← Complex numbers: past papers

AS 91577 · 2025 paper · two points on an Argand diagram, a fifth power, a modulus, comparing parts, and an argument

Question 2(c) Merit

Practises Arithmetic & the conjugate

Five separate parts: a combination of two points read off an Argand diagram, a power of a number in polar form, an equation with a modulus in it, two unknowns found by comparing real and imaginary parts, and an unknown found from an argument.

Solve the equation |5 − mi| = √5m2 for m, where m is a real number.

The 2025 paper, page 6 ↗ Marking schedule, page 3 ↗

Work it out

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

  1. 1 What is the modulus of 5 − mi?

    |5 − mi| = √(52 + (−m)2) = √(25 + m2)

  2. 2 Both sides are square roots. Square them.

    25 + m2 = 5m2

  3. 3 Solve for m. How many values are there?

    4m2 = 25, so m2 = 25/4 and m = ±5/2.

  4. 4 Does each one work in the original?

    Both do: each side is √(25 + 25/4) = 5√5/2. Only m2 appears, so the sign of m makes no difference.
Answer m = 5/2 or m = −5/2 (that is, ±2.5).

What the marker wanted

  • Achieved The equation without its roots, 25 + m2 = 5m2.
  • Merit Both values of m, the negative one as well.

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

How did it go?

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