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

Suppose that $w=r(\cos \theta +i\sin \theta)$ and $z=\rho (\cos \psi +i\sin \psi)$. Then de Moivre's formula gives $$z^n=\rho^n(\cos n\psi+i\sin n\psi).$$ 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

\[z=\sqrt[n]{r}\left[\cos\left(\frac{\theta}{n}+\frac{2k\pi}{n}\right)+i\sin\left(\frac{\theta}{n}+\frac{2k\pi}{n}\right) \right].\]

Each value of $k=0,1,2,\ldots ,n-1$ gives a different value of $z$. Any other value of $k$ merely repeats one of the values of $z$ 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
$$z=re^{i\theta}$$

Thus, the $n$th roots of a nonzero complex number $z\neq 0$ can also be expressed as \begin{eqnarray}\label{expform} z=\sqrt[n]{r}\;\mbox{exp}\left[i\left(\frac{\theta}{n}+\frac{2k\pi}{n}\right)\right] \end{eqnarray} where $k=0, 1, 2, \ldots , n-1$.

The applet below shows a geometrical 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 drag the sliders.

**Exercise:** 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 $\theta/n$.

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