Roots of Complex Numbers

Recall that if $z=x+iy$ is a nonzero complex number, then it can be written in polar form as \[z=r(\cos \theta +i \sin \theta)\] where $r=\sqrt{x^2+y^2}$ and $\theta$ is the angle, in radians, from the positive $x$-axis to the ray connecting the origin to the point $z.$

Now, de Moivre's formula establishes that if $z=r(\cos \theta +i\sin \theta)$ and $n$ is a positive integer, then \begin{eqnarray*} z^n=r^n(\cos n\theta+i\sin n\theta). \end{eqnarray*}

Let $z$ be a nonzero complex number. Using de Moivre's formula will help us to solve the equation $$w^n=z$$ for $w$ when $z$ is given.

Suppose that $z=r(\cos \theta +i\sin \theta)$ and $w=\rho (\cos \psi +i\sin \psi).$ Then de Moivre's formula gives $$w^n=\rho^n(\cos n\psi+i\sin n\psi).$$ Since $w^n=z,$ it follows that $$\rho^n=r=|w|$$ by uniqueness of the polar representation and $$n\psi = \theta +k(2\pi),$$ where $k$ is some integer. Thus, the $n$th roots of a nonzero complex number $z$ are given by

\begin{eqnarray}\label{n-roots} w_k=z^{1/n}=\sqrt[n]{r}\left[\cos\left(\frac{\theta}{n}+\frac{2k\pi}{n}\right)+i\sin\left(\frac{\theta}{n}+\frac{2k\pi}{n}\right) \right]. \end{eqnarray}

Each value of $k=0,1,2,\ldots ,n-1$ gives a different value of $w_k.$ Any other value of $k$ merely repeats one of the values of $w_k$ corresponding to $k=0,1,2,\ldots ,n-1.$ Thus there are exactly $n$th roots of a nonzero complex number.

Using Euler's formula: $$e^{i\theta}=\cos \theta +i \sin \theta,$$ the complex number $z=r(\cos \theta +i\sin \theta)$ can also be written in exponential form as $$z=re^{i\theta} = r \,\mbox{exp}(i \theta).$$

Thus, the $n$th roots of a complex number $z\neq 0$ can also be expressed as

\begin{eqnarray}\label{expform} w_k=z^{1/n}=\sqrt[n]{r}\;\mbox{exp}\left[i\left(\frac{\theta}{n}+\frac{2k\pi}{n}\right)\right] \end{eqnarray}

where $k=0, 1, 2, \ldots , n-1.$

Example: The complex number $z=i$ has three cube roots. In this case $r=1,$ and $\theta = \arg(z) = \pi/2.$ Thus the polar of the given number is \[ z = \cos \frac{\pi}{2} + i \sin\frac{\pi}{2}. \] Using (\ref{n-roots}) we then obtain

\[ w_k = \sqrt[3]{1}\left[\cos\left(\frac{\pi/2}{3}+\frac{2k\pi}{3}\right)+i\sin\left(\frac{\pi/2}{3}+\frac{2k\pi}{3}\right)\right] \]

with $k=0,1,2.$ Hence the three roots are \begin{eqnarray*} k=0,\quad w_0 &=& \cos\frac{\pi}{6} + i \sin\frac{\pi}{6} = \frac{\sqrt{3}}{2}+\frac{1}{2}i\\ k=1,\quad w_1 &=& \cos\frac{5\pi}{6} + i \sin\frac{5\pi}{6} = -\frac{\sqrt{3}}{2}+\frac{1}{2} i \\ k=2,\quad w_2 &=& \cos\frac{3\pi}{2} + i \sin\frac{3\pi}{2} = -i . \end{eqnarray*}

The applet below shows a geometrical and numerical representation of the $n$th roots of a complex number, up to $n=10.$ Drag the red point around to change the value of $z$ or activate the check box Specify values to drag the sliders. You can check the roots of the previous example.

Code

Enter the following script in GeoGebra to explore it yourself and make your own version. The symbol # indicates comments.

#Complex number
Z = 1 + ί

#Modulus of Z
r = abs(Z)

#Angle of Z
theta = arg(Z)

#Number of roots
n = Slider(2, 10, 1, 1, 150, false, true, false, false)

#Plot n-roots
nRoots = Sequence(r^(1/n) * exp( ί * ( theta/n + 2 * pi * k/n ) ), k, 0, n-1)

Exercise 1: From the exponential form (\ref{expform}) of the roots, show that all the $n$th roots lie on the circle $|z|=\sqrt[n]{r}$ about the origin and are equally spaced every $2\pi/n$ radians, starting with argument $\theta/n.$

Exercise 2: Find the four roots of $z=1+i.$ You can check your results in the previous applets.