解答本题时请勿使用计算器。 已知三个复数为 $z_1 = 1 + i$，$z_2 = \sqrt{3} - i$ 和 $z_3 = \cos \dfrac{\pi}{3} + i\sin \dfrac{\pi}{3}$。

1. Find the modulus and argument of $z_1$.

$|z_1| = \sqrt{1^2 + 1^2} = \sqrt{2}$

$\arg(z_1) = \tan^{-1}\left(\frac{1}{1}\right) = \frac{\pi}{4}$

$z_1 = \sqrt{2}\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right)$

2. Find the modulus and argument of $z_2$.

$|z_2| = \sqrt{(\sqrt{3})^2 + (-1)^2} = \sqrt{3 + 1} = 2$

Since $z_2$ is in the fourth quadrant:
$\arg(z_2) = -\tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = -\frac{\pi}{6}$

$z_2 = 2\left(\cos\left(-\frac{\pi}{6}\right) + i\sin\left(-\frac{\pi}{6}\right)\right)$

3. Identify the modulus and argument of $z_3$ and compute the modulus of the combined expression.

From $z_3 = \cos\frac{\pi}{3} + i\sin\frac{\pi}{3}$, we have $|z_3| = 1$ and $\arg(z_3) = \frac{\pi}{3}$.

Let $Z = \dfrac{z_1 z_2}{z_3}$.

$|Z| = \dfrac{|z_1| \times |z_2|}{|z_3|} = \dfrac{\sqrt{2} \times 2}{1} = 2\sqrt{2}$

4. Compute the principal argument of the combined expression.

$\arg(Z) = \arg(z_1) + \arg(z_2) - \arg(z_3)$
$\arg(Z) = \frac{\pi}{4} + \left(-\frac{\pi}{6}\right) - \frac{\pi}{3}$
$\arg(Z) = \frac{3\pi - 2\pi - 4\pi}{12} = -\frac{3\pi}{12} = -\frac{\pi}{4}$

Since $-\pi < -\frac{\pi}{4} \le \pi$, this is the principal argument.

5. Write the final complex number in the required polar form.

Therefore, the complex number in the required form is:
$\dfrac{z_1 z_2}{z_3} = 2\sqrt{2}\left(\cos\left(-\frac{\pi}{4}\right) + i\sin\left(-\frac{\pi}{4}\right)\right)$

1. Substitute $z_4$ and multiply by the complex conjugate of the denominator.

$\frac{1 + z_4}{1 - z_4} = \frac{1 + \cos\theta + i\sin\theta}{1 - \cos\theta - i\sin\theta}$
Multiply the numerator and denominator by $(1 - \cos\theta) + i\sin\theta$:
$= \frac{((1 + i\sin\theta) + \cos\theta)}{((1 - \cos\theta) - i\sin\theta)} \times \frac{((1 - \cos\theta) + i\sin\theta)}{((1 - \cos\theta) + i\sin\theta)}$

2. Expand and simplify the numerator using the Pythagorean identity.

$\text{Numerator} = ((1 + i\sin\theta) + \cos\theta)((1 + i\sin\theta) - \cos\theta)$
$= (1 + i\sin\theta)^2 - \cos^2\theta$
$= 1 + 2i\sin\theta + i^2\sin^2\theta - \cos^2\theta$
$= 1 + 2i\sin\theta - \sin^2\theta - \cos^2\theta$
$= 1 - (\sin^2\theta + \cos^2\theta) + 2i\sin\theta$
$= 1 - 1 + 2i\sin\theta = 2i\sin\theta$

3. Expand and simplify the denominator.

$\text{Denominator} = (1 - \cos\theta - i\sin\theta)(1 - \cos\theta + i\sin\theta)$
$= (1 - \cos\theta)^2 - (i\sin\theta)^2$
$= 1 - 2\cos\theta + \cos^2\theta - i^2\sin^2\theta$
$= 1 - 2\cos\theta + \cos^2\theta + \sin^2\theta$
$= 1 - 2\cos\theta + 1 = 2 - 2\cos\theta$

4. Combine the fraction and apply double-angle formulas to show the final form.

$\frac{1 + z_4}{1 - z_4} = \frac{2i\sin\theta}{2 - 2\cos\theta} = \frac{i\sin\theta}{1 - \cos\theta}$
Apply the double-angle formulas $\sin\theta = 2\sin\frac{\theta}{2}\cos\frac{\theta}{2}$ and $1 - \cos\theta = 2\sin^2\frac{\theta}{2}$:
$= \frac{i(2\sin\frac{\theta}{2}\cos\frac{\theta}{2})}{2\sin^2\frac{\theta}{2}}$
$= i\left(\frac{\cos\frac{\theta}{2}}{\sin\frac{\theta}{2}}\right) = i\cot\frac{\theta}{2}$
Therefore, $k = i$.

1. Evaluate $z_4$ at $\theta = \frac{\pi}{4}$ in Cartesian form.

$z_4 = \cos\frac{\pi}{4} + i\sin\frac{\pi}{4} = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i$

2. Substitute $z_4$ into the given expression and simplify the fraction.

$\frac{1 + z_4}{1 - z_4} = \frac{1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i}{1 - \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}i} = \frac{\sqrt{2} + 1 + i}{\sqrt{2} - 1 - i}$

3. Multiply the numerator and denominator by the complex conjugate of the denominator.

$\begin{aligned} \frac{\sqrt{2} + 1 + i}{\sqrt{2} - 1 - i} &= \frac{(\sqrt{2} + 1 + i)((\sqrt{2} - 1) + i)}{((\sqrt{2} - 1) - i)((\sqrt{2} - 1) + i)} \\ &= \frac{(\sqrt{2} + 1)(\sqrt{2} - 1) - 1 + i(\sqrt{2} - 1 + \sqrt{2} + 1)}{(\sqrt{2} - 1)^2 + 1} \\ &= \frac{(2 - 1) - 1 + 2\sqrt{2}i}{2 - 2\sqrt{2} + 1 + 1} \\ &= \frac{2\sqrt{2}i}{4 - 2\sqrt{2}} \end{aligned}$

4. Simplify the expression and rationalise the denominator to obtain the final result.

$\begin{aligned} \frac{2\sqrt{2}i}{4 - 2\sqrt{2}} &= \frac{\sqrt{2}i}{2 - \sqrt{2}} \\ &= \frac{\sqrt{2}i(2 + \sqrt{2})}{(2 - \sqrt{2})(2 + \sqrt{2})} \\ &= \frac{2\sqrt{2} + 2}{4 - 2}i \\ &= \frac{2(\sqrt{2} + 1)}{2}i \\ &= (1 + \sqrt{2})i \end{aligned}$

1. Determine the constant $k$ from part (b)(i) by simplifying the general expression.

Using half-angle identities to simplify the given expression:
$\frac{1 + z_4}{1 - z_4} = \frac{1 + \cos\theta + i\sin\theta}{1 - \cos\theta - i\sin\theta}$
$= \frac{2\cos^2\frac{\theta}{2} + 2i\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\sin^2\frac{\theta}{2} - 2i\sin\frac{\theta}{2}\cos\frac{\theta}{2}}$
$= \frac{2\cos\frac{\theta}{2}(\cos\frac{\theta}{2} + i\sin\frac{\theta}{2})}{-2i\sin\frac{\theta}{2}(\cos\frac{\theta}{2} + i\sin\frac{\theta}{2})}$
$= \frac{\cos\frac{\theta}{2}}{-i\sin\frac{\theta}{2}} = i\cot\frac{\theta}{2}$
Thus, $k = i$.

2. Equate the general result with the specific case $\theta = \frac{\pi}{4}$ from (b)(ii) to find an equation for $\cot \frac{\pi}{8}$.

Substitute $\theta = \frac{\pi}{4}$ into the simplified expression $i\cot\frac{\theta}{2}$:
$\frac{1 + z_4}{1 - z_4} = i\cot\frac{\pi}{8}$
From part (b)(ii), we are given that for $\theta = \frac{\pi}{4}$, the expression evaluates to $(1 + \sqrt{2})i$. Equating the two forms:
$i\cot\frac{\pi}{8} = (1 + \sqrt{2})i$
$\cot\frac{\pi}{8} = 1 + \sqrt{2}$

3. Solve for $\tan \frac{\pi}{8}$ by taking the reciprocal and rationalising the denominator.

$\tan\frac{\pi}{8} = \frac{1}{\cot\frac{\pi}{8}} = \frac{1}{1 + \sqrt{2}}$
Rationalising the denominator by multiplying the numerator and denominator by the conjugate $(\sqrt{2} - 1)$:
$\tan\frac{\pi}{8} = \frac{\sqrt{2} - 1}{(\sqrt{2} + 1)(\sqrt{2} - 1)}$
$= \frac{\sqrt{2} - 1}{2 - 1} = \sqrt{2} - 1$