Arguments determine a triangle

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. marks

(ii) Hence or otherwise find ∣z∣|z|. marks

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

Total marks: 4

Solution

(i)

Angles from directions

arg⁡z=−π3\arg z=-\frac\pi3, ∣w∣=2|w|=2, arg⁡w=π4\arg w=\frac\pi4 and arg⁡(w−z)=π3\arg(w-z)=\frac\pi3. W,ZW,Z represent w,zw,z. Use an Argand diagram to show ∠OWZ=π12\angle OWZ=\frac\pi{12}.

Hint

  1. Find the directions of WO→\overrightarrow{WO} and WZ→\overrightarrow{WZ}.

  2. The angle at WW is the difference of those directions.

Triangle OWZ with directed arguments
Triangle OWZ with directed argumentsScroll across to see the full diagram.

Answer · (i)

∠OWZ=π/12\angle OWZ=\pi/12; ∠OZW=π/3\angle OZW=\pi/3.

(ii)

Sine rule

Hence or otherwise find ∣z∣|z|.

Explore this idea

Hint

  1. Find the angle at ZZ, and use the sine rule in triangle OWZOWZ.

Answer · (ii)

∣z∣=(6−2)/3≈0.598|z|=(\sqrt6-\sqrt2)/\sqrt3\approx0.598.

Explore the idea

Find the length that makes the direction work

ZZ slides along the ray at −60∘-60^\circ. Only one distance makes the direction from ZZ to WW exactly 60∘60^\circ.

Distance |z|0.4
Try an example
|z|
0.4
arg(w − z)
55.4°
needed
60°
w − z
1.214 + 1.761i
  • Moves as the slider changes; a faint path shows where it goes
Key idea

Three directions pin down all three angles of the triangle, and one given length (∣w∣=2|w|=2) then fixes the size.

Ideas to try
  1. Slide ∣z∣|z| and watch arg⁡(w−z)\arg(w-z) in the readouts. It crosses 60∘60^\circ at ∣z∣≈0.598|z|\approx0.598.
  2. At that point the arc at WW reads 15∘=π1215^\circ=\frac\pi{12}.
Takeaway

Directions give angles

The argument of a difference is the direction of a side. Angles of a triangle are differences of those directions.