Prove some properties of nth roots of complex numbers

Let $z = re^{i \theta}$ be a nonzero complex number where $\theta = \arg (z)$ . Then, let $R = r^{\frac{1}{n}}$ and $\alpha = \frac{\theta}{n}$ and define another complex number $z_1 = Re^{i \alpha}$ . Now, let $\varepsilon = e^{\frac{2 \pi i}{n}}$ for $n \in \mathbb{Z}_{>0}$ .

Prove that $z_1^n = z$ . We say $z_1$ is an $n$ th root of $z$ .
Prove that $z$ has exactly $n$ distinct $n$ th roots given by
$z_1, \varepsilon z_1, \varepsilon^2 z_1, \ldots, \varepsilon^{n-1} z_1,$

and that they are equally spaced on a circle of radius $R$ .
Find the three cube roots of $i$ .
Find the four fourth roots of $i$ .
Find the four fourth roots of $-i$ .

Proof. Using the definitions of $R$ and $\alpha$ we compute,
$z_1^n = \left( Re^{i \alpha} \right)^n = (R)^n (e^{i \alpha})^n = (r^{\frac{1}{n}})^n (e^{\frac{i \theta}{n}} )^n = re^{i \theta} = z. \qquad \blacksquare$
Proof. From part (a) we know $z_1 = Re^{i \alpha}$ is an $n$ th root of $z$ . Then, if $1 \leq m < n$ we have
$\left( \varepsilon^m z_1 \right)^n = e^{\frac{2 \pi i m}{n} \cdot n} \cdot z_1^n = e^{2 \pi i m} \cdot z = z$

(since $e^{2 \pi i m} = 1$ for all $m \in \mathbb{Z}_{>0}$ ). Hence, $\varepsilon^m z_1$ is an $n$ th root of $z$ .

Then, if $k \in \mathbb{Z}_{>0}$ and $k \geq n$ we have $k = j \cdot n + m$ for $1 \leq m < n$ . This implies

$\varepsilon^k = e^{2 \pi i(nj+m)} = 1 \cdot e^{2 \pi i m} = e^{2 \pi i m},$

for $1 \leq m < n$ ; hence, there are $n-1$ distinct values of $\varepsilon^k$ . Therefore, $z$ has exactly $n$ distinct $n$ th roots. By the Fundamental Theorem of Algebra we know it cannot have more than $n$ roots; hence, we have shown that they are all of the form $\varepsilon^m z_1. \qquad \blacksquare$
For the complex number $i$ we have
$i = e^{ \frac{\pi i}{2}} \quad \implies \quad r = 1, \ \ \theta = \frac{\pi}{2}.$

Therefore, from parts (a) and (b) we have

$R = 1, \qquad \alpha = \frac{\pi}{6} \qquad \varepsilon = e^{\frac{2 \pi i}{3}}.$

Hence, the three cube roots of $i$ are

$\begin{align*} z_1 &= e^{\frac{\pi i}{6}} \\[9pt] &= \cos \frac{\pi}{6} + i \sin \frac{\pi}{6} \\[9pt] &= \frac{1}{2} \sqrt{3} + \frac{1}{2} i. \\[9pt] z_2 &= e^{\frac{\pi i}{6}} \cdot e^{\frac{2 \pi i}{3}} = e^{\frac{5 \pi i}{6}} \\[9pt] &= \cos \frac{5 \pi}{6} + i \sin \frac{5 \pi}{6} \\[9pt] &= -\frac{1}{2} \sqrt{3} + \frac{1}{2} i. \\[9pt] z_3 &= e^{ \frac{\pi i}{6}} \cdot e^{\frac{4 \pi i}{3}} = e^{\frac{3 \pi i}{2}} \\[9pt] &= \cos \frac{3 \pi}{2} + i \sin \frac{3 \pi}{2} \\[9pt] &= -i. \end{align*}$
This time we have
$R = 1, \quad \alpha = \frac{\pi}{8}, \quad \implies \quad \varepsilon = e^{\frac{\pi i}{2}}.$

Therefore the four fourth roots of $i$ are

$\begin{align*} z_1 &= e^{\frac{\pi i}{8}} \\[9pt] &= \cos \frac{\pi}{8} + i \sin \frac{\pi}{8} \\[9pt] &= \frac{1}{2} \left( \sqrt{2 + \sqrt{2}} \right) + \frac{i}{2} \left( \sqrt{2 - \sqrt{2}} \right). \\[9pt] z_2 &= e^{\frac{5 \pi}{8}} \\[9pt] &= \cos \frac{5 \pi}{8} + i \sin \frac{5 \pi}{8} \\[9pt] &= -\frac{1}{2} \sqrt{2 - \sqrt{2}} + \frac{i}{2} \sqrt{2 + \sqrt{2}}. \\[9pt] z_3 &= e^{\frac{9 \pi i}{8}} \\[9pt] &= \cos \frac{9 \pi}{8} + i \sin \frac{9 \pi}{8} \\[9pt] &= -\frac{1}{2} \sqrt{2+ \sqrt{2}} - \frac{i}{2} \sqrt{2 - \sqrt{2}}. \\[9pt] z_4 &= e^{\frac{13 \pi i}{8}} \\[9pt] &= \cos \frac{13 \pi}{8} + i \sin \frac{13 \pi}{8} \\[9pt] &= \frac{1}{2}\sqrt{2 - \sqrt{2}} - \frac{i}{2} \sqrt{2 + \sqrt{2}}. \end{align*}$
For $-i$ we have
$-i = e^{\frac{3 \pi i}{2}} \quad \implies \quad r = 1, \ \ \theta = \frac{3 \pi}{2}.$

Therefore,

$R = 1, \ \ \alpha = \frac{3 \pi}{8} \quad \implies \quad \varepsilon = e^{\pi i}{2}.$

Thus,

$\begin{align*} z_1 &= e^{\frac{3 \pi i}{8}} \\[9pt] &= \cos \frac{3 \pi}{8} + i \sin \frac{3 \pi}{8} \\[9pt] &= \frac{1}{2} \left( \sqrt{2 - \sqrt{2}} \right) + \frac{i}{2} \left( \sqrt{2 + \sqrt{2}} \right). \\[9pt] z_2 &= e^{\frac{7 \pi}{8}} \\[9pt] &= \cos \frac{7 \pi}{8} + i \sin \frac{7 \pi}{8} \\[9pt] &= -\frac{1}{2} \sqrt{2 + \sqrt{2}} + \frac{i}{2} \sqrt{2 - \sqrt{2}}. \\[9pt] z_3 &= e^{\frac{11 \pi i}{8}} \\[9pt] &= \cos \frac{11 \pi}{8} + i \sin \frac{11 \pi}{8} \\[9pt] &= \frac{1}{2} \sqrt{2 - \sqrt{2}} - \frac{i}{2} \sqrt{2 - \sqrt{2}}. \\[9pt] z_4 &= e^{\frac{15 \pi i}{8}} \\[9pt] &= \cos \frac{15 \pi}{8} + i \sin \frac{15 \pi}{8} \\[9pt] &= \frac{1}{2}\sqrt{2 + \sqrt{2}} - \frac{i}{2} \sqrt{2 - \sqrt{2}}. \end{align*}$