CP 2.10 - nth roots of a complex number

Why a complex number has several roots

How can $z^4=-16$ have four answers when a calculator's root button usually returns only one value?

First try an easier equation, $z^3=8$. The real answer is $2$, but $2\mathrm{e}^{2\pi\mathrm{i}/3}$ works too:

So does $2\mathrm{e}^{4\pi\mathrm{i}/3}$. Cubing triples each angle, and all three triples land on the positive real axis.

A root of order $n$ is any number that gives the target when raised to the power $n$. The point is to solve an equation, rather than choose one preferred value of a root symbol.

Now let the target be $a=r\mathrm{e}^{\mathrm{i}\theta}$, where $r>0$, and write a root as $z=\rho\mathrm{e}^{\mathrm{i}\phi}$, where $\rho>0$. Comparing modulus and angle gives

The full turns must be included before the division by $n$. Consecutive values of $k$ then produce different directions.

The zero target is different: $z^n=0$ has only the value $z=0$, a root of multiplicity $n$. There is no argument to divide. For a non-zero target, choosing another starting argument simply cycles the same $n$ roots.

The next check diagnoses where the full turns belong in that calculation.

Finding the complete root set

List a short run of consecutive values of $k$, convert only when needed, and check that the number of distinct answers matches the degree.

For $z^4=16$, the root modulus is $2$ and the argument step is $2\pi/4=\pi/2$. Starting at angle $0$ therefore gives the four points on the axes. Changing the target's argument rotates the whole set without changing the spacing.

Worked example 2 changes only the sign of the target. Its argument increases by $\pi$, so each root angle increases by $\pi/4$.

If a question instead requests $-\pi<\phi\le\pi$, keep the same points but replace $5\pi/4$ and $7\pi/4$ by $-3\pi/4$ and $-\pi/4$. A permitted interval chooses labels; it does not discard roots.

The final accuracy mark in a root-list question usually needs the complete set in the requested form. A correct formula with an undefined integer index leaves the set unspecified.

Your turn changes the target's modulus. The directions from worked example 2 stay fixed.

The regular polygon formed by the roots

Every root has the same modulus, and successive root arguments differ by the same angle, so the complete set has a rigid geometric shape.

The roots of $z^4=-16$ sit on a circle of radius $2$, with one point in each quadrant. Join them in angular order and you get a square. This answers the opening puzzle: a fourth power folds four evenly spaced directions onto the target direction.

[DIAGRAM: asset_name: The four fourth roots of minus sixteen; asset_slug: 9fm0-cp13-fourth-roots; caption: Equal root moduli and quarter-turn argument steps produce a square centred at the origin; description: An equal-scale Argand diagram shows the circle of radius 2 centred at the origin and the four roots at the points plus or minus root 2 in each coordinate. The four points are joined to form a square, each vertex is labelled with its exponential form and its Cartesian form, a radius is labelled 2, and a central angle between neighbouring roots is labelled pi over 2]

Here is why the picture works for any non-zero target. Write its roots as

All lie at distance $r^{1/n}$ from the origin. Their arguments advance by $2\pi/n$; equal central angles cut equal chords from a circle. For $n\ge3$, those equal sides and equal turns make a regular $n$-gon centred at the origin. For $n=2$, the two roots are opposite ends of a diameter; for $n=1$, there is just one point.

There cannot be an early repeat. If $z_j=z_k$, their argument difference $2\pi(j-k)/n$ must be a whole number of full turns. For $0\le j,k<n$, that happens only when $j=k$. Increasing $k$ by $n$ does make a full turn, so every later integer repeats one of these roots.

Even a positive target can have surprising non-real roots. The sixth roots of $64$ include $2$ and $-2$, yet none is purely imaginary: the argument steps are $\pi/3$, which never land at $\pi/2$ or $3\pi/2$. Indeed, $(2\mathrm{i})^6=-64$, with the opposite sign.

When sketching, place any axis vertices accurately and preserve equal spacing. A regular-looking polygon drawn at the wrong orientation represents the wrong roots.

Quick Recap

• For $r>0$, solve $z^n=r\mathrm{e}^{\mathrm{i}\theta}$ using $z_k=r^{1/n}\mathrm{e}^{\mathrm{i}(\theta+2k\pi)/n}$, with $k=0,1,\ldots,n-1$.
• Include the full turns before dividing by $n$; otherwise you miss the other roots.
• Give all distinct answers in the requested form and argument interval.
• The roots lie on a circle of radius $r^{1/n}$, separated by $2\pi/n$; for $n\ge3$, they form a regular $n$-gon.

The four solutions of $z^4=-16$ are $\sqrt2+\mathrm{i}\sqrt2$, $-\sqrt2+\mathrm{i}\sqrt2$, $-\sqrt2-\mathrm{i}\sqrt2$ and $\sqrt2-\mathrm{i}\sqrt2$: the four vertices of the square in the diagram.