The Tiling Collective — Teacher Answer Booklet

🔐 Student HTML 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 as complex numbers, and draw the square on an Argand diagram."

Task 1aFind all four vertices geometrically. Draw the square.

All vertices lie on the unit circle (\(|z|=1\)), equally spaced by \(\dfrac{2\pi}{4}=\dfrac{\pi}{2}\). Starting from argument \(0\):

z_1=1 \qquad z_2=i \qquad z_3=-1 \qquad z_4=-i

Square on the unit circle

Let groups find these from the diagram first. Space by 90° visually then convert to complex form. Do not mention equations yet.

Task 1bUse De Moivre's theorem to raise each vertex to the power 4. What do all four results have in common? Write a single equation all four vertices satisfy.

Hint: De Moivre's theorem states \(\bigl[r(\cos\theta+i\sin\theta)\bigr]^n = r^n(\cos n\theta+i\sin n\theta)\). Apply it with \(r=1\) and \(n=4\) to each vertex.

Each vertex has \(r=1\). Multiplying the argument by 4:

z_1^4=\cos(4\cdot 0)+i\sin(4\cdot 0)=1 \qquad z_2^4=\cos\!\left(4\cdot\tfrac{\pi}{2}\right)+i\sin\!\left(4\cdot\tfrac{\pi}{2}\right)=\cos 2\pi+i\sin 2\pi=1\] \[z_3^4=\cos 4\pi+i\sin 4\pi=1 \qquad z_4^4=\cos 6\pi+i\sin 6\pi=1

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

💡 Key insight — establish this now

The vertices of a regular \(n\)-gon inscribed in the unit circle with one vertex at \((1,0)\) are exactly the solutions of \(z^n=1\). De Moivre's theorem connects the geometry (equal spacing) to the algebra (a single equation). This is the foundation for every phase that follows.

Why does \(2\pi k\) always vanish? Because \(\cos\) and \(\sin\) are \(2\pi\)-periodic. That is also why there are exactly 4 distinct roots — \(k=4\) reproduces \(k=0\).
Unlock code: VERTEX

The Rotated Triangle

HL 1.14 · Building the geometric formula · Discovering \(z^3=-1\)

🗣 Phase 2 Prompt — Read Aloud

"A second tile is an equilateral triangle, also on the unit circle. One vertex sits at argument \(\dfrac{\pi}{3}\). Find all three vertices. Draw the triangle."

Task 2aMark all three vertices on your Argand diagram using only geometric reasoning. How did you decide where to place them?

Three equally spaced vertices means the angular gap is \(\dfrac{2\pi}{3}\). Starting from \(\alpha=\dfrac{\pi}{3}\), add \(\dfrac{2\pi}{3}\) each time:

\text{Vertex 1: }\frac{\pi}{3} \qquad \text{Vertex 2: }\frac{\pi}{3}+\frac{2\pi}{3}=\pi \qquad \text{Vertex 3: }\frac{\pi}{3}+\frac{4\pi}{3}=\frac{5\pi}{3}

Rotated equilateral triangle, spacing \(=2\pi/3\)

Boards only — no algebra yet. If stuck: "If you share one full rotation equally among 3 vertices, how much does each get?"

Task 2bWrite each vertex in polar form. Then write a formula for the argument of the \(k\)-th vertex \((k=0,1,2)\).

Each vertex has \(r=1\):

z_1=\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}=\frac{1}{2}+\frac{\sqrt{3}}{2}\,i \qquad z_2=\cos\pi+i\sin\pi=-1 \qquad z_3=\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

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

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

Task 2cUse De Moivre's theorem to raise each vertex to the power 3. What single equation do all three vertices satisfy?

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

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

"Why does \(2\pi k\) disappear?" Periodicity of \(\cos\) and \(\sin\). Also why there are exactly 3 roots: \(k=3\) gives the same point as \(k=0\).

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."

Task 3aMark all five vertices using the geometric formula. State the angular spacing and the modulus of each vertex.

Five equally spaced vertices, angular gap \(=\dfrac{2\pi}{5}\). Starting at \(\alpha=\dfrac{\pi}{5}\), modulus \(r=2\):

\begin{array}{lll} k=0: & \arg=\dfrac{\pi}{5} & z_1=2\!\left(\cos\dfrac{\pi}{5}+i\sin\dfrac{\pi}{5}\right)\approx 1.618+1.176\,i\\[9pt] k=1: & \arg=\dfrac{3\pi}{5} & z_2=2\!\left(\cos\dfrac{3\pi}{5}+i\sin\dfrac{3\pi}{5}\right)\approx -0.618+1.902\,i\\[9pt] k=2: & \arg=\pi & z_3=2(\cos\pi+i\sin\pi)=-2\\[9pt] k=3: & \arg=\dfrac{7\pi}{5} & z_4=2\!\left(\cos\dfrac{7\pi}{5}+i\sin\dfrac{7\pi}{5}\right)\approx -0.618-1.902\,i\\[9pt] k=4: & \arg=\dfrac{9\pi}{5} & z_5=2\!\left(\cos\dfrac{9\pi}{5}+i\sin\dfrac{9\pi}{5}\right)\approx 1.618-1.176\,i \end{array}

Pentagon, \(r=2\), spacing \(=2\pi/5\)

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: "How is this diagram different from Phase 2? How is it the same?"

Task 3bUse De Moivre's theorem to raise \(z_1\) to the power 5. What single equation do all five vertices satisfy?

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

Same holds for any \(z_k\) since \(\cos(\pi+2\pi k)=-1\) for all integers \(k\).

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

Watch for the modulus. Students writing \(z^5=-1\) forgot \(2^5=32\). Prompt: "What does De Moivre's say happens to the modulus when you raise to the 5th power?"
Unlock code: SPACING

Reverse-Engineer the Hexagon

HL 1.14 · Running De Moivre's backwards · Synthesis

🗣 Phase 4 Prompt — Read Aloud

"The studio found an old tile: a regular hexagon. Six vertices equally spaced on a circle. One vertex is labelled \(-\sqrt{3}+i\). Nobody recorded the equation. Figure out what equation \(z^n=w\) generated these vertices. Find \(n\) and \(w\), and verify."

Task 4aFrom the diagram: find \(n\), the circle radius, and the argument of the given vertex.

Count vertices: \(n=6\). Modulus: \(|-\sqrt{3}+i|=\sqrt{3+1}=2\). Argument (second quadrant):

\text{Reference angle}=\arctan\!\left(\frac{1}{\sqrt{3}}\right)=\frac{\pi}{6} \;\Rightarrow\; \arg(-\sqrt{3}+i)=\pi-\frac{\pi}{6}=\frac{5\pi}{6}

\(n=6\), radius \(=2\), given vertex at argument \(\dfrac{5\pi}{6}\)

Common error: stopping at \(\pi/6\) without adjusting for the quadrant.

Task 4bUse the geometric formula to find all six vertices in exact rectangular form. Draw the hexagon.

From Task 4a: \(\alpha=\dfrac{5\pi}{6}\), spacing \(=\dfrac{2\pi}{6}=\dfrac{\pi}{3}\), modulus \(=2\). Apply \(z_k=2\!\left(\cos\!\left(\dfrac{5\pi}{6}+\dfrac{\pi k}{3}\right)+i\sin\!\left(\dfrac{5\pi}{6}+\dfrac{\pi k}{3}\right)\right)\), converting to exact rectangular form using \(\cos\tfrac{\pi}{6}=\tfrac{\sqrt{3}}{2}\), \(\sin\tfrac{\pi}{6}=\tfrac{1}{2}\):

\begin{array}{llll} k=0: & \arg=\dfrac{5\pi}{6} & \cos\dfrac{5\pi}{6}=-\dfrac{\sqrt{3}}{2},\;\sin\dfrac{5\pi}{6}=\dfrac{1}{2} & z_1=-\sqrt{3}+i \;\checkmark\\[10pt] k=1: & \arg=\dfrac{7\pi}{6} & \cos\dfrac{7\pi}{6}=-\dfrac{\sqrt{3}}{2},\;\sin\dfrac{7\pi}{6}=-\dfrac{1}{2} & z_2=-\sqrt{3}-i\\[10pt] k=2: & \arg=\dfrac{3\pi}{2} & \cos\dfrac{3\pi}{2}=0,\;\sin\dfrac{3\pi}{2}=-1 & z_3=-2i\\[10pt] k=3: & \arg=\dfrac{11\pi}{6} & \cos\dfrac{11\pi}{6}=\dfrac{\sqrt{3}}{2},\;\sin\dfrac{11\pi}{6}=-\dfrac{1}{2} & z_4=\sqrt{3}-i\\[10pt] k=4: & \arg=\dfrac{13\pi}{6}\equiv\dfrac{\pi}{6} & \cos\dfrac{\pi}{6}=\dfrac{\sqrt{3}}{2},\;\sin\dfrac{\pi}{6}=\dfrac{1}{2} & z_5=\sqrt{3}+i\\[10pt] k=5: & \arg=\dfrac{15\pi}{6}\equiv\dfrac{\pi}{2} & \cos\dfrac{\pi}{2}=0,\;\sin\dfrac{\pi}{2}=1 & z_6=2i \end{array}

Hexagon \(z^6=-64\), starting from \(-\sqrt{3}+i\)

For \(k=4,5\): reduce arguments \(>\!2\pi\) before evaluating trig. E.g. \(\tfrac{13\pi}{6}=2\pi+\tfrac{\pi}{6}\equiv\tfrac{\pi}{6}\).

Rectangular form is required here for graphing. All six answers use only \(\pm\sqrt{3}\), \(\pm 1\), \(\pm 2\) — push groups to derive these exactly, not as decimals. The six points are \((\pm\sqrt{3},\pm 1)\), \((0,\pm 2)\) — a beautiful symmetric pattern that rewards exact working.

Task 4cUse De Moivre's theorem to raise \(-\sqrt{3}+i\) to the power 6. State the equation \(z^6=w\) and verify.

\left(-\sqrt{3}+i\right)^6=2^6\!\left(\cos\frac{6\times 5\pi}{6}+i\sin\frac{6\times 5\pi}{6}\right)=64(\cos 5\pi+i\sin 5\pi)

Reduce: \(5\pi\equiv\pi\pmod{2\pi}\):

w=64(\cos\pi+i\sin\pi)=-64

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

Note on labelling: the standard formula \(2\cdot\mathrm{cis}\!\left(\dfrac{\pi+2\pi k}{6}\right)\) starts at \(k=0\Rightarrow\arg=\dfrac{\pi}{6}\) (vertex \(\sqrt{3}+i\)), while the geometric formula started at \(\arg=\dfrac{5\pi}{6}\). Both produce the same six points — only the \(k\)-labelling differs.

💡 Important subtlety

The roots form a set, not an ordered sequence. Any root can be \(k=0\). Standard and geometric formulas describe the same set with different starting labels.

Unlock code: FORMULA
Most common error: leaving \(64(\cos 5\pi+i\sin 5\pi)\) without reducing. Ask: "What is \(\cos 5\pi\) exactly?"

Task 4d Write the general formula for all \(n\) solutions of \(z^n=w\)Stretch

Write \(w=|w|(\cos\theta+i\sin\theta)\) where \(\theta=\arg(w)\). The \(n\) roots:

\boxed{z_k=\sqrt[n]{|w|}\left(\cos\frac{\theta+2\pi k}{n}+i\sin\frac{\theta+2\pi k}{n}\right),\quad k=0,1,\ldots,n-1}

Equivalently, separating starting angle from spacing:

z_k=|w|^{1/n}\,\mathrm{cis}\!\left(\frac{\theta}{n}+\frac{2\pi k}{n}\right)

\(|w|^{1/n}\) — radius of the circle all roots lie on (Phase 3)
\(\theta/n\) — starting angle of the first root (Phase 4)
\(2\pi k/n\) — equal angular spacing between roots (Phases 2 & 3)

Every element of this formula has a geometric meaning discovered in a previous phase. Let the group that reaches this present it last as the closing statement of the activity.