Roots of unity for beginners 3

$\begin{aligned} x^8 & = 16 = 16(1+0i) = 16(\color{#3D99F6}{\cos 2\color{#D61F06}{n}\pi + i \sin 2\color{#D61F06}{n} \pi}) = 16\color{#3D99F6}{e^{2\color{#D61F06}{n}\pi i}} & \small \color{#3D99F6}{\text{Euler's identity}} \text{ and } \color{#D61F06}{n = 0,1,2,3,4,5,6,7} \\ \implies x & = 16^{\frac{1}{8}} e^{\frac{2n\pi}{8}i} = \sqrt{2} e^{\frac{n\pi}{4}i} = \sqrt{2}(\cos \frac{n\pi}{4} + i \sin \frac{n\pi}{4}) \\ & = \begin{cases} \sqrt{2}(\cos 0 + i \sin 0) & = \color{#D61F06}{\sqrt{2}} & \color{#D61F06}{\text{rejected}}\\ \sqrt{2}(\cos \frac{\pi}{4} + i \sin \frac{\pi}{4}) & = \color{#3D99F6}{1 + i} & \color{#3D99F6}{\text{accepted}} \\ \sqrt{2}(\cos \frac{\pi}{2} + i \sin \frac{\pi}{2}) & = \color{#D61F06}{\sqrt{2}i} & \color{#D61F06}{\text{rejected}}\\ \sqrt{2}(\cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4}) & = \color{#3D99F6}{-1 + i} & \color{#3D99F6}{\text{accepted}} \\ \sqrt{2}(\cos \pi + i \sin \pi) & = \color{#D61F06}{-\sqrt{2}} & \color{#D61F06}{\text{rejected}}\\ \sqrt{2}(\cos \frac{5\pi}{4} + i \sin \frac{5\pi}{4}) & = \color{#3D99F6}{-1 - i} & \color{#3D99F6}{\text{accepted}} \\ \sqrt{2}(\cos \frac{3\pi}{2} + i \sin \frac{3\pi}{2}) & = \color{#D61F06}{-\sqrt{2}i} & \color{#D61F06}{\text{rejected}} \\ \sqrt{2}(\cos \frac{7\pi}{4} + i \sin \frac{7\pi}{4}) & = \color{#3D99F6}{1 - i} & \color{#3D99F6}{\text{accepted}} \end{cases} \end{aligned}$

Therefore, there are $\color{#3D99F6}{\boxed{4} \text{ acceptable}}$ roots.

We have $\dfrac{{x}^{8}}{16}=1 \Rightarrow {(\dfrac{x}{\sqrt{2}})}^{8}=1$ . Now imagine a unit circle in the Argand plane: solutions for $\dfrac{x}{\sqrt{2}}$ will lie on the circumference of the circle at intervals of $\dfrac{2\pi}{8}$ , i.e. every $\dfrac{\pi}{4}$ .

Due to symmetry, it suffices to consider the roots $\dfrac{x}{\sqrt{2}}=1, ~\cos(\dfrac{\pi}{4})$ . Therefore $x=1 \times \sqrt{2}, ~\dfrac{\sqrt{2}}{2} \times \sqrt{2} \Rightarrow x= \sqrt{2}, ~1$ .

This means half of the solutions for ${(\dfrac{x}{\sqrt{2}})}^{8}=1$ are integers i.e $\large\color{#20A900}{\boxed{4}}$ solutions.

Since $x^8=16$ is of eighth degree, it has 8 roots. Due to the properties of roots of unities, we know that these 8 roots will be equally spaced out on a circle centered on the origin of the complex plane. Clearly, one of the roots will be $\sqrt[4]{16} = \sqrt{2}$ . In the polar complex plane, this root lies on the x-axis. Since there are 8 equally spaced roots, each root will be $\dfrac{2\pi}{8} =\dfrac{\pi}{4}$ radians apart. One root lies on the x-axis ( $\theta = 0$ ), therefore possible values for $\theta$ are $\dfrac{n\pi}{4}$ where $0 \geq n \geq 7$ and $n \in \mathbb{Z}$ .

Since the first solution is $\sqrt{2}$ from the origin, each additional solution will also be $\sqrt{2}$ from the origin because they are on a circle about the origin. Therefore, each solution will be in the form $\sqrt{2}(\cos{\theta}+i\sin{\theta})$ . For both the imaginary and real parts to have integer coefficients, $\sqrt{2}\cos{\theta}$ and $\sqrt{2}\sin{\theta}$ must simultaneously be integers. From this and our possible values for $\theta$ it is clear that there are four values that satisfy this condition.

N.B. I could explain more thoroughly how I know there are four satisfying values, however this problem has to do with roots of unity, not special angles or simple geometry.