← Complex numbers: past papers

AS 91577 · 2023 paper · a squared surd, no real roots, a real part of zero, a complex cubic, and a modulus equation

Question 1(d) Merit

Practises Polynomial equations with complex roots

Five separate parts: a bracket with a surd squared, a quadratic with no real roots, a proof from a real part of zero, a cubic with one complex root given, and a constant found from a modulus.

One solution of the equation z3 − 8z2 + 6z + d = 0 is z = 5 − i. If d is real, find the value of d and the other two solutions of the equation.

The 2023 paper, page 3 ↗ Marking schedule, page 1 ↗

Work it out

Try it on paper first. Stuck? Take one hint at a time.

  1. 1 d is real, so what is a second root?

    The conjugate, z = 5 + i.

  2. 2 Multiply the two factors to get a real quadratic factor.

    (z − 5 + i)(z − 5 − i) = (z − 5)2 − i2 = z2 − 10z + 26

  3. 3 The cubic is that quadratic times (z + B). Which term pins B down?

    The z2 term: B − 10 = −8, so B = 2. Check with the z term: 26 − 10B = 26 − 20 = 6 ✓

  4. 4 Now read off d, and the third root.

    Constant term: d = 26 × 2 = 52. The factor z + 2 gives z = −2.
Answer d = 52, and the other two solutions are z = 5 + i and z = −2.

What the marker wanted

  • Achieved The other two solutions, or d.
  • Merit The other two solutions and d.

In our words, from NZQA’s 2023 marking schedule.

How did it go?

Got it moves it further out. Review brings it back tomorrow.