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Roots of complex numbers (SACE Stage 2 Specialist Mathematics)

Syllabus dot point

“Find the nth roots of a complex number using polar form and de Moivre's theorem, and represent them geometrically on the Argand plane.”

8 min read

Quick answer

Every nonzero complex number has exactly nn distinct nnth roots. Write the number in polar form r(cos⁡θ+isin⁡θ)r(\cos\theta + i\sin\theta), then the roots are r1/n(cos⁡θ+2πkn+isin⁡θ+2πkn)r^{1/n}\left(\cos\dfrac{\theta + 2\pi k}{n} + i\sin\dfrac{\theta + 2\pi k}{n}\right) for k=0,1,…,n−1k = 0, 1, \ldots, n-1. They all have modulus r1/nr^{1/n}, so they lie equally spaced on a circle, separated by angle 2πn\dfrac{2\pi}{n}.

Jump to a section
  1. What this dot point is asking
  2. The formula for nth roots
  3. Geometry on the Argand plane
  4. Roots of unity
  5. Solving polynomial equations with roots
  6. A worked tip in practice
  7. Exam-style questions

What this dot point is asking

You need to find the nnth roots of a complex number using de Moivre's theorem, solve equations of the form zn=wz^{n} = w, and represent the roots geometrically on the Argand plane.

The formula for nth roots

To solve zn=wz^{n} = w, write ww in polar form w=r(cos⁡θ+isin⁡θ)w = r(\cos\theta + i\sin\theta). Because adding any whole number of full turns leaves ww unchanged, write the argument as θ+2πk\theta + 2\pi k. Taking the nnth root with de Moivre's theorem gives

The nth roots of a complex number

If w=r(cos⁡θ+isin⁡θ)w = r(\cos\theta + i\sin\theta), the nn distinct nnth roots of ww are

zk=r1/n(cos⁡θ+2πkn+isin⁡θ+2πkn),k=0,1,2,…,n−1.z_k = r^{1/n}\left(\cos\frac{\theta + 2\pi k}{n} + i\sin\frac{\theta + 2\pi k}{n}\right), \quad k = 0, 1, 2, \ldots, n-1.

Each root has modulus r1/nr^{1/n}, and successive arguments differ by 2πn\dfrac{2\pi}{n}.

Only k=0k = 0 to n−1n-1 give distinct roots; k=nk = n repeats k=0k = 0 (the angle increases by a full 2π2\pi).

Geometry on the Argand plane

Because all nn roots share the modulus r1/nr^{1/n}, they lie on a circle of radius r1/nr^{1/n} centred at the origin. The equal angular spacing of 2πn\dfrac{2\pi}{n} means they form the vertices of a regular nn-sided polygon. Sketching this circle and marking one root lets you read off the rest by rotation.

Cube roots of a complex number

Solve z3=8iz^{3} = 8i, giving roots in polar form

Step 1 - polar form of the right side. 8i8i has modulus r=8r = 8 and lies on the positive imaginary axis, so θ=π2\theta = \dfrac{\pi}{2}:

8i=8(cos⁡π2+isin⁡π2).8i = 8\left(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\right).

Step 2 - apply the roots formula with n=3n = 3. The modulus of each root is 81/3=28^{1/3} = 2, and the arguments are

π2+2πk3,k=0,1,2.\frac{\frac{\pi}{2} + 2\pi k}{3}, \quad k = 0, 1, 2.

  • k=0k = 0: argument π6\dfrac{\pi}{6}, so z0=2(cos⁡π6+isin⁡π6)z_0 = 2\left(\cos\dfrac{\pi}{6} + i\sin\dfrac{\pi}{6}\right).
  • k=1k = 1: argument π/2+2π3=5π6\dfrac{\pi/2 + 2\pi}{3} = \dfrac{5\pi}{6}, so z1=2(cos⁡5π6+isin⁡5π6)z_1 = 2\left(\cos\dfrac{5\pi}{6} + i\sin\dfrac{5\pi}{6}\right).
  • k=2k = 2: argument π/2+4π3=3π2\dfrac{\pi/2 + 4\pi}{3} = \dfrac{3\pi}{2}, equivalently −π2-\dfrac{\pi}{2}, so z2=2(cos⁡3π2+isin⁡3π2)=−2iz_2 = 2\left(\cos\dfrac{3\pi}{2} + i\sin\dfrac{3\pi}{2}\right) = -2i.

The three roots all have modulus 22 and are spaced 2π3\dfrac{2\pi}{3} apart, forming an equilateral triangle on the circle of radius 22.

Roots of unity

The special case zn=1z^{n} = 1 gives the nnth roots of unity:

zk=cos⁡2πkn+isin⁡2πkn,k=0,1,…,n−1.z_k = \cos\frac{2\pi k}{n} + i\sin\frac{2\pi k}{n}, \quad k = 0, 1, \ldots, n-1.

They lie on the unit circle, one of them is always z=1z = 1, and their sum is 00 for n≥2n \ge 2 (the vertices of a regular polygon centred at the origin balance out). If ω\omega denotes the root with smallest positive argument, the full set is 1,ω,ω2,…,ωn−11, \omega, \omega^2, \ldots, \omega^{n-1}.

Spacing shortcut

Once you have found one nnth root, the others follow by repeatedly rotating it by 2πn\dfrac{2\pi}{n} about the origin (add 2πn\dfrac{2\pi}{n} to the argument, keep the same modulus). You rarely need to recompute each root from scratch.

Solving polynomial equations with roots

Finding nnth roots is the engine for solving binomial polynomial equations such as z4+16=0z^4 + 16 = 0. Rearrange to z4=−16z^4 = -16, write −16-16 in polar form as 16cis⁡π16\operatorname{cis}\pi, and take fourth roots: the modulus of each root is 161/4=216^{1/4} = 2 and the arguments are π+2πk4\dfrac{\pi + 2\pi k}{4} for k=0,1,2,3k = 0, 1, 2, 3. This yields four complex roots that pair into conjugates, consistent with the fact that a real polynomial has complex roots in conjugate pairs. Whenever a polynomial equation reduces to zn=wz^n = w, the nnth-roots formula gives every solution directly.

A worked tip in practice

Because the roots are equally spaced, the fastest method is to compute the first root carefully and then add 2πn\dfrac{2\pi}{n} to the argument repeatedly, keeping the modulus fixed. For sketching, draw the circle of radius r1/nr^{1/n}, mark the first root at its argument, then step around by 2πn\dfrac{2\pi}{n} to place the rest at the vertices of a regular polygon. This both saves time and guards against the common error of recomputing each root from scratch and slipping on the arithmetic.

Common errors
Stopping after one root
zn=wz^{n} = w has nn solutions. Finding only the "obvious" real root loses most of the marks.
Using the wrong range of kk
Use k=0k = 0 to n−1n - 1. Going further repeats roots; starting elsewhere is fine but you must still take exactly nn consecutive values.
Forgetting to take the nth root of the modulus
Each root has modulus r1/nr^{1/n}, not rr. The argument is divided by nn and the modulus is rooted by nn.
Not reducing arguments
Final arguments should usually be given as principal values in (−π,π](-\pi, \pi]; an argument like 3π2\frac{3\pi}{2} may be expected as −π2-\frac{\pi}{2}.

Exam-style questions

Questions in the style of SACE Board exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

SACE 2022-style3 marks
Calculator-free. Solve z5=1z^5 = 1, writing your solutions in polar form.
Show worked answer →

Write the right-hand side in polar form: 1=cis⁡(2kπ)1 = \operatorname{cis}(2k\pi) for any integer kk, since adding full turns does not change the number. [1 mark]

Take fifth roots by de Moivre's theorem: z=cis⁡ ⁣(2kπ5)z = \operatorname{cis}\!\left(\dfrac{2k\pi}{5}\right) for k=0,1,2,3,4k = 0, 1, 2, 3, 4 (five distinct roots). [1 mark]

The solutions are cis⁡(0)=1\operatorname{cis}(0) = 1, cis⁡2π5\operatorname{cis}\tfrac{2\pi}{5}, cis⁡4π5\operatorname{cis}\tfrac{4\pi}{5}, cis⁡6π5\operatorname{cis}\tfrac{6\pi}{5}, cis⁡8π5\operatorname{cis}\tfrac{8\pi}{5} - equally spaced by 2π5\tfrac{2\pi}{5} around the unit circle. [1 mark]

SACE 2023-style3 marks
Calculator-free. Use de Moivre's theorem to find all solutions of z5=32z^5 = 32, giving the modulus and arguments.
Show worked answer →

Write 32=32cis⁡(2kπ)32 = 32\operatorname{cis}(2k\pi) for any integer kk, since 3232 is a positive real with argument 00 (plus full turns). [1 mark]

Take fifth roots. The modulus of each root is 321/5=232^{1/5} = 2, and the arguments are 2kπ5\dfrac{2k\pi}{5} for k=0,1,2,3,4k = 0, 1, 2, 3, 4: namely 0,2π5,4π5,6π5,8π50, \tfrac{2\pi}{5}, \tfrac{4\pi}{5}, \tfrac{6\pi}{5}, \tfrac{8\pi}{5}. [1 mark]

Writing the last two as principal values, 6π5=−4π5\tfrac{6\pi}{5} = -\tfrac{4\pi}{5} and 8π5=−2π5\tfrac{8\pi}{5} = -\tfrac{2\pi}{5}. So the solutions are 2cis⁡(0)2\operatorname{cis}(0), 2cis⁡2π52\operatorname{cis}\tfrac{2\pi}{5}, 2cis⁡4π52\operatorname{cis}\tfrac{4\pi}{5}, 2cis⁡ ⁣(−4π5)2\operatorname{cis}\!\left(-\tfrac{4\pi}{5}\right), 2cis⁡ ⁣(−2π5)2\operatorname{cis}\!\left(-\tfrac{2\pi}{5}\right). [1 mark]

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