What does it mean to extract a root from a complex number?

For real numbers, extracting a root usually raises no questions: for example, \( \sqrt{9}=3 \). On the complex plane, the same operation often produces several results.

A complex number \(w\) is called an \(n\)-th root of a number \(z\) if the following equality holds:

\[ w^n=z. \]

For example, the equation \(w^2=1\) has two solutions: \(w=1\) and \(w=-1\). In general, every nonzero complex number has exactly \(n\) distinct \(n\)-th roots.

The general formula

Let a nonzero complex number be written in exponential form:

\[ z=\rho e^{i\varphi}, \qquad \rho=|z|. \]

We look for a root in the same form:

\[ w=re^{i\theta}. \]

Then

\[ w^n=r^n e^{in\theta}. \]

The condition \(w^n=z\) leads to two equalities:

\[ r^n=\rho, \qquad n\theta=\varphi+2\pi k, \qquad k\in\mathbb Z. \]

Therefore, the modulus of every root is \( \sqrt[n]{\rho} \), and its argument is

\[ \theta_k=\frac{\varphi+2\pi k}{n}. \]

All \(n\)-th roots of \(z\) are given by the formula

\[ w_k= \sqrt[n]{\rho}\, e^{\,i\frac{\varphi+2\pi k}{n}}, \qquad k=0,1,\ldots,n-1. \]

In trigonometric form:

\[ w_k= \sqrt[n]{\rho} \left( \cos\frac{\varphi+2\pi k}{n} + i\sin\frac{\varphi+2\pi k}{n} \right). \]

Why are values from \(0\) to \(n-1\) enough?

The formula contains any integer \(k\), but no new roots appear after \(k=n-1\). Indeed,

\[ \theta_{k+n} = \frac{\varphi+2\pi(k+n)}{n} = \theta_k+2\pi. \]

Arguments that differ by \(2\pi\) determine the same point on the complex plane, since \(e^{i(\theta+2\pi)}=e^{i\theta}\). Therefore, the values begin to repeat.

Geometric interpretation

All roots found have the same modulus:

\[ |w_k|=\sqrt[n]{\rho}. \]

Therefore, they lie on the same circle centered at the origin. The arguments of neighbouring roots differ by

\[ \theta_{k+1}-\theta_k=\frac{2\pi}{n}. \]

Thus, the roots are distributed evenly around the circle and form the vertices of a regular \(n\)-gon:

  • two roots lie at opposite ends of a diameter;
  • three roots form an equilateral triangle;
  • four roots form a square;
  • five roots form a regular pentagon.

Example: square roots of \(i\)

Example

Let us write the number \(i\) in exponential form:

\[ i=e^{i\pi/2}. \]

Here \(\rho=1\), \(\varphi=\frac{\pi}{2}\), and \(n=2\). Substituting these values into the general formula gives

\[ w_k=e^{\,i\frac{\pi/2+2\pi k}{2}}, \qquad k=0,1. \]

For \(k=0\):

\[ w_0=e^{i\pi/4} = \frac{1+i}{\sqrt2}. \]

For \(k=1\):

\[ w_1=e^{i5\pi/4} = -\frac{1+i}{\sqrt2}. \]

Therefore, the solutions of the equation \(w^2=i\) are

\[ \left\{ \frac{1+i}{\sqrt2}, -\frac{1+i}{\sqrt2} \right\}. \]

Roots of unity

A special and very important case is the set of solutions to the equation

\[ z^n=1. \]

Since \(1=e^{i2\pi m}\), where \(m\in\mathbb Z\), all roots of unity have the form

\[ z_k=e^{i\frac{2\pi k}{n}}, \qquad k=0,\ldots,n-1. \]

They lie on the unit circle. For example, the fourth roots of unity are

\[ 1,\qquad i,\qquad -1,\qquad -i. \]

Example: \(\sqrt{1-i\sqrt3}\)

Example

Let

\[ z=1-i\sqrt3. \]

Let us find the modulus:

\[ |z| = \sqrt{1^2+(-\sqrt3)^2} = 2. \]

The argument of the number is \(-\frac{\pi}{3}\), so

\[ z=2e^{-i\pi/3}. \]

For the square root, we obtain

\[ w_k= \sqrt2\, e^{\,i\frac{-\pi/3+2\pi k}{2}}, \qquad k=0,1. \]

The first root is:

\[ w_0= \sqrt2\,e^{-i\pi/6} = \frac{\sqrt3-i}{\sqrt2}. \]

The second root is opposite to the first:

\[ w_1= -\frac{\sqrt3-i}{\sqrt2}. \]

Algorithm for finding roots

  1. Write the number in the form \(z=a+bi\).
  2. Find the modulus \( \rho=\sqrt{a^2+b^2} \).
  3. Find the argument \( \varphi=\arg z \).
  4. Write the number in the form \(z=\rho e^{i\varphi}\).
  5. Calculate \( \sqrt[n]{\rho} \).
  6. Substitute \(k=0,1,\ldots,n-1\) into the root formula.
\[ \sqrt[n]{\rho e^{i\varphi}} = \left\{ \sqrt[n]{\rho}\, e^{\,i\frac{\varphi+2\pi k}{n}} \;\middle|\; k=0,\ldots,n-1 \right\}. \]