Complex numbers — Question 16
Original QCE Vault practice · 6 marks
Q16 · Practice questionComplex familiar6 marks
QUESTION 16 (6 marks)
Let ω be a non-real cube root of unity. a)Prove that 1+ω+ω2=0. b)Without converting ω to Cartesian form at the start, determine the exact value of (2+ω)6. WORKED SOLUTION
Practice marking scheme
6 marksANSWER(a) 1+ω+ω2=0; (b) −27. Worked solution
(a) Since ω3=1 and ω=1, 0=ω3−1=(ω−1)(ω2+ω+1). Dividing by ω−1 gives 1+ω+ω2=0. (b) Use ω2=−1−ω first: (2+ω)2=4+4ω+ω2=3(1+ω). Also 1+ω=−ω2. Hence (2+ω)6=27(1+ω)3=27(−ω2)3=−27ω6=−27. Factors ω3−1 and uses ω=1. Uses the identity to simplify (2+ω)2. Raises to the third power using the cube-root relation.
Obtains the exact result −27. Practice question aligned to the current QCAA syllabus; review the worked solution and marking criteria.
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