Exercises — Roots of unity — cube roots, nth roots, geometric interpretation

A 6th root of unity is any $z$ with $z^6=1$, i.e. $z=e^{i2\pi k/6}=e^{i\pi k/3}$ for $k=0,\dots ,5$. The six allowed angles are multiples of $\frac{\pi }{3}=60^{\circ }$: $0,60,120,180,240,300$ degrees.

• $1=e^{i0}$ → angle $0$. Yes.
• $i=e^{i\pi /2}$ → angle $90^{\circ }$, not a multiple of $60^{\circ }$. No.
• $e^{i\pi /3}$ → angle $60^{\circ }$. Yes (this is $\omega$).
• $e^{i2\pi /5}$ → angle $72^{\circ }$ (a 5th root). No.
• $-1=e^{i\pi }$ → angle $180^{\circ }$. Yes.

Answer: $1,e^{i\pi /3},-1$ are 6th roots of unity.

For $n=3$ the spacing is $\frac{2\pi }{3}=120^{\circ }$, starting at angle $0$. So the arguments are $0,\frac{2\pi }{3}(120^{\circ }),\frac{4\pi }{3}(240^{\circ }).$ Three equal-length spokes $120^{\circ }$ apart form an equilateral triangle inscribed in the unit circle (a regular polygon with one vertex at $1$).

What we do: apply $z_k=e^{i2\pi k/5}$ for $k=0,1,2,3,4$. Why: $z^5=1$ matches modulus ($r=1$) and argument ($5\theta =2\pi k$), so $\theta =\frac{2\pi k}{5}=72^{\circ }k$. $z_0=e^{i0},z_1=e^{i2\pi /5},z_2=e^{i4\pi /5},z_3=e^{i6\pi /5},z_4=e^{i8\pi /5}.$ Sum: this is a Geometric Series with ratio $\omega =e^{i2\pi /5}\ne 1$: $S=\frac{{\omega }^5-1}{\omega -1}=\frac{1-1}{\omega -1}=0.$ Why the numerator is $0$: ${\omega }^5=1$ by definition of a 5th root. The five vectors are symmetric around the circle, so they cancel at the origin. ✓

What we do: write the right side in polar form. $-16$ sits on the negative real axis, so its angle is $\pi$ and its modulus is $16$: $-16=16e^{i\pi }=16e^{i(\pi +2\pi k)}.$ Why include $+2\pi k$: angles repeat every full turn, and we need all four roots. Match modulus: $r^4=16\Rightarrow r=2$. Match argument: $4\theta =\pi +2\pi k\Rightarrow \theta =\frac{\pi }{4}+\frac{\pi k}{2}$, $k=0,1,2,3$. $z=2e^{i\pi /4},2e^{i3\pi /4},2e^{i5\pi /4},2e^{i7\pi /4}.$ In rectangular form, using $e^{i\pi /4}=\frac{\sqrt{2}}{2}(1+i)$: $z=\sqrt{2}(1+i),\sqrt{2}(-1+i),\sqrt{2}(-1-i),\sqrt{2}(1-i).$ These are the corners of a square of radius $2$, rotated $45^{\circ }$ from the axes (see figure).

What we do: group to use $1+\omega +{\omega }^2=0$. Why: any expression in $\omega$ collapses once you spot that identity. $2+3\omega +3{\omega }^2=2+3(\omega +{\omega }^2).$ From $1+\omega +{\omega }^2=0$ we get $\omega +{\omega }^2=-1$. So $2+3(-1)=-1.$ Answer: $-1$.

What we do: recognise the Geometric Series and sum it. Why: the first-term $1$ plus a chain of powers of $z$ is exactly a geometric sum with ratio $z$. $1+z+\cdots +z^6=\frac{z^7-1}{z-1}.$ Since $z^7=1$ the numerator is $0$, and since $z\ne 1$ the denominator is nonzero, so the whole sum is $0$. Second part: subtract the first term: $z+z^2+\cdots +z^6=(1+z+\cdots +z^6)-1=0-1=-1.$ Answer: $-1$.

Rectangular form: $\omega =\cos 120^{\circ }+i\sin 120^{\circ }=-\frac{1}{2}+\frac{\sqrt{3}}{2}i.$ Conjugation flips the sign of the imaginary part (a mirror across the real axis on the Argand Diagram): $\overline{\omega }=-\frac{1}{2}-\frac{\sqrt{3}}{2}i.$ Via the identity: ${\omega }^2=e^{i4\pi /3}=\cos 240^{\circ }+i\sin 240^{\circ }=-\frac{1}{2}-\frac{\sqrt{3}}{2}i.$ They match. Why they must: $\overline{e^{i\theta }}=e^{-i\theta }$, and $e^{-i2\pi /3}=e^{i(2\pi -2\pi /3)}=e^{i4\pi /3}={\omega }^2$. Conjugating rotates by $-120^{\circ }$, which lands on the same point as $+240^{\circ }$. ✓

What we do: list the 6th roots, then remove those that are also cube roots. Why: "$z^6=1$ but $z^3\ne 1$" describes roots that need the full 6th-power to return home — the primitive-flavoured leftovers. The 6th roots are $e^{i\pi k/3}$, angles $0,60,120,180,240,300$ degrees: $1,e^{i\pi /3},e^{i2\pi /3},-1,e^{i4\pi /3},e^{i5\pi /3}.$ The cube roots of unity ($z^3=1$) are the angles that are multiples of $120^{\circ }$: $\{1,e^{i2\pi /3},e^{i4\pi /3}\}$. Remove these. Remaining: $e^{i\pi /3},-1,e^{i5\pi /3}\text{i.e. angles }60^{\circ },180^{\circ },300^{\circ }.$ These are exactly the roots of $z^3=-1$ (check: their angles tripled give $180^{\circ },540^{\circ },900^{\circ }\equiv 180^{\circ }$, so $z^3=e^{i\pi }=-1$). ✓

What we do: connect the product of $(1-\text{root})$ factors to the polynomial $z^2+z+1$. Why: the two complex cube roots are exactly the roots of $z^2+z+1=0$, so that polynomial factors as $(z-\omega )(z-{\omega }^2)$. $z^2+z+1=(z-\omega )(z-{\omega }^2).$ Set $z=1$: $1^2+1+1=(1-\omega )(1-{\omega }^2)\Rightarrow 3=(1-\omega )(1-{\omega }^2).$ Answer: $3$. (General fact: ${\prod }_{k=1}^{n-1}(1-{\omega }^k)=n$, the same idea with $\frac{z^n-1}{z-1}$ evaluated at $z=1$.)

Setup (see figure): two adjacent vertices are $z_0=1$ and $z_1=e^{i2\pi /n}$, separated by the central angle $\frac{2\pi }{n}$. The side is the straight-line distance between them (a chord). What we do: drop the chord into an isosceles triangle with two unit sides and apex angle $\frac{2\pi }{n}$. Why: the chord of a unit circle subtending angle $\alpha$ has length $2\sin (\alpha /2)$ — split the isosceles triangle down the middle into two right triangles, each with hypotenuse $1$ and opposite side $\sin (\alpha /2)$. With $\alpha =\frac{2\pi }{n}$: $\text{side}=2\sin \left(\frac{\pi }{n}\right).$

• $n=3$: side $=2\sin 60^{\circ }=2\cdot \frac{\sqrt{3}}{2}=\sqrt{3}$.
• $n=6$: side $=2\sin 30^{\circ }=2\cdot \frac{1}{2}=1$ (a regular hexagon's side equals its radius — a classic fact). ✓

What we do: substitute rectangular values. Why: the coefficients $1,2,3$ are unequal, so no cancellation identity applies directly — compute numerically. $\omega =-\frac{1}{2}+\frac{\sqrt{3}}{2}i$, ${\omega }^2=-\frac{1}{2}-\frac{\sqrt{3}}{2}i$. $S=1+2(-\frac{1}{2}+\frac{\sqrt{3}}{2}i)+3(-\frac{1}{2}-\frac{\sqrt{3}}{2}i)$ Real part: $1-1-\frac{3}{2}=-\frac{3}{2}.$ Imaginary part: $\sqrt{3}-\frac{3\sqrt{3}}{2}=-\frac{\sqrt{3}}{2}.$ $S=-\frac{3}{2}-\frac{\sqrt{3}}{2}i,\overline{S}=-\frac{3}{2}+\frac{\sqrt{3}}{2}i.$ $|S{|}^2=S\overline{S}=(\frac{3}{2}{)}^2+(\frac{\sqrt{3}}{2}{)}^2=\frac{9}{4}+\frac{3}{4}=3.$ Answer: $|S{|}^2=3$, so $|S|=\sqrt{3}$.

What we do: the squares of the roots are $\{({\omega }^k{)}^2\}=\{{\omega }^{2k}\}$. Why analyse this: squaring doubles every angle, so we're really summing ${\sum }_k{\omega }^{2k}$, another geometric series in ratio ${\omega }^2$. $T={\sum }_{k=0}^{n-1}{\omega }^{2k}={\sum }_{k=0}^{n-1}({\omega }^2{)}^k.$ Case A (${\omega }^2\ne 1$): geometric sum with ratio ${\omega }^2$: $T=\frac{({\omega }^2{)}^n-1}{{\omega }^2-1}=\frac{({\omega }^n{)}^2-1}{{\omega }^2-1}=\frac{1-1}{{\omega }^2-1}=0.$ since ${\omega }^n=1$. So $T=0$ whenever ${\omega }^2\ne 1$. Case B (${\omega }^2=1$): this happens when $e^{i4\pi /n}=1$, i.e. $n\mid 2$, i.e. $n=2$ (given $n\ge 2$). Then $\omega =e^{i\pi }=-1$, its square is $1$, and each of the $n=2$ terms equals $1$: $T=\underset{n=2\text{ terms}}{\underbrace{1+1}}=2.$ Answer: the sum of squares is $0$ for all $n\ge 3$; the exception is $n=2$, where it equals $2$. Geometry: for $n\ge 3$, doubling all angles just permutes the roots among themselves (still symmetric ⇒ cancel). For $n=2$, both roots $\pm 1$ square to $+1$, collapsing onto a single point that can't cancel.

What we do: recognise this as ${\prod }_{k=1}^4(1+{\omega }^k)={\prod }_{k=1}^4(1-(-{\omega }^k))$, and use the factorisation of $z^5-1$. Why: the roots of $z^5-1$ are $1,\omega ,{\omega }^2,{\omega }^3,{\omega }^4$, so $z^5-1=(z-1)(z-\omega )(z-{\omega }^2)(z-{\omega }^3)(z-{\omega }^4).$ Divide by $(z-1)$: $\frac{z^5-1}{z-1}=1+z+z^2+z^3+z^4=(z-\omega )(z-{\omega }^2)(z-{\omega }^3)(z-{\omega }^4).$ Now set $z=-1$ (because $1+{\omega }^k=-(-1-{\omega }^k)=-((-1)-{\omega }^k)$, and each factor $(-1-{\omega }^k)=-(1+{\omega }^k)$): ${\prod }_{k=1}^4((-1)-{\omega }^k)=1+(-1)+(-1{)}^2+(-1{)}^3+(-1{)}^4=1-1+1-1+1=1.$ The left side is ${\prod }_{k=1}^4(-1-{\omega }^k)={\prod }_{k=1}^4-(1+{\omega }^k)=(-1{)}^4{\prod }_{k=1}^4(1+{\omega }^k)=P.$ Therefore $P=1$. (Sanity pattern: ${\prod }_{k=1}^{n-1}(1+{\omega }^k)$ equals $1$ for odd $n$ — matches $n=5$.)

What we do: the centre of mass is $\frac{1}{n}{\sum }_{k=0}^{n-1}{\omega }^k$. Why the average and not the sum: centre of mass = average position of equal point-masses. From the Geometric Series sum in the parent note, ${\sum }_{k=0}^{n-1}{\omega }^k=\frac{{\omega }^n-1}{\omega -1}=\frac{0}{\omega -1}=0$ (valid since $\omega \ne 1$ for $n\ge 2$). Hence the average is $\frac{0}{n}=0$. The centre of mass is the origin. $■$ Physical reading: place $n$ equal weights on the corners of a regular $n$-gon; by the rotational symmetry (a $\frac{2\pi }{n}$ turn maps the figure onto itself), the balance point can only be the fixed centre. Any nonzero centre of mass would have to rotate to a different point under that symmetry — impossible, so it must be $0$.

L1.1 ::: $1,e^{i\pi /3},-1$ L1.2 ::: angles $0,120^{\circ },240^{\circ }$; equilateral triangle L2.1 ::: sum $=0$ L2.2 ::: $\sqrt{2}(\pm 1\pm i)$ (four values) L2.3 ::: $-1$ L3.1 ::: $-1$ L3.2 ::: $\overline{\omega }={\omega }^2=-\frac{1}{2}-\frac{\sqrt{3}}{2}i$ L3.3 ::: $e^{i\pi /3},-1,e^{i5\pi /3}$ L4.1 ::: $3$ L4.2 ::: side $=2\sin (\pi /n)$; $\sqrt{3}$ and $1$ L4.3 ::: $|S{|}^2=3$ L5.1 ::: $0$ for $n\ge 3$, $=2$ for $n=2$ L5.2 ::: $1$ L5.3 ::: centre of mass $=0$