← Complex numbers: past papers

AS 91577 · 2024 paper · a quotient in terms of k, equal roots, a modulus from a quotient, a complex cubic, and a circle locus

Question 2(d) Merit

Practises Polynomial equations with complex roots

Five separate parts: a quotient written as a + bi, a quadratic with equal roots, the modulus of an unknown found from a quotient, a cubic with one complex root given, and a locus that is a circle.

One solution of the equation 2z3 + dz2 + 140z − 200 = 0 is z = 6 − 2i. If d is real, find the value of d and the other two solutions of the equation.

The 2024 paper, page 7 ↗ Marking schedule, page 4 ↗

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 = 6 + 2i.

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

    (z − 6 + 2i)(z − 6 − 2i) = (z − 6)2 − (2i)2 = z2 − 12z + 36 + 4 = z2 − 12z + 40

  3. 3 The cubic is that quadratic times (Az + B). Match the z3 terms and the constant terms.

    z3: A = 2. Constant: 40B = −200, so B = −5. The cubic is (2z − 5)(z2 − 12z + 40).

  4. 4 Check with the z term, then read off d from the z2 term.

    z term: 2 × 40 + (−5)(−12) = 80 + 60 = 140 ✓
    z2 term: 2 × (−12) + (−5) = −29, so d = −29.

  5. 5 What is the third root?

    2z − 5 = 0, so z = 5/2.
Answer d = −29, and the other two solutions are z = 6 + 2i and z = 5/2.

What the marker wanted

  • Achieved The value of d, or the other two solutions.
  • Merit The value of d and both other solutions, found by algebra.

In our words, from NZQA’s 2024 marking schedule.

How did it go?

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