证明 $\frac{\text{e}^{\text{i}\theta}}{1 - \text{e}^{\text{i}\theta}}$ 可以表示为 $k\left(\text{i}\cot\frac{\theta}{2} - 1\right)$ 的形式，其中 $k$ 是一个待确定的实常数。 将复数 $\text{i}$ 表示为三种等价的 $r\text{e}^{\text{i}\theta}$ 形式，其中 $r > 0$ 且 $-3\pi < \theta \le 3\pi$。 由此求方程 $\left(\frac{w}{w+1}\right)^3 - \text{i} = 0$ 的根，并将你的答案表示为 $k(\text{i}\cot\phi - 1)$ 的形式，其中 $-\frac{\pi}{2} < \phi \le \frac{\pi}{2}$。

1. Factor out the half-angle exponential $\text{e}^{\text{i}\frac{\theta}{2}}$ from both the numerator and the denominator.

$\frac{\text{e}^{\text{i}\theta}}{1 - \text{e}^{\text{i}\theta}} = \frac{\text{e}^{\text{i}\frac{\theta}{2}} \cdot \text{e}^{\text{i}\frac{\theta}{2}}}{\text{e}^{\text{i}\frac{\theta}{2}} \left(\text{e}^{-\text{i}\frac{\theta}{2}} - \text{e}^{\text{i}\frac{\theta}{2}}\right)}$\n$= \frac{\text{e}^{\text{i}\frac{\theta}{2}}}{\text{e}^{-\text{i}\frac{\theta}{2}} - \text{e}^{\text{i}\frac{\theta}{2}}}$

2. Apply Euler's formula to convert the exponential terms into trigonometric form.

Numerator:
$\text{e}^{\text{i}\frac{\theta}{2}} = \cos\frac{\theta}{2} + \text{i}\sin\frac{\theta}{2}$
Denominator:
$\text{e}^{-\text{i}\frac{\theta}{2}} - \text{e}^{\text{i}\frac{\theta}{2}} = \left(\cos\frac{\theta}{2} - \text{i}\sin\frac{\theta}{2}\right) - \left(\cos\frac{\theta}{2} + \text{i}\sin\frac{\theta}{2}\right) = -2\text{i}\sin\frac{\theta}{2}$
Substitute these back into the fraction:
$\frac{\text{e}^{\text{i}\theta}}{1 - \text{e}^{\text{i}\theta}} = \frac{\cos\frac{\theta}{2} + \text{i}\sin\frac{\theta}{2}}{-2\text{i}\sin\frac{\theta}{2}}$

3. Separate the fraction and simplify to obtain the required format and determine $k$.

$= \frac{\cos\frac{\theta}{2}}{-2\text{i}\sin\frac{\theta}{2}} + \frac{\text{i}\sin\frac{\theta}{2}}{-2\text{i}\sin\frac{\theta}{2}}$
Since $\frac{1}{\text{i}} = -\text{i}$ and $\frac{\cos x}{\sin x} = \cot x$:
$= \frac{\text{i}}{2}\cot\frac{\theta}{2} - \frac{1}{2}$
$= \frac{1}{2}\left(\text{i}\cot\frac{\theta}{2} - 1\right)$
By comparing with the given form $k\left(\text{i}\cot\frac{\theta}{2} - 1\right)$, we have $k = \frac{1}{2}$.

1. Determine the modulus and principal argument of $\text{i}$.

$r = |\text{i}| = 1$
$\arg(\text{i}) = \frac{\pi}{2}$

2. Find all valid arguments within the specified range $-3\pi < \theta \le 3\pi$.

The general argument is $\theta = \frac{\pi}{2} + 2k\pi$, where $k \in \mathbb{Z}$.
For $k = -1$: $\theta = \frac{\pi}{2} - 2\pi = -\frac{3\pi}{2}$
For $k = 0$: $\theta = \frac{\pi}{2}$
For $k = 1$: $\theta = \frac{\pi}{2} + 2\pi = \frac{5\pi}{2}$
(Checking boundaries: $-3\pi < -\frac{3\pi}{2}, \frac{\pi}{2}, \frac{5\pi}{2} \le 3\pi$)

3. Write the three equivalent exponential forms.

The three forms $r\text{e}^{\text{i}\theta}$ are:
$\text{e}^{-\text{i}\frac{3\pi}{2}}, \text{e}^{\text{i}\frac{\pi}{2}}, \text{e}^{\text{i}\frac{5\pi}{2}}$

1. Briefly establish the results from parts (i) and (ii) required to solve the target.

From (i), finding $k$:
$\frac{\text{e}^{\text{i}\theta}}{1 - \text{e}^{\text{i}\theta}} = \frac{\text{e}^{\text{i}\theta/2}}{\text{e}^{-\text{i}\theta/2} - \text{e}^{\text{i}\theta/2}} = \frac{\text{e}^{\text{i}\theta/2}}{-2\text{i}\sin(\frac{\theta}{2})} = \frac{1}{2}\left(\text{i}\cot\frac{\theta}{2} - 1\right)$
Thus, $k = \frac{1}{2}$.

From (ii), expressing $\text{i}$ in $r\text{e}^{\text{i}\theta}$ form within $-3\pi < \theta \le 3\pi$:
$\text{i} = \text{e}^{-\text{i}\frac{3\pi}{2}}, \text{e}^{\text{i}\frac{\pi}{2}}, \text{e}^{\text{i}\frac{5\pi}{2}}$

2. Take the cube root of both sides to find the possible expressions for the base.

Given the equation:
$\left(\frac{w}{w+1}\right)^3 = \text{i}$

Taking the cube root using the three forms of $\text{i}$ from (ii):
$\frac{w}{w+1} = \text{e}^{\text{i}\theta}$
where $\theta = \frac{1}{3}\left(-\frac{3\pi}{2}\right), \frac{1}{3}\left(\frac{\pi}{2}\right), \frac{1}{3}\left(\frac{5\pi}{2}\right)$
So $\theta = -\frac{\pi}{2}, \frac{\pi}{6}, \frac{5\pi}{6}$

3. Solve for w algebraically.

$w = w\text{e}^{\text{i}\theta} + \text{e}^{\text{i}\theta}$
$w(1 - \text{e}^{\text{i}\theta}) = \text{e}^{\text{i}\theta}$
$w = \frac{\text{e}^{\text{i}\theta}}{1 - \text{e}^{\text{i}\theta}}$

Using the identity proven in part (i), we can rewrite this as:
$w = \frac{1}{2}\left(\text{i}\cot\frac{\theta}{2} - 1\right)$

4. Substitute the angles into the expression to find the final roots and check the domain of phi.

Letting $\phi = \frac{\theta}{2}$:
For $\theta = -\frac{\pi}{2}$, $\phi = -\frac{\pi}{4}$, so $w = \frac{1}{2}\left(\text{i}\cot\left(-\frac{\pi}{4}\right) - 1\right)$
For $\theta = \frac{\pi}{6}$, $\phi = \frac{\pi}{12}$, so $w = \frac{1}{2}\left(\text{i}\cot\left(\frac{\pi}{12}\right) - 1\right)$
For $\theta = \frac{5\pi}{6}$, $\phi = \frac{5\pi}{12}$, so $w = \frac{1}{2}\left(\text{i}\cot\left(\frac{5\pi}{12}\right) - 1\right)$

All $\phi$ values $\left\{-\frac{\pi}{4}, \frac{\pi}{12}, \frac{5\pi}{12}\right\}$ strictly satisfy $-\frac{\pi}{2} < \phi \le \frac{\pi}{2}$.