Solving Geometric Problems
The roots of unity turn regular-polygon geometry into algebra — distances, sums of position vectors and areas can all be found without trigonometry.
Remember from the previous lesson: if $\omega=e^{\frac{2\pi}{n}i}$ is a primitive $n$th root of unity, then $1+\omega+\omega^2+\cdots+\omega^{n-1}=0$. This lesson uses that fact geometrically.
Solving geometric problems
Because the $n$th roots of unity are evenly spaced points on the unit circle, they're a powerful tool for problems about regular polygons: distances between vertices, sums of vectors to vertices, and areas can often be computed algebraically using the roots-sum-to-zero property and the geometric series formula, instead of trigonometry alone.
Since $729=3^6$, the vertices are $z_k=3e^{\frac{2k\pi}6 i}=3\omega^k$ for $k=0,\ldots,5$, where $\omega=e^{\frac\pi3 i}$ is a 6th root of unity.
Each vertex has $|z_k|=3$ (all vertices lie on a circle of radius $3$, as expected for a regular hexagon centred at the origin), so the distance from the origin to each vertex is $3$, and the sum of squares of these distances is $6\times3^2=54$.
The sum of the position vectors is $\sum_{k=0}^5z_k=3\sum_{k=0}^5\omega^k=3(1+\omega+\omega^2+\omega^3+\omega^4+\omega^5)=3\times0=0$, using the roots-of-unity sum property, since $\omega$ is a primitive 6th root of unity.
The roots-sum-to-zero property applies specifically to the roots of $z^n=1$. For a general equation $z^n=w$ with $w\neq1$, the roots still sum to zero whenever the equation has no $z^{n-1}$ term (by the standard sum-of-roots result for polynomials), which is automatically true here since $z^n-w=0$ has no $z^{n-1}$ term — so in fact the roots of $z^n=w$ always sum to zero too, for any $w$ and any $n\geq2$.
Exam-style questions
Question 1 6 marks
The vertices of a regular hexagon, centred at the origin, are represented by the six 6th roots of $z^6=64$. One vertex is at $z=2$. (a) Write down the other five vertices in the form $x+iy$. (b) Show that the sum of the position vectors of all six vertices is zero.
Since $64=2^6$, the vertices are $z_k=2e^{\frac{k\pi}3 i}$ for $k=0,\ldots,5$, i.e. $2\omega^k$ where $\omega=e^{\frac\pi3 i}$. (M1)
$z_0=2$, $z_1=2\left(\cos\dfrac\pi3+i\sin\dfrac\pi3\right)=1+\sqrt3i$, $z_2=-1+\sqrt3i$, $z_3=-2$, $z_4=-1-\sqrt3i$, $z_5=1-\sqrt3i$. (M1, A1, A1)
Sum $=2(1+\omega+\omega^2+\omega^3+\omega^4+\omega^5)=2\times0=0$ by the roots-of-unity sum property, since $\omega$ is a primitive 6th root of unity. (M1, A1)
Question 2 7 marks
The vertices of a square, centred at the origin, are represented by the four 4th roots of $z^4=-64$. Find the exact area of the square.
$-64=64e^{i\pi}$, or more generally $64e^{i(\pi+2k\pi)}$. (M1)
$\rho^4=64\Rightarrow\rho=64^{1/4}=2\sqrt2$; $4\phi=\pi+2k\pi\Rightarrow\phi=\dfrac\pi4+\dfrac{k\pi}2$, giving $\phi=\dfrac\pi4,\dfrac{3\pi}4,-\dfrac\pi4,-\dfrac{3\pi}4$. (M1, A1)
Vertices: $2+2i,\;-2+2i,\;-2-2i,\;2-2i$. (A1)
All four vertices lie at radius $2\sqrt2$ from the origin, spaced $\dfrac\pi2$ apart, so they form a square with diagonal $2\times2\sqrt2=4\sqrt2$. Side length $=\dfrac{4\sqrt2}{\sqrt2}=4$, so the area is $4^2=16$. (M1, A1)