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

1. Determine the position of point $Q$ representing $w$.

We are given $w = i\sqrt{3}z$.
The modulus is $|w| = |i\sqrt{3}| |z| = \sqrt{3}(1) = \sqrt{3}$.
The argument is $\arg(w) = \arg(i) + \arg(z) = \frac{\pi}{2} + \theta$.
Thus, $Q$ is obtained by rotating $P$ anticlockwise by $90^\circ$ about the origin, and scaling its distance from the origin by a factor of $\sqrt{3}$. Since $P$ is in the second quadrant, $Q$ will lie in the third quadrant, with $OQ \perp OP$ and $OQ = \sqrt{3}OP$.

2. Determine the position of point $R$ representing $z-w$.

The complex number $z-w$ corresponds to the position vector $\vec{OR} = \vec{OP} - \vec{OQ}$.
Rearranging this vector equation gives $\vec{OQ} + \vec{OR} = \vec{OP}$.
By the parallelogram law of vector addition, the vectors $\vec{OQ}$ and $\vec{OR}$ form the adjacent sides of a parallelogram with the diagonal being $\vec{OP}$.
Therefore, points $O, R, P,$ and $Q$ form a parallelogram $ORPQ$ in that order.

3. Show the geometrical relationship between $P, Q,$ and $R$ on the diagram.

To clearly show the relationship on the diagram, the student should:

• Draw the line segments $OR, RP, PQ,$ and $QO$ to construct the parallelogram $ORPQ$.
• Use parallel arrows on $OR$ and $QP$, and on $OQ$ and $RP$ to indicate they are parallel.
• Add a right-angle symbol (square) at $\angle POQ$ to signify that the side $OQ$ is perpendicular to the diagonal $OP$.

1. Determine the lengths of $OP$ and $OQ$ and the angle between them.

$|OP| = |z| = 1$

$|OQ| = |w| = |i\sqrt{3}z| = \sqrt{3}|z| = \sqrt{3}$

Since $w = i\sqrt{3}z$, multiplying $z$ by $i$ corresponds to an anticlockwise rotation of $\frac{\pi}{2}$ radians on the Argand diagram. Thus, $\angle POQ = 90^\circ$.

2. Deduce the shape of quadrilateral $ORPQ$ and calculate its area.

Using vector addition on the Argand diagram, $\vec{OR} + \vec{OQ} = (z-w) + w = z = \vec{OP}$.
This shows that $ORPQ$ is a parallelogram with adjacent sides $OR$ and $OQ$, and diagonal $OP$.

The area of the parallelogram is twice the area of the right-angled triangle $OPQ$:

$\text{Area of } ORPQ = 2 \times \text{Area of } \triangle OPQ$
$= 2 \times \left( \frac{1}{2} \times |OP| \times |OQ| \right)$
$= 1 \times \sqrt{3} = \sqrt{3}$

1. Express $z$ in polar form using its modulus and argument

Given $|z| = 1$ and $\arg(z) = \theta = \dfrac{3\pi}{4}$,
$z = 1 \left( \cos \dfrac{3\pi}{4} + i \sin \dfrac{3\pi}{4} \right)$

2. Evaluate the exact trigonometric ratios to find the Cartesian form

Since $\cos \dfrac{3\pi}{4} = -\dfrac{\sqrt{2}}{2}$ and $\sin \dfrac{3\pi}{4} = \dfrac{\sqrt{2}}{2}$,
$z = -\dfrac{\sqrt{2}}{2} + \dfrac{\sqrt{2}}{2}i$
(Alternatively, $-\dfrac{1}{\sqrt{2}} + \dfrac{1}{\sqrt{2}}i$)

1. Substitute $z$ from part (c) into the expression for $z-w$.

From part (c), we found $z = -\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i$.
$z - w = z - i\sqrt{3}z = z(1 - i\sqrt{3})$
$z - w = \left(-\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i\right)(1 - i\sqrt{3})$

2. Expand the product of the two complex numbers.

$z - w = -\frac{\sqrt{2}}{2} + \frac{\sqrt{6}}{2}i + \frac{\sqrt{2}}{2}i - \frac{\sqrt{6}}{2}i^2$
$= -\frac{\sqrt{2}}{2} + \frac{\sqrt{6}}{2}i + \frac{\sqrt{2}}{2}i + \frac{\sqrt{6}}{2}$
$= \left(\frac{\sqrt{6}}{2} - \frac{\sqrt{2}}{2}\right) + \left(\frac{\sqrt{6}}{2} + \frac{\sqrt{2}}{2}\right)i$

3. Factorise the real and imaginary parts to match the required form and identify $k$.

$= \frac{\sqrt{2}}{2}(\sqrt{3} - 1) + \frac{\sqrt{2}}{2}(\sqrt{3} + 1)i$
$= \frac{\sqrt{2}}{2}\bigl[ (\sqrt{3} - 1) + (\sqrt{3} + 1)i \bigr]$
Comparing this with $z - w = k\bigl[ (\sqrt{3}-1) + (\sqrt{3}+1)i \bigr]$, we have $k = \frac{\sqrt{2}}{2}$. (Alternatively, $k = \frac{1}{\sqrt{2}}$)

1. Find the exact argument of $z-w$ using the product of complex numbers.

$z-w = z - i\sqrt{3}z = z(1-i\sqrt{3})$
$\arg(z-w) = \arg(z) + \arg(1-i\sqrt{3})$
Since $\arg(z) = \frac{3\pi}{4}$ and $\arg(1-i\sqrt{3}) = \tan^{-1}\left(\frac{-\sqrt{3}}{1}\right) = -\frac{\pi}{3}$,
$\arg(z-w) = \frac{3\pi}{4} - \frac{\pi}{3} = \frac{9\pi}{12} - \frac{4\pi}{12} = \frac{5\pi}{12}$

2. Equate the tangent of the argument to the ratio of the imaginary and real parts from part (d) to show the result.

From part (d), $z-w = k\bigl[ (\sqrt{3}-1) + (\sqrt{3}+1)i \bigr]$.
Because $k>0$, both the real and imaginary parts are positive, meaning $z-w$ lies in the first quadrant.
For a complex number in the first quadrant, $\tan(\arg(z-w)) = \frac{\text{Im}(z-w)}{\text{Re}(z-w)}$.
$\tan\left(\frac{5\pi}{12}\right) = \frac{k(\sqrt{3}+1)}{k(\sqrt{3}-1)}$
$\tan\left(\frac{5\pi}{12}\right) = \frac{\sqrt{3}+1}{\sqrt{3}-1}$