 = z − (−1)
5. a − b = −2
6. 2a = 2, so a = 1add the equations
7. b = 4 − 1 = 3substitute back
8. a = 1, b = 3

Divide by z − 1

• Divide by the known factor
• Discriminant < 0: complex

1. (z³ + z² + 3z − 5) ÷ (z − 1)
2. z²(z − 1) = z³ − z²leaves 2z² + 3z − 5
3. 2z(z − 1) = 2z² − 2zleaves 5z − 5
4. 5(z − 1) = 5z − 5leaves 0
5. p(z) = (z − 1)(z² + 2z + 5)
6. b² − 4ac = 2² − 4(1)(5)
7. = 4 − 20 = −16< 0: complex roots

The complex pair

• Complete the square
• √−4 = 2i
• Check a root

1. z² + 2z + 5 = 0
2. z² + 2z = −5
3. z² + 2z + 1 = −5 + 1add 1 to both sides
4. (z + 1)² = −4
5. z + 1 = ±√−4 = ±2i√−4 = 2i
6. z = −1 ± 2i
7. roots 1, −1 + 2i, −1 − 2i
8. (−1 + 2i)² = −3 − 4icheck
9. −3 − 4i − 2 + 4i + 5 = 0

A cubic from one complex root

• Pair gives z² − 4z + 5
• Compare constants
• Expand, then compare

1. (z² − 4z + 5)(z + c)the other factor
2. 5c = 15, so c = 3compare constants
3. (z² − 4z + 5)(z + 3)
4. = z³ + 3z² − 4z² − 12z + 5z + 15
5. = z³ − z² − 7z + 15collect terms
6. a = −1, b = −7compare coefficients
7. z + 3 = 0, so z = −3
8. roots 2 + i, 2 − i, −3

Or substitute z = 2 + i

• Substitute z = 2 + i
• Expand the powers first

1. (2 + i)² = 4 + 4i + i²expand
2. = 3 + 4ii² = −1
3. (2 + i)³ = (3 + 4i)(2 + i)
4. = 6 + 3i + 8i + 4i²expand
5. = 2 + 11ii² = −1
6. (2 + 11i) + a(3 + 4i)
7. + b(2 + i) + 15 = 0

Equate real and imaginary parts

• Real part = 0
• Imaginary part = 0
• Double, then subtract

1. (2 + 11i) + a(3 + 4i)
2. + b(2 + i) + 15 = 0
3. real: 2 + 3a + 2b + 15 = 0
4. 17 + 3a + 2b = 0
5. imaginary: 11 + 4a + b = 0
6. 22 + 8a + 2b = 0double the second
7. 5 + 5a = 0, so a = −1subtract the first
8. b = −11 − 4(−1) = −7substitute back

Forming a polynomial from its roots

• Add the conjugate root
• Multiply the factors

1. roots 3, 1 − 2i, 1 + 2ireal coefficients
2. sum = 2, product = 1 − 4i² = 5
3. factor z² − 2z + 5
4. (z² − 2z + 5)(z − 3)
5. = z³ − 3z² − 2z² + 6z + 5z − 15
6. = z³ − 5z² + 11z − 15

The answers: questions 1 and 2

1. 1. p(2) = 2³ + k(2²) + 2(2) − 1remainder theorem
2. = 8 + 4k + 4 − 1 = 11 + 4k
3. 11 + 4k = 3, so k = −2the remainder is 3
4. 2. z² − 4z + 4 = −13 + 4add 4 to both sides
5. (z − 2)² = −9
6. z − 2 = ±√−9 = ±3i√−9 = 3i
7. z = 2 ± 3i

The answers: question 3

1. 3. roots 1 + 2i and 1 − 2ireal coefficients
2. sum 2, product 1 + 4 = 5
3. factor z² − 2z + 5
4. × (z + c): 5c = −10, c = −2compare constants
5. (z² − 2z + 5)(z − 2)
6. = z³ − 2z² − 2z² + 4z + 5z − 10
7. = z³ − 4z² + 9z − 10collect terms
8. a = −4, b = 9
9. roots 1 + 2i, 1 − 2i, 2

The answers: question 4

1. 4. roots −1, 2 + 3i, 2 − 3ireal coefficients
2. sum 4, product 4 − 9i² = 13
3. factor z² − 4z + 13
4. (z² − 4z + 13)(z + 1)
5. = z³ + z² − 4z² − 4z + 13z + 13
6. = z³ − 3z² + 9z + 13