How do I solve equations like x⁴ + 1 = 0 using roots of unity?

In one line

Rewrite the equation as \(x^n=\) a complex number, express that number in polar form with \(2k\pi\) added, and take the \(n\)th roots.

Explained simply

The equation \(x^4+1=0\) has no real solution, because a real number to the fourth power is never negative. But it has four complex solutions. The trick is that \(-1\) is just a point on the unit circle at angle \(\pi\), so its fourth roots are four points on the same circle, \(\pi/2\) apart.

Technically, the \(n\)th roots of unity are the solutions of \(x^n=1\): \(x=\cos\dfrac{2k\pi}{n}+i\sin\dfrac{2k\pi}{n}\), \(k=0,\dots,n-1\). If \(\alpha=e^{2\pi i/n}\), they are \(1,\alpha,\alpha^2,\dots,\alpha^{n-1}\) and their sum is 0. For \(n=3\): \(1,\omega,\omega^2\) with \(\omega=\dfrac{-1+i\sqrt3}{2}\), \(1+\omega+\omega^2=0\), \(\omega^3=1\).

How it works

1. Bring the equation to the form \(x^n=w\). If it factorises (e.g. \(x^7+x^4+x^3+1=(x^4+1)(x^3+1)\)), solve each factor.
2. Write \(w=r[\cos(\theta+2k\pi)+i\sin(\theta+2k\pi)]\). For \(w=-1\): \(\cos(2k+1)\pi+i\sin(2k+1)\pi\).
3. Take the \(n\)th root and list \(k=0,\dots,n-1\).
4. Simplify each root to \(a+ib\) form if the angles are standard.

Key formula

\(x^n=1\Rightarrow x=e^{2k\pi i/n}\); \(\ x^n=-1\Rightarrow x=e^{(2k+1)\pi i/n}\), \(k=0,\dots,n-1\).

Sum of all \(n\)th roots of unity \(=0\) (for \(n\ge2\)); product \(=(-1)^{n-1}\).

Quick example

Solve \(x^4+1=0\). \(x^4=-1=\cos(2k+1)\pi+i\sin(2k+1)\pi\), so \(x=\cos\dfrac{(2k+1)\pi}4+i\sin\dfrac{(2k+1)\pi}4\).

• \(k=0\): \(\frac{1+i}{\sqrt2}\)
• \(k=1\): \(\frac{-1+i}{\sqrt2}\)
• \(k=2\): \(\frac{-1-i}{\sqrt2}\)
• \(k=3\): \(\frac{1-i}{\sqrt2}\)

Check: \(\left(\frac{1+i}{\sqrt2}\right)^2=\frac{2i}{2}=i\) and \(i^2=-1\). Correct.

Common mistakes

• Writing \(-1=\cos2k\pi+i\sin2k\pi\); that is \(+1\). For \(-1\) use \((2k+1)\pi\).
• Cancelling a common factor such as \(x^3+1\) without solving it, which loses roots.
• Giving fewer than \(n\) roots for a degree-\(n\) equation.

How MU asks it

Typical 5/6-mark questions: "Solve \(x^4-x^3+x^2-x+1=0\)" (multiply by \(x+1\) to get \(x^5+1=0\), then discard \(x=-1\)), "Solve \(x^7+x^4+x^3+1=0\)", or "If \(\omega\) is a cube root of unity, prove \((1-\omega+\omega^2)^6=64\)". Show the factorisation, the general root, all values of \(k\), and state which root is rejected and why.