How to find all nth roots of a complex number

Taking a root reverses the power rule for complex numbers. We take a root of the modulus and divide the possible angles by the root index. Keeping the full family of angles gives all the distinct roots.

Use the nth-root formula

Let $w = R (\cos φ + i \sin φ)$ with $R > 0$ . Its $n$ distinct nth roots are

z_{k} = R^{1 / n} (\cos θ_{k} + i \sin θ_{k})

θ_{k} = \frac{φ + 2 π k}{n}

Use $k = 0, 1, \dots, n - 1$ . The root angles are spaced by $2 π / n$ . In degrees, replace $2 π k$ with $360^{°} k$ . Zero is a special case: its only distinct nth root is zero.

Find the fourth roots of negative 81

Write $- 81 = 81 (\cos 180^{°} + i \sin 180^{°})$ . The root modulus is $81^{1 / 4} = 3$ , and the angles are

θ_{k} = \frac{180^{°} + 360^{°} k}{4}

Four distinct angles
k	Angle
$0$	$45^{°}$
$1$	$135^{°}$
$2$	$225^{°}$
$3$	$315^{°}$

Write the roots in rectangular form

Each root has the form $3 (\cos θ_{k} + i \sin θ_{k})$ . Evaluating sine and cosine gives

\begin{array}{l} z_{0} = \frac{3 \sqrt{2}}{2} + \frac{3 \sqrt{2}}{2} i \\ z_{1} = - \frac{3 \sqrt{2}}{2} + \frac{3 \sqrt{2}}{2} i \\ z_{2} = - \frac{3 \sqrt{2}}{2} - \frac{3 \sqrt{2}}{2} i \\ z_{3} = \frac{3 \sqrt{2}}{2} - \frac{3 \sqrt{2}}{2} i \end{array}

The four roots have modulus 3 and arguments 45°, 135°, 225°, and 315°. Their directions are 90° apart.

Check a root with De Moivre’s theorem

Raising the root with angle $45^{°}$ to the fourth power gives modulus $3^{4} = 81$ and angle $180^{°}$ , so its fourth power is $- 81$ . Each remaining root adds a full turn after multiplication of its angle by four, giving the same result.

Example: Solve z⁴ = 16

The modulus of each root is 2. Dividing the arguments 0, 2π, 4π, and 6π by four gives four equally spaced directions.

The four roots of sixteen
Angle	Complex solution
$0$	$2$
$\frac{π}{2}$	$2 i$
$π$	$- 2$
$\frac{3 π}{2}$	$- 2 i$

The solutions 2, 2i, −2, and −2i are equally spaced around the origin, each at distance 2.

Check the solutions in the original equation

The real solutions satisfy $2^{4} = (- 2)^{4} = 16$ . For the imaginary solutions, $(2 i)^{4} = (- 2 i)^{4} = 16 i^{4} = 16$ . All four therefore satisfy the equation.

For a nonzero right side in $z^{n} = w$ , this same process gives $n$ distinct complex solutions: take the nth root of the modulus and divide the full family of arguments by $n$ .

Learn mathKrista KingAugust 18, 2021math, learn online, online course, online math, equilibrium solutions