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AS 91577 · 2023 paper · a squared surd, no real roots, a real part of zero, a complex cubic, and a modulus equation

Question 1(c) Merit

Practises Equations, arguments, and comparing components

Five separate parts: a bracket with a surd squared, a quadratic with no real roots, a proof from a real part of zero, a cubic with one complex root given, and a constant found from a modulus.

If z = p + qi and w = a + bi and Re(z/w) = 0, show that ap = −bq.

The 2023 paper, page 2 ↗ Marking schedule, page 1 ↗

Work it out

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

  1. 1 Write z/w as a + bi: multiply top and bottom by the conjugate of w.

    ((p + qi)(a − bi))/((a + bi)(a − bi)) = (ap − bpi + aqi − bqi2)/(a2 + b2) = ((ap + bq) + (aq − bp)i)/(a2 + b2)

  2. 2 Which part is the real part?

    Re(z/w) = (ap + bq)/(a2 + b2)

  3. 3 Set it to zero. When is a fraction zero?

    When its top is zero (the bottom, a2 + b2, is not zero, since w ≠ 0). So ap + bq = 0, which is ap = −bq, as required.
Answer Re(z/w) = (ap + bq)/(a2 + b2) = 0, so ap + bq = 0 and ap = −bq.

What the marker wanted

  • Achieved The quotient multiplied out over a2 + b2, with i2 replaced by −1.
  • Merit The proof finished: the real part set to zero, giving ap = −bq.

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

How did it go?

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