解方程 $z^4 = \sqrt{3} - i$，将根写成 $re^{i\theta}$ 的形式，其中 $r > 0$ 且 $-\pi < \theta \le \pi$。 在阿冈图上表示这些根，并写出这些根所在的几何图形的直角坐标方程。

1. Express the complex number $\sqrt{3} - i$ in exponential form.

Let $w = \sqrt{3} - i$.
Modulus $r = \sqrt{(\sqrt{3})^2 + (-1)^2} = \sqrt{3 + 1} = 2$
Argument $\theta = \tan^{-1}\left(\frac{-1}{\sqrt{3}}\right) = -\frac{\pi}{6}$
So, $w = 2e^{-i\frac{\pi}{6}}$.

To find all roots, introduce the general argument by adding multiples of $2\pi$:
$z^4 = 2e^{i\left(-\frac{\pi}{6} + 2k\pi\right)}$ for $k \in \mathbb{Z}$.

2. Take the 4th root to find the general expression for $z$.

$z = \left[2e^{i\left(-\frac{\pi}{6} + 2k\pi\right)}\right]^{\frac{1}{4}}$
$z = 2^{\frac{1}{4}} e^{i\left(-\frac{\pi}{24} + \frac{k\pi}{2}\right)}$

3. Substitute integer values of $k$ to find the four distinct roots within the principal argument range $-\pi < \theta \le \pi$.

For $k = 0$:
$z_1 = 2^{\frac{1}{4}} e^{-i\frac{\pi}{24}}$

For $k = 1$:
$z_2 = 2^{\frac{1}{4}} e^{i\left(-\frac{\pi}{24} + \frac{12\pi}{24}\right)} = 2^{\frac{1}{4}} e^{i\frac{11\pi}{24}}$

For $k = 2$:
$z_3 = 2^{\frac{1}{4}} e^{i\left(\frac{11\pi}{24} + \frac{12\pi}{24}\right)} = 2^{\frac{1}{4}} e^{i\frac{23\pi}{24}}$

For $k = -1$:
$z_4 = 2^{\frac{1}{4}} e^{i\left(-\frac{\pi}{24} - \frac{12\pi}{24}\right)} = 2^{\frac{1}{4}} e^{-i\frac{13\pi}{24}}$

All four roots have arguments within the interval $(-\pi, \pi]$.

1. Show the roots on an Argand diagram.

From part (i), the roots are of the form $z = 2^{\frac{1}{4}} e^{i\theta}$ where $\theta = -\frac{\pi}{24}, \frac{11\pi}{24}, \frac{23\pi}{24}, -\frac{13\pi}{24}$.

[Argand diagram sketch]

• Draw and label the Real (Re) and Imaginary (Im) axes.
• Plot the four points representing the roots.
• Each point is at a distance of $r = 2^{\frac{1}{4}}$ from the origin.
• The points are equally spaced by $\frac{\pi}{2}$ radians ($90^\circ$), with the first root positioned slightly below the positive Real axis (at an angle of $-\frac{\pi}{24}$).

2. Deduce the geometrical shape the roots lie on and state its cartesian equation.

Since all four roots have a constant modulus of $|z| = 2^{\frac{1}{4}}$, they are equidistant from the origin. Thus, the geometrical shape they lie on is a circle centered at the origin $(0,0)$.

The radius of the circle is $r = 2^{\frac{1}{4}}$.
The Cartesian equation of a circle centered at the origin is $x^2 + y^2 = r^2$.
$x^2 + y^2 = (2^{\frac{1}{4}})^2$
$x^2 + y^2 = \sqrt{2}$