← Complex numbers: past papers

AS 91577 · 2023 paper · dividing in polar form, a modulus in k, an argument, cube roots, and a tangent to a circle

Question 2(d) Merit

Practises Roots of complex numbers

Five separate parts: a quotient of two numbers in polar form, the modulus of a difference, an argument found from an equation, the cube roots of a negative number, and a line that touches a circular locus.

Solve the equation z3 + 64m12 = 0, where m is a real constant. Write your solution(s) in polar form, in terms of m.

The 2023 paper, page 6 ↗ Marking schedule, page 3 ↗

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. 1 Rearrange to z3 = …. Where is that number on an Argand diagram?

    z3 = −64m12. m12 is never negative, so (for m ≠ 0) this is on the negative real axis.

  2. 2 Write it in polar form.

    z3 = 64m12 cis(π)

  3. 3 The cube root of the modulus?

    ∛64m12 = 4m4, since 43 = 64 and (m4)3 = m12.

  4. 4 Three roots: add 2π n to the argument and divide by 3, for n = 0, 1 and 2.

    θ = (π + 2π n)/3: π/3, π, and 5π/3, which is −π/3 in the range −π < θ ≤ π.

  5. 5 Check the easiest one.

    4m4 cis(π) = −4m4, and (−4m4)3 = −64m12 ✓
Answer z = 4m4 cis(π/3), z = 4m4 cis(π) and z = 4m4 cis(−π/3).

What the marker wanted

  • Achieved Any one of the three solutions.
  • Merit All three, with the working that gets them. The schedule takes the third argument as 5π/3 or −π/3, and other equivalent forms.

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

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Common mistake

Starting from the wrong angle. NZQA’s 2023 report, in its general comments on the paper, says students need to get the argument right when they first write the equation in polar form, because every root after a wrong first angle is wrong too. A negative real number has argument π.

How did it go?

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