2.5 - Complex roots and roots of unity | Cambridge A-Level Further Mathematics

A nonzero complex number has several distinct nth roots, arranged at equal angles on a circle. You will find every branch of positive and negative rational powers, locate the roots of unity, and use their geometry and polynomial factorisation to evaluate exact sums and products.

All roots from a rational power

A calculator may display one cube root, but how do you find all the solutions of $w^3=z$ ? If $z=R(\cos\theta+i\sin\theta)\ne0$ , then $w$ must have modulus $R^{1/3}$ . Its argument $\phi$ must satisfy $3\phi=\theta+2k\pi$ , not just $3\phi=\theta$ . The three values from $k=0,1,2$ are separated by $2\pi/3$ .

The same argument works for every positive integer $n$ . Write $w=\rho(\cos\phi+i\sin\phi)$ and compare moduli and arguments in $w^n=z$ :

\rho^n=R,\qquad n\phi=\theta+2k\pi.

Choosing $k=0,1,\ldots,n-1$ gives every distinct root once. Increasing $k$ by $n$ adds a full turn to $\phi$ and repeats a root.

All branches of a rational power

Let $m/n$ be in lowest terms, with integers $m\ne0$ and $n>0$ . For nonzero $z$ , the full set of values of $z^{m/n}$ means the solutions of

w^n=z^m.

For $m<0$ , this is $w^n=1/z^{-m}$ , so the base and all resulting values must be nonzero. We are finding the full set, not choosing a single principal value.

Reducing the fraction matters. The power $z^{2/4}$ is treated as $z^{1/2}$ ; solving $w^4=z^2$ without reduction would generally introduce two extra values. Rules that select or combine individual branches need care, so use the defining equation instead of assuming that fractional powers obey every real-power manipulation.

Every rational-power branch

If $z=R(\cos\theta+i\sin\theta)$ , $R>0$ , and $m/n$ is reduced as above, then

w_k=R^{m/n}\left(\cos\frac{m\theta+2k\pi}{n} +i\sin\frac{m\theta+2k\pi}{n}\right), \qquad k=0,1,\ldots,n-1.

There are $n$ distinct values. For a positive rational exponent, zero has the sole value zero; a negative power of zero is undefined.

Not in the formula booklet: learn how to obtain all the branches.

Why does taking all nth roots of $z$ and then raising them to $m$ give the same set? Their angles are $(m\theta+2mj\pi)/n$ . Since $m$ and $n$ are coprime, the residues of $mj$ modulo $n$ run through all $n$ possibilities. This also explains the lowest-terms condition.

Worked example 1: a positive rational exponent

Question. Let $z=8(\cos(\pi/4)+i\sin(\pi/4))$ . Find all values of $z^{2/3}$ in Cartesian form.

Plan. Solve $w^3=z^2$ , using all three possible arguments.

The modulus is $8^{2/3}=4$ . The arguments are

\frac{2(\pi/4)+2k\pi}{3} =\frac\pi6+\frac{2k\pi}{3},\qquad k=0,1,2.

Thus the arguments are $\pi/6,5\pi/6,3\pi/2$ .

Answer. The three values are

2\sqrt3+2i,\qquad -2\sqrt3+2i,\qquad -4i.

Check. Here $z^2=64i$ . Direct cubing gives $(2\sqrt3+2i)^3=64i$ and $(-2\sqrt3+2i)^3=64i$ ; also $(-4i)^3=64i$ . All three are distinct, so the degree-three equation has no further roots.

Change only the exponent's sign. The moduli become reciprocals and each individual angle is reflected in the real axis.

Worked example 2: a negative rational exponent

Question. For the same $z=8(\cos(\pi/4)+i\sin(\pi/4))$ , find all values of $z^{-2/3}$ in Cartesian form.

Plan. Solve $w^3=z^{-2}$ , or equivalently take the reciprocals of the three values in the previous example.

The modulus is $8^{-2/3}=1/4$ . The arguments are

\frac{-2(\pi/4)+2k\pi}{3} =-\frac\pi6+\frac{2k\pi}{3},\qquad k=0,1,2,

namely $-\pi/6,\pi/2,7\pi/6$ .

Answer. The three values are

\frac{\sqrt3-i}{8},\qquad \frac i4,\qquad \frac{-\sqrt3-i}{8}.

Check. Cartesian reciprocation gives

\frac1{2\sqrt3+2i}=\frac{\sqrt3-i}{8},\qquad \frac1{-2\sqrt3+2i}=\frac{-\sqrt3-i}{8},\qquad \frac1{-4i}=\frac i4.

Their cubes must therefore be $1/(64i)=-i/64=z^{-2}$ , independently checking both magnitude and sign.

Reciprocation reflects an argument and inverts a modulus. It does not simply negate a complex value.

[DIAGRAM: asset_name: Reciprocal sets of three branches; asset_slug: cam9231-rational-branches; caption: The positive and negative two-thirds powers have reciprocal moduli and reflected arguments; description: Two separate Argand panels with explicitly different scales. The left shows the radius-four roots at arguments pi over six, five pi over six and three pi over two. The right shows the radius-one-quarter reciprocal roots at arguments minus pi over six, pi over two and seven pi over six. Matching colours link reciprocal pairs; a note states that the scales differ.]
The positive and negative two-thirds powers have reciprocal moduli and reflected arguments

The number of branches is determined by the reduced denominator, not by the numerator's sign.

Watch out

Dividing just one chosen argument by $n$ gives only one root. Include $2k\pi$ before division and list exactly $n$ distinct values. Then adjust each angle into any interval the question specifies.

Keep a negative exponent but change the base and its argument.

The nth roots of unity

The nth roots of unity are the solutions of $w^n=1$ , so their moduli are all $1$ and their arguments are equally spaced.

Write $1=\cos0+i\sin0$ , retaining its other arguments $2k\pi$ . The roots are

1,\quad \cos\frac{2\pi}{n}+i\sin\frac{2\pi}{n},\quad\ldots,\quad \cos\frac{2(n-1)\pi}{n}+i\sin\frac{2(n-1)\pi}{n}.

For $n\ge3$ , these are the vertices of a regular n-sided polygon on the unit circle, with one vertex at $1$ . Adjacent roots differ in argument by $2\pi/n$ .

A primitive root of unity

A root is primitive if its successive integer powers generate all the nth roots. The number

\omega=\cos\frac{2\pi}{n}+i\sin\frac{2\pi}{n}

is primitive: the complete list is $1,\omega,\omega^2,\ldots,\omega^{n-1}$ .

Not every root is primitive. For example, $-1$ is a sixth root of unity, but its powers give only $1$ and $-1$ .

This notation is useful because $\omega^n=1$ : reduce integer exponents modulo $n$ . In particular, $\omega^{-1}=\omega^{n-1}$ , so conjugate roots form reflected pairs on the unit circle. A negative real root occurs only when $n$ is even.

Worked example 3: five equally spaced roots

Question. Find all fifth roots of unity in polar form, with arguments in $[0,2\pi)$ , and describe their positions on an Argand diagram.

Plan. Divide the arguments $2k\pi$ of unity by $5$ , taking five distinct residues.

|w|=1,\qquad \arg w=\frac{2k\pi}{5},\qquad k=0,1,2,3,4.

The arguments are $0,2\pi/5,4\pi/5,6\pi/5,8\pi/5$ .

Answer. The roots are $\cos(2k\pi/5)+i\sin(2k\pi/5)$ for $k=0,1,2,3,4$ . They form a regular pentagon on the unit circle, with a vertex at $1$ , and angular spacing $2\pi/5$ .

Check. Raising any listed value to the fifth power multiplies its angle to $2k\pi$ , giving $1$ . No two listed angles differ by a full turn, so the five roots are distinct.

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2.5 - Complex roots and roots of unity

All roots from a rational power

The nth roots of unity

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