AS 91577 · 2024 paper · a factor, a power in polar form, a surd equation, a locus gradient, and a line a quotient never meets

Question 1(e) Merit to Excellence

Practises Equations, arguments, and comparing components

Five separate parts: a factor of a cubic, a power of a number in polar form, an equation with two square roots, the gradient of a straight-line locus, and a proof that a quotient never lies on y = x.

Consider the complex numbers u = 2 + 3ki and v = 4 + 5ki, where k is a real constant.

Show that the complex number w = u/v will not lie on the line y = x in the Argand diagram, for any value of k.

The 2024 paper, page 4 ↗ Marking schedule, page 2 ↗

Work it out

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

1 Write w as a + bi: multiply top and bottom by the conjugate of 4 + 5ki.
w = ((2 + 3ki)(4 − 5ki))/((4 + 5ki)(4 − 5ki)) = (8 − 10ki + 12ki − 15k²i²)/(16 + 25k²) = ((8 + 15k²) + 2ki)/(16 + 25k²)
2 On the line y = x, what is true of the real and imaginary parts?
They are equal. The denominators match, so that would need 8 + 15k² = 2k, which is 15k² − 2k + 8 = 0.
3 Is there a real k that solves it? Look at the discriminant.
b² − 4ac = (−2)² − 4(15)(8) = 4 − 480 = −476
4 What does a negative discriminant say?
The quadratic has no real solutions, so no real k makes the two parts equal. w is never on the line y = x, whatever k is.

Answer w on y = x would need 15k² − 2k + 8 = 0, whose discriminant is −476 < 0. No real k does it, so w never lies on y = x.

What the marker wanted

Achieved w over a real denominator: ((8 + 15k²) + 2ki)/(16 + 25k²).
Merit The quadratic 15k² − 2k + 8 = 0, from setting the real part equal to the imaginary part.
Excellence The conclusion, with a valid, clear reason (the negative discriminant): E8. The conclusion with the reason incomplete, or a full answer with one minor error: E7.

In our words, from NZQA’s 2024 marking schedule.

How did it go?

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