Solve the equation $iz^5=-32$, giving your roots in the form $re^{i\theta}$, where $r>0$ and $-\pi<\theta\le\pi$. Sketch on an Argand diagram the points $P_1, P_2, P_3, P_4$ and $P_5$ representing these roots, where $P_1$ represents the root with the smallest argument and $P_1\;P_2\;P_3\;P_4\;P_5$ is a polygon described in an anticlockwise sense. Find the area of $P_1\;P_2\;P_3\;P_4\;P_5$. On an Argand diagram sketch clearly the locus of $P$ where $P$ represents the complex number $z$ such that $|z-2-2i|\le 1$ and $\arg(z-1)=\arg(1+3i)$. Find the range of values of $\arg(z-3-2i)$, given that $|z-2-2i|\le 1$ and $\arg(z-1)=\arg(1+3i)$.

1. Rearrange the equation and express the right-hand side in exponential form.

$iz^5 = -32 \implies z^5 = \frac{-32}{i} = 32i$
$32i = 32e^{i\frac{\pi}{2}}$

2. Formulate the general expression for the roots.

$z = \left( 32e^{i\left(\frac{\pi}{2} + 2k\pi\right)} \right)^{\frac{1}{5}} = 2e^{i\left(\frac{\pi}{10} + \frac{2k\pi}{5}\right)}, \quad k \in \mathbb{Z}$

3. Find the 5 specific roots in the required principal argument range.

Evaluating for $k = -2, -1, 0, 1, 2$ to keep $-\pi < \theta \le \pi$:
$k = -2 \implies z = 2e^{-i\frac{7\pi}{10}}$
$k = -1 \implies z = 2e^{-i\frac{3\pi}{10}}$
$k = 0 \implies z = 2e^{i\frac{\pi}{10}}$
$k = 1 \implies z = 2e^{i\frac{\pi}{2}}$
$k = 2 \implies z = 2e^{i\frac{9\pi}{10}}$

4. Sketch the Argand diagram and identify points $P_1$ to $P_5$.

The 5 points lie on a circle of radius 2 centered at the origin, separated by angles of $\frac{2\pi}{5}$.
$P_1$ has the smallest argument, so $P_1 = 2e^{-i\frac{7\pi}{10}}$ (in the 3rd quadrant).
Following an anticlockwise direction:
$P_2 = 2e^{-i\frac{3\pi}{10}}$ (in the 4th quadrant)
$P_3 = 2e^{i\frac{\pi}{10}}$ (in the 1st quadrant)
$P_4 = 2e^{i\frac{\pi}{2}}$ (on the positive imaginary axis)
$P_5 = 2e^{i\frac{9\pi}{10}}$ (in the 2nd quadrant)
(A regular pentagon is sketched on the Argand diagram with vertices at these points)

5. Calculate the area of the polygon $P_1P_2P_3P_4P_5$.

The regular pentagon can be divided into 5 identical isosceles triangles radiating from the origin. Each triangle has side lengths of 2 and an included angle of $\frac{2\pi}{5}$.
$\text{Area} = 5 \times \left( \frac{1}{2}(2)(2)\sin\left(\frac{2\pi}{5}\right) \right)$
$\text{Area} = 10\sin\left(\frac{2\pi}{5}\right) \approx 9.51 \text{ units}^2$

1. Identify the two loci and formulate equations to find their intersection.

The inequality $|z - (2+2i)| \le 1$ represents the region on and inside a circle centred at $2+2i$ with radius $1$.

The equation $\arg(z-1) = \arg(1+3i)$ represents a half-line (ray) starting from the point $(1,0)$ with a direction vector corresponding to $1+3i$. This ray lies on the line passing through $(1,0)$ with gradient $3$, so its Cartesian equation is $y = 3(x-1)$ for $x > 1$.

To sketch the locus, we must find the intersection of this ray with the boundary of the circle, $(x-2)^2 + (y-2)^2 = 1$.

2. Solve for the endpoints of the line segment representing the locus of $P$.

Substitute $y = 3x - 3$ into the circle's boundary equation:
$(x-2)^2 + (3x-5)^2 = 1$
$x^2 - 4x + 4 + 9x^2 - 30x + 25 = 1$
$10x^2 - 34x + 28 = 0$
$5x^2 - 17x + 14 = 0$
$(5x-7)(x-2) = 0$

Solving gives $x = 1.4$ or $x = 2$.
When $x = 1.4$, $y = 3(1.4) - 3 = 1.2$.
When $x = 2$, $y = 3(2) - 3 = 3$.
Both points satisfy $x > 1$ and $y > 0$, so they lie on the half-line.
The locus of $P$ is the line segment connecting $1.4+1.2i$ and $2+3i$. (The sketch should show this specific line segment bounded within the circle).

3. Evaluate the argument of $z-3-2i$ at the endpoints of the locus.

Let $w = z - (3+2i)$. We need the range of $\arg(w)$ as $z$ varies along the segment.

At the endpoint $z_1 = 2+3i$:
$w_1 = (2+3i) - (3+2i) = -1 + i$
$\arg(w_1) = \pi - \tan^{-1}(1) = \frac{3\pi}{4}$

At the endpoint $z_2 = 1.4+1.2i$:
$w_2 = (1.4+1.2i) - (3+2i) = -1.6 - 0.8i$
Since this point is in the third quadrant:
$\arg(w_2) = -\pi + \tan^{-1}\left(\frac{0.8}{1.6}\right) = -\pi + \tan^{-1}\left(\frac{1}{2}\right)$

4. Determine the full range of $\arg(z-3-2i)$ by checking the transition of the angle along the segment.

As $z$ moves along the segment from $2+3i$ to $1.4+1.2i$, the vector $w$ moves along a straight line from $-1+i$ to $-1.6-0.8i$.

This path crosses the negative real axis when the imaginary part of $w$ is zero. This occurs at $y = 2$.
On the line $y = 3x - 3$, when $y = 2$, $x = \frac{5}{3}$.
At this intercept, $w = \left(\frac{5}{3}-3\right) + 0i = -\frac{4}{3}$, which has an argument of exactly $\pi$.

Thus, the argument monotonically increases from $\frac{3\pi}{4}$ up to $\pi$ in the second quadrant. Upon crossing the negative real axis, the principal argument wraps to $-\pi$ and increases to $-\pi + \tan^{-1}\left(\frac{1}{2}\right)$ in the third quadrant.

The range of values is therefore:
$\frac{3\pi}{4} \le \arg(z-3-2i) \le \pi$ or $-\pi < \arg(z-3-2i) \le -\pi + \tan^{-1}\left(\frac{1}{2}\right)$