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Original question & working

Arguments determine a triangle · source reference

[2020/DHS/Prelim/I/3 modified] · Question pp. 4 · Working pp. 19, 20

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The working and teaching notes are included in the interactive lesson.

The angle of w−z belongs to the vector ZW, not WZ.

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Full original question, working & notes

Question

arg⁡z=−π/3\arg z=-\pi/3, ∣w∣=2|w|=2, arg⁡w=π/4\arg w=\pi/4 and arg⁡(w−z)=π/3\arg(w-z)=\pi/3. Let W and Z represent w,zw,z respectively.

(i) Use an Argand diagram with reference to origin O to show ∠OWZ=π/12\angle OWZ=\pi/12 radians. [2]

(ii) Hence or otherwise find ∣z∣|z|. [2]

[2020/DHS/Prelim/I/3 modified]

Source working

Printed working

(i) Let W and Z represent w and z. ZW→\overrightarrow{ZW} represents w−zw-z.

∠OWZ=π/3−π/4=π/12(shown).\angle OWZ=\pi/3-\pi/4=\pi/12\quad\text{(shown)}.

(ii) The source states ∠OZW=π/3\angle OZW=\pi/3 “(angle of equilateral △\triangle)”.

2sin⁡∠OZW=OZsin⁡∠OWZ,2sin⁡(π/3)=OZsin⁡(π/12).\frac2{\sin\angle OZW}=\frac{OZ}{\sin\angle OWZ},\qquad\frac2{\sin(\pi/3)}=\frac{OZ}{\sin(\pi/12)}. OZ=0.59772,∣z∣=0.598(3 s.f.).OZ=0.59772,\qquad |z|=0.598\quad(3\text{ s.f.}).

Qualification

OWZ is not equilateral: its angle at W is π/12\pi/12. The angle at Z is nevertheless π/3\pi/3, because ZO has argument 2π/32\pi/3 and ZW has argument π/3\pi/3. The sine-rule answer remains correct.

Diagram labels

O, W, Z, x, y, length OW 22, origin angle π/4\pi/4, clockwise OZ angle π/3\pi/3, direction ZW π/3\pi/3 at the axis and at Z, with a horizontal auxiliary ray through Z.

Printed answer key

(ii) ∣z∣=0.598|z|=0.598 (3 s.f.).

Native diagrams

Triangle OWZ with directed arguments

Provenance

2025 TJC H2 Math (9758) JC2 MSM: COMPLEX NUMBERS. Question pages 4; solution pages 19, 20. These are original document page numbers, retained as text metadata.