B10 - Finding all complex roots

All distinct nth roots

How many numbers have cube equal to $8$? The real answer $2$ is only one possibility. A number with modulus $2$ and argument $2\pi/3$ also cubes to $8$, because its argument triples to a full turn.

Suppose $w^n=r\mathrm{e}^{\mathrm{i}\theta}$, where $r>0$ and $n$ is a positive integer. Write $w=\rho\mathrm{e}^{\mathrm{i}\phi}$, with $\rho>0$. Comparing moduli gives $\rho^n=r$, hence $\rho=r^{1/n}$. Comparing arguments requires every full-turn alternative:

Why stop at $n-1$? Increasing $k$ by $n$ adds $2\pi$ to $\phi$, repeating the same root. Within the listed range, the angle difference between two roots is $2\pi(k-j)/n$, which cannot be a full turn unless $k=j$. This proves both completeness and distinctness.

Change only the sign of the target, from $8$ to $-8$. The root modulus stays $2$, while the starting root angle shifts by $\pi/3$.

Choosing $-\pi$ instead of $\pi$ for the target’s argument gives the same three roots in another order. A choice of initial argument cannot change the solutions of the equation.

For $w^n=0$, the only distinct root is $w=0$: a nonzero number has nonzero modulus, so its positive power cannot be zero. As a polynomial root, zero has multiplicity $n$, but it is one point, not $n$ distinct points. This is why the nonzero condition belongs in the theorem.

Rotate the target by another quarter-turn.

Arguments and regular-polygon geometry

Every root has the same modulus $r^{1/n}$, and successive roots differ in argument by $2\pi/n$. Thus, for $n\ge3$, they form 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 one point.

The full angle list is useful even if a question asks only for a diagram. It fixes the polygon’s orientation. Changing the target argument by $\delta$ rotates its root polygon by $\delta/n$, while leaving the radius unchanged.

For a principal-argument list in $-\pi<\arg w\le\pi$, first construct all roots, then subtract $2\pi$ from angles greater than $\pi$. Do not replace $\pi$ by $-\pi$ in this interval.

Change only the target’s argument, from zero to $\pi$. The square rotates through $\pi/4$.

[DIAGRAM: asset_name: Fourth roots before and after a rotation; asset_slug: aqa7367-fourth-root-squares; caption: Changing the target from 16 to minus 16 rotates the root square through pi over four; description: Two equal-scale Argand panels show the roots of w to the fourth equals 16 and w to the fourth equals minus 16. Both circles have radius 2. The first square has vertices 2, 2i, minus 2 and minus 2i. The second has vertices with coordinates plus or minus root two in each direction. Every root is labelled by its argument, including negative principal arguments below the real axis. The annotation states that adjacent arguments differ by pi over two and both side lengths are two root two.]

The opening cube equation has the same geometry with three rather than four roots: $w^3=8$ gives an equilateral triangle of radius $2$, with one vertex at $2$. Its other two vertices are the non-real answers that a real-only search misses.

Change the modulus while keeping the rotated-square structure.

Quick Recap

• For a nonzero target, take its positive real modulus root and divide every full-turn argument alternative by $n$.
• Values $k=0,1,\ldots,n-1$ give exactly $n$ distinct roots.
• Root arguments differ by $2\pi/n$; for $n\ge3$, the roots form a regular polygon centred at zero.
• Zero has only one distinct nth root, zero, even though its polynomial multiplicity is $n$.

Three numbers cube to $8$: $2$ and $-1\pm\sqrt3\mathrm{i}$. They are the vertices of an equilateral triangle in the Argand diagram.