AS 91577 · 2023 paper · a remainder, an unknown from a quotient, a surd equation, comparing parts, and a pair of conditions
Question 3(e) Merit to Excellence
Practises Equations, arguments, and comparing components
Five separate parts: a remainder, an unknown found from a quotient, an equation with two square roots, two unknowns found from a reciprocal, and a constant pinned down by two conditions at once.
If z + 2i = iz + k, where k is a real number, and w/z = 2 + 2i, where Im(w) = 8, find the value of k.
Work it out
Try it on paper first. Stuck? Take one hint at a time.
-
1 Find z from the first equation: collect the z terms.
z − iz = k − 2i, so z(1 − i) = k − 2i and z = (k − 2i)/(1 − i). -
2 Divide with the conjugate.
z = ((k − 2i)(1 + i))/2 = ((k + 2) + (k − 2)i)/2 -
3 Now w: multiply z by 2 + 2i.
w = ((k + 2) + (k − 2)i)/2 × 2(1 + i) = ((k + 2) + (k − 2)i)(1 + i) -
4 Expand, and sort into real and imaginary parts.
= (k + 2) + (k + 2)i + (k − 2)i − (k − 2) = 4 + 2ki -
5 Use Im(w) = 8, then check.
2k = 8, so k = 4. Then z = 3 + i: z + 2i = 3 + 3i and iz + 4 = 3i − 1 + 4 = 3 + 3i ✓, and w = (3 + i)(2 + 2i) = 4 + 8i ✓
What the marker wanted
- Achieved z on its own: z = (k − 2i)/(1 − i).
- Merit w written in terms of k with the fraction cleared (step 3).
- Excellence k = 4, clearly justified: E8. Reaching w = 4 + 2ki without finishing, or a full answer with one minor error: E7.
In our words, from NZQA’s 2023 marking schedule.
How did it go?
Got it moves it further out. Review brings it back tomorrow.