NCEA Level 3 Calculus 91577 Complex free practice questions with answers

Question 1 (9 marks) [Arithmetic, polar form and quadratics]

(a) Write (−2 + 2i)(4 + 6i) ÷ (5 − 2i) in the form a + bi, where a and b are real numbers (exact fractions if needed).

(a) Answer:

• Evidence level: Achieved (multiply complex numbers, then divide using the conjugate).
• Numerator: (−2 + 2i)(4 + 6i) = −20 − 4i.
• Multiply top and bottom by the conjugate of the denominator, 5 + 2i: denominator = 5² + 2² = 29.
• Numerator × conjugate = (−20 − 4i)(5 + 2i) = −92 − 60i.
• Divide both parts by 29:

Dividing complex numbers: multiply numerator and denominator by the conjugate of the denominator so the denominator becomes the real number a² + b².

(b) Write z = 3 − 6i in polar form r cis θ, where r > 0 and −π < θ ≤ π. Give r exactly and θ in radians to 3 significant figures.

(b) Answer: z = √45 cis(−1.11)

• Evidence level: Achieved (find the modulus and the argument, taking care over the quadrant).
• Modulus r = √((3)² + (−6)²) = √45 = 6.708.
• The point (3, −6) is in the fourth quadrant, so use θ = arctan(|b/a|) and adjust: θ = −1.11 radians.
• z = √45 cis(−1.11).

Polar form: r = √(a² + b²); sketch the point to choose the correct quadrant for θ (your calculator's arctan only gives the first or fourth quadrant).

(c) Solve 2z² + 50 = 0 for z, giving the solutions in the form a + bi.

(c) Answer: z = 5i or z = −5i

• Evidence level: Achieved (use the quadratic formula when the discriminant is negative).
• Δ = b² − 4ac = 0 − 400 = −400.
• √Δ = √(−400) = 20i, because √(−n) = i√n.
• z = (0 ± 20i) ÷ 4.
• z = 5i or z = −5i.

A negative discriminant gives a conjugate pair: √(−n) = i√n, so the roots are (−b ± i√|Δ|) ÷ 2a.

−92

29 − 60i

29

−92

29 − 60i

29 .