解答本题时请勿使用计算器。
复数 $z$ 和 $w$ 具有相同的模 $r$，且它们的辐角分别为 $\alpha$ 和 $\beta$。

1. Express $z$ and $w$ in trigonometric form and write their sum.

$z = r(\cos\alpha + i\sin\alpha)$
$w = r(\cos\beta + i\sin\beta)$
$z + w = r(\cos\alpha + \cos\beta) + i r(\sin\alpha + \sin\beta)$

2. Apply sum-to-product trigonometric identities (factor formulae).

Using the factor formulae:
$\cos\alpha + \cos\beta = 2\cos\dfrac{\alpha + \beta}{2}\cos\dfrac{\alpha - \beta}{2}$
$\sin\alpha + \sin\beta = 2\sin\dfrac{\alpha + \beta}{2}\cos\dfrac{\alpha - \beta}{2}$

3. Substitute the identities into the sum and factor out the common term.

$z + w = r\left[ 2\cos\dfrac{\alpha + \beta}{2}\cos\dfrac{\alpha - \beta}{2} + i \left( 2\sin\dfrac{\alpha + \beta}{2}\cos\dfrac{\alpha - \beta}{2} \right) \right]$
$z + w = 2r\cos\dfrac{\alpha - \beta}{2} \left( \cos\dfrac{\alpha + \beta}{2} + i\sin\dfrac{\alpha + \beta}{2} \right)$

4. Deduce the expressions for $|z + w|$ and $\arg(z + w)$.

Comparing the expression to the standard polar form $R(\cos\theta + i\sin\theta)$:
$|z + w| = \left| 2r\cos\dfrac{\alpha - \beta}{2} \right|$
$\arg(z + w) = \dfrac{\alpha + \beta}{2}$ (assuming $\cos\dfrac{\alpha - \beta}{2} > 0$)

1. Identify the modulus and arguments from the given complex numbers.

From $z = 2e^{i\pi/12}$ and $w = 2e^{i5\pi/12}$, we have:
$r = 2$
$\alpha = \dfrac{\pi}{12}$
$\beta = \dfrac{5\pi}{12}$

2. Substitute the parameters into the expression for magnitude and evaluate.

Using the result from part (a):
$|z + w| = \left| 2r\cos\left( \dfrac{\alpha - \beta}{2} \right) \right|$
$|z + w| = \left| 2(2)\cos\left( \dfrac{\frac{\pi}{12} - \frac{5\pi}{12}}{2} \right) \right|$
$= \left| 4\cos\left( -\dfrac{4\pi}{24} \right) \right|$
$= 4\cos\left(-\dfrac{\pi}{6}\right)$
$= 4\left(\dfrac{\sqrt{3}}{2}\right) = 2\sqrt{3}$

3. Substitute the parameters into the expression for argument and evaluate.

Since $\cos(-\frac{\pi}{6}) > 0$, the magnitude factor is positive, so the argument is simply the angle from the polar form in part (a):
$\arg(z + w) = \dfrac{\alpha + \beta}{2}$
$\arg(z + w) = \dfrac{\frac{\pi}{12} + \frac{5\pi}{12}}{2}$
$= \dfrac{\frac{6\pi}{12}}{2} = \dfrac{\pi}{4}$

1. Find the modulus of $v$ using the properties of the modulus of products and quotients.

$|v| = \left|\frac{z^2}{w}\right| = \frac{|z|^2}{|w|}$
$|v| = \frac{2^2}{2} = 2$

2. Find the argument of $v$ using the properties of arguments.

$\arg(v) = \arg\left(\frac{z^2}{w}\right) = 2\arg(z) - \arg(w)$
$\arg(v) = 2\left(\frac{\pi}{12}\right) - \frac{5\pi}{12} = \frac{2\pi}{12} - \frac{5\pi}{12} = -\frac{3\pi}{12}$
$\arg(v) = -\frac{\pi}{4}$

1. Determine the modulus and argument of points $P$ and $Q$ from parts (b) and (c).

From part (b),
$|z+w| = 2(2)\cos\left(\frac{\pi/12 - 5\pi/12}{2}\right) = 4\cos\left(-\frac{\pi}{6}\right) = 2\sqrt{3}$
$\arg(z+w) = \frac{\pi/12 + 5\pi/12}{2} = \frac{\pi}{4}$

From part (c),
$|v| = \frac{|z|^2}{|w|} = \frac{2^2}{2} = 2$
$\arg(v) = 2\arg(z) - \arg(w) = 2\left(\frac{\pi}{12}\right) - \frac{5\pi}{12} = -\frac{\pi}{4}$

2. Find the angle $\angle POQ$ subtended at the origin.

$\angle POQ = \arg(z+w) - \arg(v)$
$\angle POQ = \frac{\pi}{4} - \left(-\frac{\pi}{4}\right) = \frac{\pi}{2}$

3. Calculate the area of $\triangle OPQ$.

$\text{Area of } \triangle OPQ = \frac{1}{2} |OP| |OQ| \sin(\angle POQ)$
$= \frac{1}{2} |z+w| |v| \sin\left(\frac{\pi}{2}\right)$
$= \frac{1}{2} (2\sqrt{3}) (2) (1)$
$= 2\sqrt{3}$