Complex polynomials: 7 exam questions solved step by step

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What this lesson covers

  • Real polynomials have complex roots in conjugate pairs, so one complex root of a cubic gives a second for free.
  • The factor and remainder theorems turn a known root into an equation for an unknown coefficient.
Key idea

7 questions, two theorems

  • 1. Remainder on ÷ (z − a) is p(a).
  • 2. Factor: p(a) = 0.
  • 3. Real coefficients: conjugate pairs.
  • 4. Every expansion by hand.
Your turn

Question 1 · Achieved

  • p(z) = 2z³ − z² + kz + 6
  • The remainder on dividing by z + 2 is −4.
  • Find k.
Worked example

Q1: remainder −4 by z + 2

  • Remainder theorem.
  • z + 2: use z = −2.
  1. z + 2 = z − (−2): use p(−2)remainder theorem
  2. p(−2) = 2(−2)³ − (−2)² + k(−2) + 6
  3. = −16 − 4 − 2k + 6
  4. = −14 − 2k
  5. −14 − 2k = −4the remainder is −4
  6. −2k = 10
  7. k = −5
  8. p(−2) = −16 − 4 + 10 + 6 = −4
Your turn

Question 2 · Achieved

  • Solve z² + 4z + 29 = 0.
Worked example

Q2: z² + 4z + 29 = 0

  • Discriminant < 0: complex.
  • √−100 = 10i.
  1. a = 1, b = 4, c = 29
  2. b² − 4ac = 16 − 116 = −100< 0: complex roots
  3. √−100 = 10i
  4. z = (−4 ± 10i)/2quadratic formula
  5. z = −2 ± 5i
  6. (−2 + 5i)² = −21 − 20i
  7. −21 − 20i − 8 + 20i + 29 = 0
Your turn

Question 3 · Merit

  • Show that z = 2 is a root of
  • z³ + 6z − 20 = 0,
  • then find the other two roots.
Worked example

Q3: the factor z − 2

  • Factor theorem.
  • Compare coefficients.
  1. p(2) = 2³ + 6(2) − 20factor theorem
  2. = 8 + 12 − 20 = 0
  3. so z − 2 is a factor
  4. p(z) = (z − 2)(z² + bz + c)the other factor
  5. −2c = −20, so c = 10constants
  6. b − 2 = 0, so b = 2z² terms: 0z²
  7. c − 2b = 10 − 4 = 6z terms
  8. p(z) = (z − 2)(z² + 2z + 10)
Worked example

Q3: the other two roots

  • Complete the square.
  • √−9 = 3i.
  1. z² + 2z + 10 = 0
  2. z² + 2z + 1 = −10 + 1complete the square
  3. (z + 1)² = −9
  4. z + 1 = ±3i√−9 = 3i
  5. z = −1 ± 3i
  6. roots 2, −1 + 3i, −1 − 3i
Your turn

Question 4 · Merit

  • z = 3 − 2i is a root of
  • z³ + az² + bz + 26 = 0, a and b real.
  • Find a, b and the other roots.
Worked example

Q4: one complex root

  • Real coefficients: pair it.
  • Then compare constants.
  1. so 3 + 2i is a rootreal coefficients
  2. sum = (3 − 2i) + (3 + 2i) = 6
  3. product = 9 − 4i² = 13i² = −1
  4. factor z² − 6z + 13
  5. (z² − 6z + 13)(z + c)the other factor
  6. 13c = 26, so c = 2compare constants
  7. z³ + 2z² − 6z² − 12z + 13z + 26
  8. = z³ − 4z² + z + 26collect terms
  9. a = −4, b = 1; roots 3 ± 2i, −2
Your turn

Question 5 · Merit

  • p(z) = z³ + az² + bz + 6
  • z + 1 is a factor of p(z), and the
  • remainder on dividing by z − 2 is 12.
  • Find a and b.
Worked example

Q5: two unknown constants

  • One theorem each.
  • Then solve the pair.
  1. p(−1) = −1 + a − b + 6 = 0factor theorem
  2. a − b = −5
  3. p(2) = 8 + 4a + 2b + 6 = 12remainder theorem
  4. 4a + 2b = −2, so 2a + b = −1halve
  5. 3a = −6, so a = −2add the equations
  6. b = a + 5 = 3substitute back
  7. a = −2, b = 3
  8. p(2) = 8 − 8 + 6 + 6 = 12
  9. p(−1) = −1 − 2 − 3 + 6 = 0
Your turn

Question 6 · Excellence

  • z = 1 + i is a root of
  • z⁴ − 2z³ + 6z² − 8z + 8 = 0.
  • Find all four roots.
Worked example

Q6: the quadratic factors

  • Pair gives one factor.
  • Compare for the other.
  1. 1 − i is a root tooreal coefficients
  2. sum 2, product 1 − i² = 2
  3. factor z² − 2z + 2
  4. (z² − 2z + 2)(z² + pz + q)the other factor
  5. 2q = 8, so q = 4constants
  6. p − 2 = −2, so p = 0z³ terms
  7. q − 2p + 2 = 6z² terms
  8. 2p − 2q = −8z terms
  9. (z² − 2z + 2)(z² + 4) = 0
Worked example

Q6: all four roots

  • Solve the second factor.
  • Check a new root.
  1. (z² − 2z + 2)(z² + 4) = 0
  2. z² + 4 = 0, z² = −4
  3. z = ±2i√−4 = 2i
  4. roots 1 + i, 1 − i, 2i, −2i
  5. (2i)⁴ = 16, (2i)³ = −8i
  6. (2i)² = −4
  7. 16 + 16i − 24 − 16i + 8 = 0
Your turn

Question 7 · Excellence

  • Solve z² − (3 + i)z + 2 + 2i = 0.
  • Explain why the roots are not
  • a conjugate pair.
Worked example

Q7: the discriminant

  • b and c are complex.
  • Expand by hand.
  1. a = 1, b = −(3 + i), c = 2 + 2i
  2. b² = (3 + i)² = 9 + 6i + i²expand
  3. = 8 + 6ii² = −1
  4. 4ac = 4(2 + 2i) = 8 + 8i
  5. b² − 4ac = 8 + 6i − 8 − 8i
  6. = −2i
Worked example

Q7: the square root of −2i

  • Let it be x + iy.
  • Equate both parts.
  1. (x + iy)² = −2i, x, y real
  2. (x² − y²) + 2xyi = −2iexpand, i² = −1
  3. x² − y² = 0real parts
  4. 2xy = −2, so xy = −1imaginary parts
  5. y = x: x² = −1x is real
  6. y = −x: x² = 1, x = ±1
  7. √(−2i) = ±(1 − i)
  8. (1 − i)² = 1 − 2i + i² = −2i
Worked example

Q7: the roots, and why

  • Quadratic formula.
  • Pairs need real b and c.
  1. b² − 4ac = −2i = (1 − i)²
  2. z = ((3 + i) ± (1 − i))/2quadratic formula
  3. z = (3 + i + 1 − i)/2 = 2
  4. z = (3 + i − 1 + i)/2 = 1 + i
  5. z = 2 or z = 1 + i
  6. sum 3 + i, product 2 + 2i
  7. b and c are not realso no conjugate pair
Summary

The method that solved all seven

  • Substitute, pair the roots, compare coefficients, and check.
StepHow
Remainderp(a) when ÷ (z − a)
Factorp(a) = 0
Real coeffsroots in conjugate pairs
Pair a ± biz² − (sum)z + product
Other factorcompare coefficients
Complex b, cformula; no pairs

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