Question 3(d) Merit

Five separate parts: a fraction of surds simplified, a complex number squared and plotted, an equation in a number and its conjugate, the fourth roots of a negative number, and a cube of x + 1/x.

Solve the equation z⁴ + 81k⁸ = 0, where k is a real constant. Give your solution(s) in polar form, in terms of k.

The 2024 paper, page 10 ↗ Marking schedule, page 5 ↗

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 z⁴ = …. Where is that number on an Argand diagram?

   z⁴ = −81k⁸. k⁸ is never negative, so (for k ≠ 0) this is on the negative real axis.

2. 2 Write it in polar form.

   The modulus is 81k⁸ and the argument is π: z⁴ = 81k⁸ cis(π).

3. 3 The fourth root of the modulus?

   ⁴√81k⁸ = 3k², since 3⁴ = 81 and (k²)⁴ = k⁸.

4. 4 Four roots: add 2π n to the argument, then divide by 4, for n = 0 to 3.

   θ = (π + 2π n)/4: π/4, 3π/4, 5π/4 and 7π/4, which are −3π/4 and −π/4 in the range −π < θ ≤ π.

5. 5 Check one.

   (3k² cisπ/4)⁴ = 81k⁸ cis(π) = −81k⁸ ✓. The four roots sit a quarter-turn apart on a circle of radius 3k². (If k = 0, the only solution is z = 0.)

What the marker wanted

• Achieved Any one of the four solutions.
• Merit All four. The schedule takes the arguments as π/4, 3π/4, 5π/4, 7π/4 or with the last two as −3π/4, −π/4.

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

Common mistake

How did it go?

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