← 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(b) Achieved

Practises Arithmetic & the conjugate

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.

If z = 1 + ki and w = 7 − ki, find |z − w|, giving your answer in terms of k.

The 2023 paper, page 5 ↗ Marking schedule, page 3 ↗

Work it out

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

  1. 1 Subtract the real parts, then the imaginary parts.

    z − w = (1 − 7) + (k − (−k))i = −6 + 2ki

  2. 2 Now the modulus.

    |z − w| = √((−6)2 + (2k)2) = √(36 + 4k2)

  3. 3 Can it be simplified?

    Take out the 4: √(4(9 + k2)) = 2√(9 + k2).
Answer |z − w| = √(36 + 4k2) = 2√(9 + k2)

What the marker wanted

  • Achieved The correct expression in terms of k.

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

How did it go?

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