Cube roots of unity

Finding the roots. Factorise $x^3-1$ using the standard identity $a^3-b^3=(a-b)(a^2+ab+b^2)$ with $a=x,\ b=1$:

where: the first factor gives $x=1$ (the familiar real cube root), and the second factor $x^2+x+1=0$ supplies the other two roots. Solving it by the quadratic formula,

So the three cube roots of unity are

Squaring the expression for $\omega$ does indeed produce the expression for $\omega^2$ — that check is what justifies calling the third root "$\omega^2$" rather than inventing a new symbol.

Two identities that do all the work. Because $\omega$ satisfies $\omega^2+\omega+1=0$ (it is a root of that factor), and because $\omega$ is a cube root of 1 by construction,

where: the first says powers of $\omega$ repeat every three steps; the second says the three roots always cancel when added. Together they are the only two facts you ever substitute — every simplification problem in this topic reduces to spotting where one of them applies. It is worth checking $1+\omega+\omega^2=0$ once by direct substitution rather than trusting the factorisation alone: $1+\dfrac{-1+i\sqrt3}{2}+\dfrac{-1-i\sqrt3}{2}=1+\dfrac{(-1+i\sqrt3)+(-1-i\sqrt3)}{2}=1+\dfrac{-2}{2}=1-1=0$. The $i\sqrt3$ terms cancel exactly, leaving a purely real sum, which is itself a small confirmation that the identity is not a rounding artefact but an exact algebraic fact.

Geometric picture. Each root has modulus 1, so all three sit on the unit circle in the Argand plane, spaced $120^\circ$ apart: $1$ at $0^\circ$, $\omega$ at $120^\circ$, $\omega^2$ at $240^\circ$. They form an equilateral triangle inscribed in the circle — three equally spaced unit vectors always cancel when summed, which is the geometric reason $1+\omega+\omega^2=0$. This picture also explains multiplication by $\omega$: multiplying any complex number by $\omega$ rotates it by $120^\circ$ about the origin without changing its distance from the origin, since $\omega$ itself has modulus 1 and argument $120^\circ$. Multiplying by $\omega$ three times in a row therefore rotates a point by $360^\circ$, back to where it started — a geometric restatement of $\omega^3=1$.

Powers cycle with period 3. Since $\omega^3=1$, any power $\omega^n$ collapses to $1$, $\omega$, or $\omega^2$ depending on $n\bmod 3$. For instance $\omega^7=\omega^6\cdot\omega=(\omega^3)^2\cdot\omega=\omega$, and $\omega^{100}=\omega^{99}\cdot\omega=(\omega^3)^{33}\cdot\omega=\omega$.

Conjugates and polar form. Compare the explicit values: $\omega=\frac{-1+i\sqrt3}{2}$ and $\omega^2=\frac{-1-i\sqrt3}{2}$ differ only in the sign of the imaginary part, so $\omega^2=\overline{\omega}$, the complex conjugate of $\omega$. This is exactly what the picture predicts — reflecting a point at $120^\circ$ across the real axis lands you at $-120^\circ$, which is the same location as $240^\circ$. In polar form, $\omega=\cos120^\circ+i\sin120^\circ$ and $\omega^2=\cos240^\circ+i\sin240^\circ=\cos(-120^\circ)+i\sin(-120^\circ)$, confirming the conjugate relationship without any extra algebra. Knowing $\omega^2=\bar\omega$ is a fast way to double-check a computed value of $\omega^2$: just flip the sign on the imaginary part of $\omega$.

Simplify $\omega^{15}+\omega^{20}+\omega^{25}$, then evaluate $(1+\omega-\omega^2)(1-\omega+\omega^2)$.

Step 1. Reduce each exponent modulo 3, since $\omega^3=1$. $\omega^{15}=(\omega^3)^5=1$. For $\omega^{20}$, write $20=18+2$, so $\omega^{20}=(\omega^3)^6\cdot\omega^2=\omega^2$. For $\omega^{25}$, write $25=24+1$, so $\omega^{25}=(\omega^3)^8\cdot\omega=\omega$.

Step 2. Add the reduced terms. $\omega^{15}+\omega^{20}+\omega^{25}=1+\omega^2+\omega=1+\omega+\omega^2=0$, using the key identity directly.

Step 3. For the second expression, first rewrite each bracket using $1+\omega=-\omega^2$ and $1+\omega^2=-\omega$ (both immediate consequences of $1+\omega+\omega^2=0$). $1+\omega-\omega^2=(1+\omega)-\omega^2=-\omega^2-\omega^2=-2\omega^2$, and $1-\omega+\omega^2=(1+\omega^2)-\omega=-\omega-\omega=-2\omega$.

Step 4. Multiply the two simplified brackets. $(-2\omega^2)(-2\omega)=4\omega^3=4(1)=4$, using $\omega^3=1$ in the last step. So $\omega^{15}+\omega^{20}+\omega^{25}=\mathbf{0}$ and $(1+\omega-\omega^2)(1-\omega+\omega^2)=\mathbf{4}$.

Competitive stretch

The same reduction trick handles arbitrarily large exponents without ever computing a decimal power: for any integer $n$, $\omega^n$ depends only on $n\bmod 3$, so $\omega^{2026}=\omega^{2025}\cdot\omega=(\omega^3)^{675}\cdot\omega=\omega$ — a fact that turns exam questions with huge exponents into a one-line remainder calculation. The same idea extends to sums: $1+\omega^n+\omega^{2n}$ equals $3$ whenever $n$ is a multiple of $3$ (because every term collapses to $1$), and equals $0$ otherwise (because the three terms are just $1,\omega,\omega^2$ in some order, which always sum to zero). Recognising which case a question falls into — check whether the exponent is a multiple of $3$ — is often the entire solution.

• Treating $\omega$ as "just some complex number" without fixing which root it is — by convention $\omega=\dfrac{-1+i\sqrt3}{2}$, the root in the upper half of the Argand plane, and $\omega^2$ is its conjugate.
• Reducing exponents by dividing instead of taking the remainder — $\omega^{20}$ needs $20\bmod 3=2$, giving $\omega^2$, not $\omega^{20/3}$.
• Misremembering the identity as $1+\omega+\omega^2=1$; the sum of all three cube roots is always $0$, never $1$.
• Leaving $\omega+\omega^2$ unsimplified in the middle of a calculation instead of replacing it with $-1$ immediately, which is usually the step that unlocks the rest of the problem.
• Forgetting that $\omega^2=\bar\omega$ only holds for this specific pair — it is a consequence of $\omega$ having modulus 1 and the two roots being reflections in the real axis, not a general rule about squaring complex numbers.

Ask the tutor: If a problem with a huge exponent on $\omega$ or a tangled sum of $1,\omega,\omega^2$ has you stuck, ask the AI tutor to walk through reducing the exponent mod 3 or substituting $1+\omega+\omega^2=0$ step by step alongside your own working.