The complex number $z$ satisfies the following inequalities:
$|z|^2 \le 4 \quad \text{and} \quad -\frac{\pi}{6} \le \arg(z + \sqrt{3} - \mathrm{i}) \le 0.$

1. Interpret the modulus inequality $|z|^2 \le 4$.

$|z|^2 \le 4 \implies |z| \le 2$
This defines the region inside and on the boundary of a circle centred at the origin $(0,0)$ with radius $2$.

2. Interpret the argument inequality.

$-\frac{\pi}{6} \le \arg(z - (-\sqrt{3} + \mathrm{i})) \le 0$
This defines a region bounded by two half-lines emanating from the point $A(-\sqrt{3}, 1)$.
The first half-line has an angle of $0$ (horizontal, extending to the right).
The second half-line has an angle of $-\frac{\pi}{6}$ (directed downwards to the right).

3. Determine the position of the starting point and the intersections of the half-lines with the circle.

The modulus of $A(-\sqrt{3}, 1)$ is $\sqrt{(-\sqrt{3})^2 + 1^2} = 2$, so $A$ lies exactly on the circle $|z| = 2$.
The horizontal half-line $y=1$ (for $x \ge -\sqrt{3}$) intersects $x^2+y^2=4$ at $x^2+1=4 \implies x=\sqrt{3}$. This gives the intersection point $(\sqrt{3}, 1)$.
The half-line at angle $-\frac{\pi}{6}$ has gradient $\tan(-\frac{\pi}{6}) = -\frac{1}{\sqrt{3}}$.
Its equation is $y - 1 = -\frac{1}{\sqrt{3}}(x + \sqrt{3}) \implies y = -\frac{x}{\sqrt{3}}$.
This line passes through the origin $(0,0)$ and intersects the circle when $x^2 + \frac{x^2}{3} = 4 \implies x^2 = 3$.
Since it extends to the right, $x = \sqrt{3}$, which gives $y = -1$. This corresponds to the point $(\sqrt{3}, -1)$.

4. Describe the required sketch of region $R$.

The sketch must include:

• A circle (or arc) centred at $O$ with radius $2$.
• The starting point marked at $-\sqrt{3} + \mathrm{i}$.
• A horizontal half-line from $-\sqrt{3} + \mathrm{i}$ ending at $\sqrt{3} + \mathrm{i}$.
• A half-line from $-\sqrt{3} + \mathrm{i}$ passing through the origin $O$ and ending at $\sqrt{3} - \mathrm{i}$.
• The region $R$ shaded, bounded by these two segments and the minor arc of the circle between $\sqrt{3} + \mathrm{i}$ and $\sqrt{3} - \mathrm{i}$.

1. Interpret $|z - 2\mathrm{i}|$ geometrically and find the minimum value by identifying the closest point in region $R$.

Let $z = x + \mathrm{i}y$. The expression $|z - 2\mathrm{i}|$ represents the distance from the point $(0, 2)$ to a point $z$ in region $R$.
From the inequalities defining $R$, the upper boundary is given by $\arg(z + \sqrt{3} - \mathrm{i}) = 0$, which is the horizontal line segment $y = 1$ for $-\sqrt{3} \le x \le \sqrt{3}$.
The perpendicular distance from $(0, 2)$ to the line $y = 1$ is $2 - 1 = 1$.
Since the foot of the perpendicular, $(0, 1)$, lies on this boundary segment (as $-\sqrt{3} \le 0 \le \sqrt{3}$), it is the closest point in $R$.
Thus, the minimum possible value of $|z - 2\mathrm{i}|$ is exactly $1$.

2. Set up an algebraic expression for the distance squared to evaluate the maximum value.

To find the maximum distance, we can express the square of the distance algebraically:
$|z - 2\mathrm{i}|^2 = x^2 + (y - 2)^2$
$|z - 2\mathrm{i}|^2 = x^2 + y^2 - 4y + 4$
For any point $z$ in the region $R$, it must lie within the disc $|z|^2 \le 4$, so we know that $x^2 + y^2 \le 4$.

3. Determine the lowest point in $R$ to bound $y$, and hence calculate the maximum distance.

The maximum of $|z - 2\mathrm{i}|^2$ will occur when $y$ is as small as possible.
The lowest point in $R$ is the intersection of the lower ray $\arg(z + \sqrt{3} - \mathrm{i}) = -\frac{\pi}{6}$ and the circle $x^2 + y^2 = 4$.
The ray passes through $(-\sqrt{3}, 1)$ with gradient $\tan(-\frac{\pi}{6}) = -\frac{1}{\sqrt{3}}$, giving the line $x + \sqrt{3}y = 0$.
Substituting $x = -\sqrt{3}y$ into $x^2 + y^2 = 4$ yields $3y^2 + y^2 = 4 \implies y^2 = 1$.
Since the ray extends in the positive real direction, the intersection is at $y = -1$ (the point $S(\sqrt{3}, -1)$).
Thus, for all $z \in R$, we have $y \ge -1$, which means $-4y \le 4$.
Substituting the geometric bounds $x^2 + y^2 \le 4$ and $-4y \le 4$ into our expression:
$|z - 2\mathrm{i}|^2 \le 4 + 4 + 4 = 12$
The maximum possible value of $|z - 2\mathrm{i}|$ is exactly $\sqrt{12} = 2\sqrt{3}$.

1. Express the roots of the given equation in exponential form.

The equation $z^{100} = 2^{100}$ can be rewritten as $z^{100} = 2^{100}e^{\mathrm{i}2k\pi}$ for $k \in \mathbb{Z}$.
Solving for $z$ gives:
$z = 2e^{\mathrm{i}\frac{2k\pi}{100}} = 2e^{\mathrm{i}\frac{k\pi}{50}}$
All roots have a modulus of $2$, meaning they lie exactly on the circular boundary of the region $|z|^2 \le 4$.

2. Determine the range of arguments for the part of the circle $|z| = 2$ that lies in region $R$.

The region $R$ is bounded by the circle $|z| = 2$ and the two rays from $z_0 = -\sqrt{3} + \mathrm{i}$.
For the horizontal ray $\arg(z + \sqrt{3} - \mathrm{i}) = 0$, the line is $y = 1$ (for $x \ge -\sqrt{3}$).
Substituting into $x^2 + y^2 = 4$ gives $x^2 + 1 = 4 \implies x = \sqrt{3}$. This gives the intersection $z = \sqrt{3} + \mathrm{i}$, where $\arg(z) = \frac{\pi}{6}$.

The ray $\arg(z + \sqrt{3} - \mathrm{i}) = -\frac{\pi}{6}$ is a line with gradient $\tan(-\frac{\pi}{6}) = -\frac{1}{\sqrt{3}}$ passing through $(-\sqrt{3}, 1)$, which simplifies to the line $y = -\frac{1}{\sqrt{3}}x$.
Substituting into $x^2 + y^2 = 4$ gives $x^2 + \frac{1}{3}x^2 = 4 \implies x = \sqrt{3}$ (since $x>0$), giving the intersection $z = \sqrt{3} - \mathrm{i}$ with $\arg(z) = -\frac{\pi}{6}$.

Thus, any root lying in the main body of $R$ must be on the arc where:
$-\frac{\pi}{6} \le \arg(z) \le \frac{\pi}{6}$
(Note: The vertex point $-\sqrt{3} + \mathrm{i}$ is also in $R$ and on $|z|=2$, but its argument is $\frac{5\pi}{6}$, which is not of the form $\frac{k\pi}{50}$, so it is not a root.)

3. Solve for the possible values of $k$ to find the number of roots.

Substitute $\arg(z) = \frac{k\pi}{50}$ into the inequality:
$-\frac{\pi}{6} \le \frac{k\pi}{50} \le \frac{\pi}{6}$
$-\frac{50}{6} \le k \le \frac{50}{6}$
$-8.33 \le k \le 8.33$

Since $k$ must be an integer, $k \in \{-8, -7, \dots, 0, \dots, 8\}$.
The number of valid roots is $8 - (-8) + 1 = 17$.