New Zealand · Scholarship

NCEA Scholarship Calculus free practice questions with answers

Read question 1 and its full worked answer from paper 1 of our 20-paper NCEA Scholarship Calculus set, then download the 2-page PDF. These are original practice questions, not official past papers. Page updated 3 November 2026.

Question 1

Question One (8 marks)

(a) Solve z³ = −8 and give all three solutions exactly, in the form a + bi.

[2 marks]

(b) Let ω = cos( Explain why 1 + ω + ω² + ω³ + ω⁴ = 0, and use it to show that cos( [3 marks]

(c) Hence find the exact value of cos( [3 marks]

2π

5 ) + i sin(2π

5 ).

2π

5 ) + cos(4π

5 ) = −1

2 .

2π

5 ). Show how you decide between the possible values.

Worked answer

Question 1 (8 marks) [Complex numbers and roots of unity]

(a) Solve z³ = −8 and give all three solutions exactly, in the form a + bi.

(a) Answer: z = −2, z = 1 + i√3, z = 1 − i√3

  • Entry step: worth marks in the 1-4 range of the 8-point question score.
  • Write −8 = 8 cis(π + 2kπ), so z = 2 cis( • k = 0: 1 + i√3; k = 1: −2; k = 2: 1 − i√3. [1 mark]

Roots of a complex number: take the root of the modulus and divide the argument (plus 2kπ) by n; the n roots are evenly spaced around a circle.

(b) Let ω = cos( cos( (b) Answer: cos( • Scholarship band (5-6): the part needs your own method and accurate algebra.

  • By de Moivre ω⁵ = cis(2π) = 1, so 0 = ω⁵ − 1 = (ω − 1)(1 + ω + ω² + ω³ + ω⁴). Since ω ≠ 1, the second factor is 0. [1 mark]
  • Take real parts: 1 + cos( • cos( + cos( Sums of roots of unity vanish: multiply out (ω − 1) times the geometric sum, then take real parts and use symmetry to pair the angles.

(c) Hence find the exact value of cos( (c) Answer: cos( • Outstanding band (7-8): the part needs insight and a complete, efficient argument.

  • Let c = cos( • c = (−2 ± √20) ÷ 8 = • belongs to cos( Two expressions for the same unknown: a double-angle identity turns the sum from the roots of unity into a quadratic you can solve exactly; use the quadrant to reject the wrong root. π + 2kπ

3

) for k = 0, 1, 2. [1 mark]

2π

5 ) + i sin(2π

5 ). Explain why 1 + ω + ω² + ω³ + ω⁴ = 0, and use it to show that

2π

5 ) + cos(4π

5 ) = −1

2 .

2π

5 ) + cos(4π

5 ) = −1

2

2π

5 ) + cos(4π

5 ) + cos(6π

5 ) + cos(8π

5 ) = 0. [1 mark]

6π

5 ) = cos(4π

5 ) and cos(8π

5 ) = cos(2π

5 ) (symmetry about the real axis), so 1 + 2[cos(2π

5 )

4π

5 )] = 0, giving the result. [1 mark]

2π

5 ). Show how you decide between the possible values.

2π

5 ) = √5 − 1

4

2π

5 ). Then cos(4π

5 ) = 2c² − 1 (double angle). Substituting into part (b): c + 2c² − 1 = −1

2 , so 4c² + 2c − 1 = 0. [1 mark]

−1 ± √5

4

. [1 mark]

2π

5 is in the first quadrant, so c > 0 and c = √5 − 1

4

≈ 0.309. (The other root is negative and 4π

5 ).) [1 mark]

Download the free sample: 3 questions and answers (PDF)

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Questions about this sample

What is in the NCEA Scholarship Calculus free sample?

The first 3 questions from paper 1 of the set, in the real exam layout, followed by the step-by-step worked answers. It is a 2-page PDF; this page shows question 1 with its answer.

How many full NCEA Scholarship Calculus papers are there?

20 original papers for New Zealand Year 13 students preparing for NZ Scholarship Calculus (standard 93202). Each is four questions of 8 points (32 in all) in three hours, on complex numbers, equations, differentiation, integration, differential equations and series, with a worked solution for every part showing the score band, with a worked solution for every question. Papers start at A$11.99 and download instantly as a PDF.

Are these official past papers?

No. The sample is taken from our own original practice paper and is not an official past paper or affiliated with any exam authority.