Do not use a calculator in answering this question. The complex number $z$ has modulus 1 and argument $\theta$, where $\frac{\pi}{2} < \theta < \pi$, and the complex number $w$ is given by $w = \mathrm{i}\sqrt{3}z$. The point $P$ on the Argand diagram represents $z$.

1. Determine the position of $Q$ based on the relationship $w = \mathrm{i}\sqrt{3}z$.

Since $w = \mathrm{i}\sqrt{3}z$, the modulus is $|w| = |\mathrm{i}\sqrt{3}| |z| = \sqrt{3}|z|$.
The argument is $\arg(w) = \arg(\mathrm{i}) + \arg(z) = \frac{\pi}{2} + \theta$.
Thus, to plot $Q$, we rotate the segment $OP$ by $90^\circ$ ($\frac{\pi}{2}$ radians) anticlockwise and scale its length by a factor of $\sqrt{3}$. Because $P$ is in the 2nd quadrant, adding $90^\circ$ places $Q$ in the 3rd quadrant.

2. Determine the position of $R$ using vector addition/subtraction.

The point $R$ represents $z - w$. In vector form, $\vec{OR} = \vec{OP} - \vec{OQ}$.
We can rearrange this as $\vec{OP} = \vec{OR} + \vec{OQ}$, meaning that $P$ is the fourth vertex of a parallelogram formed by $O, R,$ and $Q$.
Alternatively, observe that $\vec{QP} = \vec{OP} - \vec{OQ} = z - w = \vec{OR}$. Therefore, $R$ is obtained by drawing a vector from $O$ that is parallel and equal in length to $QP$.

3. State and visually indicate the required geometrical relationship.

Because $\vec{OR} = \vec{QP}$ and $\vec{OQ} = \vec{RP}$, the four points $O, R, P,$ and $Q$ form a parallelogram ($ORPQ$).
To 'show clearly' this relationship on the Argand diagram, the line segments $OR, RP, PQ,$ and $QO$ should be drawn (often with dashed lines for $RP$ and $QP$) to explicitly display the parallelogram.

1. Deduce the shape of the quadrilateral $ORPQ$.

Let the position vectors in the Argand diagram be $\vec{OP} = z$, $\vec{OQ} = w$, and $\vec{OR} = z - w$.
We find the vector $\vec{QP}$:
$\vec{QP} = \vec{OP} - \vec{OQ} = z - w$
Since $\vec{OR} = z - w = \vec{QP}$, the line segments $OR$ and $QP$ are parallel and equal in length. Therefore, the quadrilateral $ORPQ$ is a parallelogram.

2. Evaluate the lengths of the line segments $OP$ and $OQ$, and the angle between them.

Using the given $|z| = 1$ and $w = \mathrm{i}\sqrt{3}z$:
$|OP| = |z| = 1$
$|OQ| = |w| = |\mathrm{i}\sqrt{3}z| = \sqrt{3}|z| = \sqrt{3}$
The factor of $\mathrm{i}$ in $w = \mathrm{i}\sqrt{3}z$ indicates that the vector $\vec{OQ}$ is rotated $\frac{\pi}{2}$ radians ($90^\circ$) anticlockwise from $\vec{OP}$.
Thus, $OP \perp OQ$.

3. Calculate the area of the parallelogram.

Treating $OQ$ as the base of the parallelogram $ORPQ$:
Since $OP \perp OQ$, the perpendicular height of the parallelogram from the line $OQ$ to the opposite side $RP$ is exactly the length $|OP|$.
$\text{Area of } ORPQ = \text{base} \times \text{height} = |OQ| \times |OP|$
$\text{Area} = \sqrt{3} \times 1 = \sqrt{3}$

1. Substitute the given modulus and argument into the trigonometric form of a complex number

A complex number $z$ with modulus $r$ and argument $\theta$ is given by $z = r(\cos\theta + \mathrm{i}\sin\theta)$.

Given $|z| = 1$ (from the main question stem) and $\theta = \frac{3\pi}{4}$,
$z = 1\left(\cos\frac{3\pi}{4} + \mathrm{i}\sin\frac{3\pi}{4}\right)$

2. Evaluate the exact trigonometric values to express $z$ in Cartesian form

Since calculators are not allowed, we use the exact values for the trigonometric ratios:
$\cos\frac{3\pi}{4} = -\frac{\sqrt{2}}{2}$
$\sin\frac{3\pi}{4} = \frac{\sqrt{2}}{2}$

Substituting these in gives:
$z = -\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}\mathrm{i}$

1. Find the cartesian form of $z$ using the given modulus and argument.

From part (c), we are given $\theta = \frac{3\pi}{4}$ and $|z| = 1$.
$z = \cos\frac{3\pi}{4} + \mathrm{i}\sin\frac{3\pi}{4}$
$z = -\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}\mathrm{i}$

2. Substitute $z$ into the expression for $z - w$ and expand.

Substitute $w = \mathrm{i}\sqrt{3}z$ into $z - w$:
$z - w = z - \mathrm{i}\sqrt{3}z = z(1 - \mathrm{i}\sqrt{3})$
$z - w = \left( -\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}\mathrm{i} \right)(1 - \mathrm{i}\sqrt{3})$
Expanding the product:
$= -\frac{\sqrt{2}}{2} + \mathrm{i}\frac{\sqrt{6}}{2} + \mathrm{i}\frac{\sqrt{2}}{2} - \mathrm{i}^2\frac{\sqrt{6}}{2}$
$= -\frac{\sqrt{2}}{2} + \frac{\sqrt{6}}{2} + \left( \frac{\sqrt{6}}{2} + \frac{\sqrt{2}}{2} \right)\mathrm{i}$

3. Factorise the expanded expression to match the required form and determine $k$.

$z - w = \left( \frac{\sqrt{6} - \sqrt{2}}{2} \right) + \left( \frac{\sqrt{6} + \sqrt{2}}{2} \right)\mathrm{i}$
Extracting $\frac{\sqrt{2}}{2}$ as a common factor:
$z - w = \frac{\sqrt{2}}{2} \left[ (\sqrt{3} - 1) + (\sqrt{3} + 1)\mathrm{i} \right]$
Comparing this to $k\left[\left(\sqrt{3} - 1\right) + \left(\sqrt{3} + 1\right)\mathrm{i}\right]$, we can see that the expression matches the required form with:
$k = \frac{\sqrt{2}}{2}$

1. Find the argument of $z - w$ algebraically by factorising.

$z - w = z - \mathrm{i}\sqrt{3}z = z(1 - \mathrm{i}\sqrt{3})$
$\arg(z - w) = \arg(z) + \arg(1 - \mathrm{i}\sqrt{3})$
From part (c), $\arg(z) = \theta = \frac{3\pi}{4}$. The complex number $1 - \mathrm{i}\sqrt{3}$ lies in the fourth quadrant:
$\arg(1 - \mathrm{i}\sqrt{3}) = \tan^{-1}\left(\frac{-\sqrt{3}}{1}\right) = -\frac{\pi}{3}$
$\arg(z - w) = \frac{3\pi}{4} - \frac{\pi}{3} = \frac{9\pi}{12} - \frac{4\pi}{12} = \frac{5\pi}{12}$

2. Find the argument of $z - w$ using the Cartesian form from part (d).

From part (d), $z - w = k\left[(\sqrt{3} - 1) + (\sqrt{3} + 1)\mathrm{i}\right]$, where $k = \frac{\sqrt{2}}{2}$.
Since $k > 0$, $\sqrt{3} - 1 > 0$, and $\sqrt{3} + 1 > 0$, the complex number $z - w$ lies in the first quadrant.
$\arg(z - w) = \tan^{-1}\left(\frac{k(\sqrt{3} + 1)}{k(\sqrt{3} - 1)}\right) = \tan^{-1}\left(\frac{\sqrt{3} + 1}{\sqrt{3} - 1}\right)$

3. Equate the two expressions for the argument to prove the identity.

Equating the two values for $\arg(z - w)$:
$\frac{5\pi}{12} = \tan^{-1}\left(\frac{\sqrt{3} + 1}{\sqrt{3} - 1}\right)$
Taking the tangent of both sides:
$\tan\frac{5\pi}{12} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \quad \text{(shown)}$