← Complex numbers: past papers

AS 91577 · 2025 paper · a remainder, a quadratic in k, a discriminant proof, a cube root, and a hyperbola

Question 1(e) Merit to Excellence

Practises Loci of complex numbers

Five separate parts: a remainder, a quadratic solved in terms of a constant, a proof that a quadratic always has real roots, the cube roots of an imaginary number, and a locus that turns out to be a hyperbola.

The locus of a complex number z is described by |z − 5i| − |z + 5i| = 4.

Find the Cartesian equation of the locus of z, giving your answer in the form ay2 − bx2 = k, where a, b and k are constants.

The 2025 paper, page 4 ↗ Marking schedule, page 2 ↗

Work it out

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

  1. 1 Put z = x + yi. What is each modulus?

    z − 5i = x + (y − 5)i and z + 5i = x + (y + 5)i, so
    √(x2 + (y − 5)2) − √(x2 + (y + 5)2) = 4

  2. 2 Squaring it as it stands leaves a product of two roots. Move one root to the other side first.

    √(x2 + (y − 5)2) = 4 + √(x2 + (y + 5)2)

  3. 3 Square both sides. The right side is a bracket squared, so it has three terms.

    x2 + y2 − 10y + 25 = 16 + 8√(x2 + (y + 5)2) + x2 + y2 + 10y + 25
    The x2, y2 and 25 cancel: −20y − 16 = 8√(x2 + (y + 5)2), or −5y − 4 = 2√(x2 + (y + 5)2).

  4. 4 One root is left, on its own. Square again.

    25y2 + 40y + 16 = 4(x2 + y2 + 10y + 25) = 4x2 + 4y2 + 40y + 100
    21y2 − 4x2 = 84

  5. 5 Squaring can add points. Is all of that curve on the locus?

    No. In step 3 the right side is never negative, so −5y − 4 ≥ 0, and on this curve that leaves only its lower half, y ≤ −2. That fits the picture: those are the points 4 units nearer −5i than 5i. The question asks only for the equation.
Answer 21y2 − 4x2 = 84, so a = 21, b = 4 and k = 84. The locus is the lower half (y ≤ −2) of that hyperbola.

What the marker wanted

  • Achieved The first squaring done correctly, after z = x + yi and moving one root across (step 3).
  • Merit Both square roots gone: the second squaring, on the way to the equation.
  • Excellence The equation of the locus, clearly justified: E8. With one minor error: E7. Any fully simplified equivalent is accepted.

In our words, from NZQA’s 2025 marking schedule.

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Common mistake

Squaring each term on its own. NZQA’s 2025 report says many students tried to get rid of the two roots by squaring each of them separately, but (√A − √B)2 is A − 2√AB + B, not A − B.

How did it go?

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