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

Question 1(d) Merit

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.

Solve the equation z³ + 8m²⁷i = 0, where m is a positive real constant. Write your solution(s) in polar form, in terms of m.

The 2025 paper, page 3 ↗ Marking schedule, page 2 ↗

See it move: The Argand plane, and what a power does to a point

Work it out

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

1 Rearrange to z³ = …. Where is that number on an Argand diagram?
z³ = −8m²⁷i. Since m > 0, it is on the negative imaginary axis, straight down from the origin.
2 Write it in polar form: the modulus, then the argument.
The modulus is 8m²⁷. Pointing straight down, the argument is −π/2.
z³ = 8m²⁷ cis(−π/2)
3 Take the cube root of the modulus.
∛8m²⁷ = 2m⁹, since 2³ = 8 and (m⁹)³ = m²⁷.
4 A cubic has three roots. Add 2π n to the argument before dividing by 3, for n = 0, 1 and 2.
θ = (−π/2 + 2π n)/3
n = 0: −π/6. n = 1: π/2. n = 2: 7π/6, which is −5π/6 in the range −π < θ ≤ π.
5 Check the easiest one, 2m⁹ cis(π/2).
That is z = 2m⁹ i, and z³ = 8m²⁷i³ = −8m²⁷i, so z³ + 8m²⁷i = 0 ✓. The three roots sit 2π/3 apart on a circle of radius 2m⁹.

Answer z = 2m⁹ cis(−π/6), z = 2m⁹ cis(π/2) and z = 2m⁹ cis(−5π/6).

What the marker wanted

Achieved Any one of the three solutions.
Merit All three, with the working that gets them. The schedule takes the arguments as −π/6, π/2, −5π/6 or as 11π/6, π/2, 7π/6.

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

Common mistake

Starting from the wrong angle. NZQA’s 2025 report says students often missed that the argument here is negative and used +π/2, which makes every root after it wrong. A quick sketch of −8m²⁷i shows it points down.

How did it go?

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