The Tiling Collective — Teacher Answers

🔐 Student Activity Unlock Codes (case-insensitive)

Phase 1 → 2

VERTEX

Phase 2 → 3

POLYGON

Phase 3 → 4

SPACING

Phase 4 → Done

FORMULA

🎨 Oral Launch — Read Aloud to Class

"A design studio creates geometric tile patterns. Every pattern is built from a regular polygon whose vertices are equally spaced around a circle centred at the origin of the complex plane. Each vertex is a complex number. Your job today: find those numbers, understand why they always come out the way they do, and reverse-engineer a pattern you're given."

The Square Tile

HL 1.12–1.13 · Polar form · Discovering \(z^4=1\) via De Moivre's theorem

🗣 Phase 1 Prompt — Read Aloud

"The studio's first tile is a square. Its four vertices sit equally spaced on a circle of radius 1, centred at the origin. One vertex is at \((1,\,0)\). Find all four vertices and draw the square on your whiteboard."

Task 1a Find all four vertices. Draw the square.

Equally spaced on the unit circle ⟹ angular spacing \(\dfrac{2\pi}{4}=\dfrac{\pi}{2}\). Starting at argument 0:

z_0=1 \qquad z_1=\cos\tfrac{\pi}{2}+i\sin\tfrac{\pi}{2}=i \qquad z_2=\cos\pi+i\sin\pi=-1 \qquad z_3=\cos\tfrac{3\pi}{2}+i\sin\tfrac{3\pi}{2}=-i

Vertices: \(1,\;i,\;-1,\;-i\)

Most groups will place these correctly by symmetry. Push them to state why the spacing is \(\pi/2\) before moving on.

Task 1b Raise each vertex to the power 4. Write the equation.

Applying De Moivre's with \(r=1,\;n=4\):

1^4=1 \qquad i^4=(\cos\tfrac{\pi}{2}+i\sin\tfrac{\pi}{2})^4=\cos2\pi+i\sin2\pi=1\] \[(-1)^4=(\cos\pi+i\sin\pi)^4=\cos4\pi+i\sin4\pi=1 \qquad (-i)^4=1

All four vertices satisfy \(z^4=1\).

Key insight

The four vertices are the four 4th roots of unity. This is the core equation students will generalise across all four phases.

Unlock code: VERTEX

The Rotated Triangle

HL 1.13–1.14 · Building the general spacing formula · \(z^3=w\)

🗣 Phase 2 Prompt — Read Aloud

"The studio's next tile is an equilateral triangle. Same unit circle, but this time one vertex is at argument \(\pi/3\) — not at \((1,0)\). Find all three vertices, write a formula for them, and find the equation \(z^3=w\)."

Task 2a Mark all three vertices geometrically. Draw the triangle.

Angular spacing for an equilateral triangle: \(\dfrac{2\pi}{3}\). Starting at \(\alpha=\dfrac{\pi}{3}\):

\text{arg}(z_0)=\frac{\pi}{3} \qquad \text{arg}(z_1)=\frac{\pi}{3}+\frac{2\pi}{3}=\pi \qquad \text{arg}(z_2)=\frac{\pi}{3}+\frac{4\pi}{3}=\frac{5\pi}{3}

Groups should place vertices purely from geometric reasoning here — "three equally spaced points, first one at \(\pi/3\)." No formula needed yet.

Task 2b Write the vertices in polar form. Build the formula.

Each vertex has \(r=1\):

z_0=\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}=\frac{1}{2}+\frac{\sqrt{3}}{2}\,i \qquad z_1=\cos\pi+i\sin\pi=-1 \qquad z_2=\cos\frac{5\pi}{3}+i\sin\frac{5\pi}{3}=\frac{1}{2}-\frac{\sqrt{3}}{2}\,i

Pattern — start at \(\alpha=\tfrac{\pi}{3}\), add \(\tfrac{2\pi}{3}\) each step:

\text{arg of }k\text{-th vertex}=\frac{\pi}{3}+\frac{2\pi k}{3}, \quad k=0,1,2

General formula: \(\displaystyle z_k=\cos\!\left(\frac{\pi}{3}+\frac{2\pi k}{3}\right)+i\sin\!\left(\frac{\pi}{3}+\frac{2\pi k}{3}\right),\quad k=0,1,2\)

Make sure every group writes this formula explicitly before moving to 2c.

Task 2c Raise to the power 3. Find the equation.

Apply De Moivre's with \(r=1,\;n=3\) to \(z_k=\mathrm{cis}\!\left(\tfrac{\pi}{3}+\tfrac{2\pi k}{3}\right)\):

z_k^3=\cos\!\left(3\cdot\frac{\pi}{3}+3\cdot\frac{2\pi k}{3}\right)+i\sin\!\left(\cdots\right)=\cos(\pi+2\pi k)+i\sin(\pi+2\pi k)=-1

All three vertices satisfy \(z^3=-1\).

Key insight

The \(2\pi k\) term vanishes due to periodicity. This is why there are exactly 3 distinct roots: \(k=3\) returns to \(k=0\).

Unlock code: POLYGON

The Pentagon — Bigger Circle

HL 1.14 · Extending the geometric formula to \(r\neq 1\)

🗣 Phase 3 Prompt — Read Aloud

"The studio scales up. The next tile is a regular pentagon, but its vertices lie on a circle of radius 2. One vertex is at \(2\!\left(\cos\dfrac{\pi}{5}+i\sin\dfrac{\pi}{5}\right)\). Find all five vertices. Draw the pentagon. Find the equation \(z^5=w\)."

Task 3a Mark all five vertices. Draw the pentagon.

Angular spacing: \(\dfrac{2\pi}{5}\). Starting at \(\alpha=\dfrac{\pi}{5}\), all vertices have \(r=2\):

\text{arg}(z_k)=\frac{\pi}{5}+\frac{2\pi k}{5}, \quad k=0,1,2,3,4

All 5 vertices: modulus \(2\), arguments \(\dfrac{\pi}{5}+\dfrac{2\pi k}{5}\) for \(k=0,1,2,3,4\)

The angular formula is identical to Phase 2. Only \(r\) has changed. Ask groups to articulate this before continuing.

Task 3b Raise to the power 5. Find the equation.

Apply De Moivre's with \(r=2,\;n=5\) to \(z_1=2\,\mathrm{cis}\!\left(\tfrac{\pi}{5}\right)\):

z_1^5=2^5\!\left(\cos\!\left(5\cdot\frac{\pi}{5}\right)+i\sin\!\left(5\cdot\frac{\pi}{5}\right)\right)=32(\cos\pi+i\sin\pi)=32\cdot(-1)=-32

All five vertices satisfy \(z^5=-32\).

The modulus of \(w\) is \(32=2^5=r^n\). The radius of the circle is \(|w|^{1/n}=32^{1/5}=2\). ✓

Key insight

When \(r\neq 1\), the modulus of \(w\) is \(r^n\). This is the step most groups find surprising — the formula still works, just with the modulus scaled.

Unlock code: SPACING

Reverse-Engineer the Hexagon

HL 1.14 · Synthesis · Running De Moivre's backwards from a given vertex

🗣 Phase 4 Prompt — Read Aloud

"The studio found an old tile: a regular hexagon. One vertex is labelled \(-\sqrt{3}+i\). Using only this information, reconstruct the full hexagon and find the equation \(z^6=w\)."

Task 4a Find \(n\), the radius, and the argument of the given vertex.

It's a hexagon, so \(n=6\). Convert \(-\sqrt{3}+i\) to polar:

r=|-\sqrt{3}+i|=\sqrt{3+1}=2 \qquad \arg(-\sqrt{3}+i)=\pi-\arctan\!\left(\frac{1}{\sqrt{3}}\right)=\pi-\frac{\pi}{6}=\frac{5\pi}{6}

\(n=6\), radius \(=2\), \(\arg(z_0)=\dfrac{5\pi}{6}\)

Students often struggle with \(\arg\) in Q2. Encourage them to draw the point first and use the reference angle \(\pi/6\).

Task 4b Find all six vertices in exact rectangular form. Draw the hexagon.

Angular spacing \(\dfrac{2\pi}{6}=\dfrac{\pi}{3}\). Arguments: \(\dfrac{5\pi}{6}+\dfrac{\pi k}{3}\) for \(k=0,1,2,3,4,5\). All have \(r=2\).

\begin{align} z_0&=2\!\left(\cos\tfrac{5\pi}{6}+i\sin\tfrac{5\pi}{6}\right)=-\sqrt{3}+i\\ z_1&=2\!\left(\cos\tfrac{7\pi}{6}+i\sin\tfrac{7\pi}{6}\right)=-\sqrt{3}-i\\ z_2&=2\!\left(\cos\tfrac{3\pi}{2}+i\sin\tfrac{3\pi}{2}\right)=-2i\\ z_3&=2\!\left(\cos\tfrac{11\pi}{6}+i\sin\tfrac{11\pi}{6}\right)=\sqrt{3}-i\\ z_4&=2\!\left(\cos\tfrac{\pi}{6}+i\sin\tfrac{\pi}{6}\right)=\sqrt{3}+i\\ z_5&=2\!\left(\cos\tfrac{\pi}{2}+i\sin\tfrac{\pi}{2}\right)=2i \end{align}

Students must use exact trig values. Accept arguments written as \(\tfrac{5\pi}{6},\;\tfrac{7\pi}{6},\;\tfrac{3\pi}{2},\;\tfrac{11\pi}{6},\;\tfrac{\pi}{6},\;\tfrac{\pi}{2}\) (or equivalently with \(-\tfrac{\pi}{6}\) etc.).

Task 4c Find \(w\). State and verify the equation \(z^6=w\).

Raise \(z_0=-\sqrt{3}+i=2\,\mathrm{cis}\!\left(\tfrac{5\pi}{6}\right)\) to the power 6:

z_0^6=2^6\!\left(\cos\!\left(6\cdot\frac{5\pi}{6}\right)+i\sin\!\left(6\cdot\frac{5\pi}{6}\right)\right)=64(\cos5\pi+i\sin5\pi)=64(-1)=-64

The equation is \(z^6=-64\).

Verification: \(|w|=64=2^6=r^n\) ✓ \(\arg(w)=5\pi\equiv\pi\pmod{2\pi}\) ✓ (odd multiple of \(\pi\) gives \(-1\) direction).

Unlock code: FORMULA

Task 4d — Extension Write the general formula for all \(n\) roots.

\(\displaystyle z_k = |w|^{1/n}\left(\cos\!\left(\frac{\arg(w)}{n}+\frac{2\pi k}{n}\right)+i\sin\!\left(\frac{\arg(w)}{n}+\frac{2\pi k}{n}\right)\right),\quad k=0,1,\ldots,n-1\)

Geometric meaning of each part:

\(|w|^{1/n}\) — radius of the circle (modulus of each root)
\(\arg(w)/n\) — argument of the first root (\(k=0\))
\(2\pi k/n\) — equal angular spacing between consecutive roots
\(k=0,\ldots,n-1\) — exactly \(n\) distinct roots before the pattern repeats

📋 Consolidation — select 2–3 boards to debrief

Phase 1: Why does \(z^4=1\) have exactly 4 roots? (fundamental theorem of algebra)
Phase 2: Why does the starting angle \(\alpha\) not affect the spacing — only the position?
Phase 3: How does \(|w|^{1/n}\) give the radius? Connect to inverse of \(r^n\).
Phase 4: How did you decide the argument from the rectangular form? Argand diagram first.
General: Why are there exactly \(n\) roots and no more? Periodicity of \(\cos\) and \(\sin\).