Let \(n \in \mathbb{N}\).
We say that \(b\) is an **\(n\)th root** of \(a\) if \(b^n = a\).
When \(n = 2\), \(b\) is called a **square root** of \(a\).
When \(n = 3\), \(b\) is called a **cube root** of \(a\).

It is well known that every positive real number has two square roots. For example, the square roots of \(4\) are \(2\) and \(-2\) because \(2^2 = 4\) and \((-2)^2 = 4\). We now turn to finding \(n\)th roots for complex numbers.

Consider the complex numbers \(z = 2\operatorname{cis}\frac{\pi}{6}\) and \(w = 2\operatorname{cis}\frac{5\pi}{6}\). By De Moivre's Formula, we get \[z^3 = 2^3 \operatorname{cis}\left(3\cdot \frac{\pi}{6}\right) = 8 \operatorname{cis}\frac{\pi}{2}\] and \[w^3 = 2^3 \operatorname{cis}\left(3\cdot \frac{5\pi}{6}\right) = 8 \operatorname{cis}\frac{5\pi}{2} = 8 \operatorname{cis}\left(2\pi + \frac{\pi}{2}\right) = 8 \operatorname{cis}\frac{\pi}{2}.\] Even though \(z\neq w\), we have \(z^3 = w^3 = 8 \operatorname{cis} \frac{\pi}{2}\). Hence, \(z\) and \(w\) are distinct cube roots of \(8 \operatorname{cis} \frac{\pi}{2}\). (Check that \(2\operatorname{cis}\frac{3\pi}{2}\) is also a cube root of \(8 \operatorname{cis} \frac{\pi}{2}\).)

Observe that the argument of \(w\), which is \(\frac{5\pi}{6}\), is \(\frac{2\pi}{3}\) more than the argument of \(z\), which is \(\frac{\pi}{6}\). Even if we did not know the actual arguments of \(z\) and \(w\), we could tell from this observation that the argument of \(w^3\) is \(2\pi\) more than the argument of \(z^3\) since three times \(\frac{2\pi}{3}\) is \(2\pi\). As complex numbers in polar form having the same modulus but with arguments differing by integer multiplies of \(2\pi\) are the same, we see that the cubes of \(z\) and \(w\) are the same complex number.

In general, with \(n \in \mathbb{N}\) and \(r > 0\), the \(n\)th power of each of the following \(n\) complex numbers is equal to \(r \operatorname{cis} \theta\): \[ r^{\frac{1}{n}}\operatorname{cis}\frac{\theta}{n}, r^{\frac{1}{n}}\operatorname{cis}\left(\frac{2\pi}{n}+\frac{\theta}{n}\right), r^{\frac{1}{n}}\operatorname{cis}\left(\frac{4\pi}{n}+\frac{\theta}{n}\right), \ldots, r^{\frac{1}{n}}\operatorname{cis}\left(\frac{2(n-1)\pi}{n}+\frac{\theta}{n}\right). \] These are all the \(n\)th roots of \(r \operatorname{cis} \theta\).

Find all cube roots of \(-8\).

The polar form of \(-8\) is \(8 \operatorname{cis} \pi\). Hence, applying the generic list above with \(\theta = \pi\) and \(n = 3\), we obtain the cube roots \[2\operatorname{cis}\frac{\pi}{3},~ 2\operatorname{cis}\left(\frac{2\pi}{3}+\frac{\pi}{3}\right),~ 2\operatorname{cis}\left(\frac{4\pi}{3} + \frac{\pi}{3}\right).\] Simplifying gives \[2\operatorname{cis}\frac{\pi}{3},~ 2\operatorname{cis}\pi,~ 2\operatorname{cis}\frac{5\pi}{3}.\] Converting to rectangular form gives \[1+\sqrt{3}i,~ -2,~ 1-\sqrt{3}i\]

Find all square roots of \(17-14i\). Express the answers in rectangular form with real and imaginary parts rounded to 4 decimal places.

Find all 4th roots of \(-1\). Express the answers in rectangular form with real and imaginary parts rounded to 4 decimal places.