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

Question 1(e) Merit to Excellence

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.

The complex numbers u and v are u = 3 + i and v = 1 + 2i.

Determine the possible value(s) of the real constant k if |u/v + k| = √(k + 2).

The 2023 paper, page 4 ↗ Marking schedule, page 2 ↗

Work it out

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

1 Start with u/v on its own: multiply top and bottom by the conjugate of 1 + 2i.
((3 + i)(1 − 2i))/((1 + 2i)(1 − 2i)) = (3 − 6i + i − 2i²)/5 = (5 − 5i)/5 = 1 − i
2 Add k. Which part does it join?
k is real, so it joins the real part: u/v + k = (1 + k) − i.
3 Write out the modulus and set it equal to √(k + 2).
√((1 + k)² + 1) = √(k + 2)
4 Square both sides and solve.
k² + 2k + 2 = k + 2, so k² + k = 0, k(k + 1) = 0, and k = 0 or k = −1.
5 Do both work? Check each in the original.
k = 0: |1 − i| = √2 = √(0 + 2) ✓
k = −1: |−i| = 1 = √(−1 + 2) ✓
Both are valid.

Answer k = 0 or k = −1.

What the marker wanted

Achieved u/v written as 1 − i.
Merit The modulus written out: √((1 + k)² + 1) = √(k + 2).
Excellence Both k = 0 and k = −1, each clearly justified: E8. Leaving k = 0 out, or a full answer with one minor error: E7.

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

Common mistake

Throwing away k = 0. It can look as if zero does not count, but it satisfies the equation, and the schedule marks leaving it out as E7, one short of full marks.

How did it go?

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