Find the set of values of $\alpha$ for which the expression $\dfrac{2 - i\sin 2\alpha}{1 + 2i\sin 2\alpha}$ is purely real. If $z$ and $w$ are different non-zero complex numbers and $|w| = 1$, find $\left|\dfrac{3(w - z)}{1 - \overline{z}w}\right|$.

1. Multiply the numerator and the denominator by the complex conjugate of the denominator to express the fraction in standard Cartesian form.

$\frac{2 - i\sin 2\alpha}{1 + 2i\sin 2\alpha} = \frac{2 - i\sin 2\alpha}{1 + 2i\sin 2\alpha} \times \frac{1 - 2i\sin 2\alpha}{1 - 2i\sin 2\alpha}$
$= \frac{2 - 4i\sin 2\alpha - i\sin 2\alpha + i^2(2\sin^2 2\alpha)}{1^2 - (2i\sin 2\alpha)^2}$
$= \frac{2 - 2\sin^2 2\alpha - 5i\sin 2\alpha}{1 + 4\sin^2 2\alpha}$

2. Isolate the imaginary part and set it to zero, as the expression is purely real.

$\text{Im}\left( \frac{2 - i\sin 2\alpha}{1 + 2i\sin 2\alpha} \right) = \frac{-5\sin 2\alpha}{1 + 4\sin^2 2\alpha}$
For the expression to be purely real:
$\frac{-5\sin 2\alpha}{1 + 4\sin^2 2\alpha} = 0$
$\implies \sin 2\alpha = 0$

3. Solve the resulting trigonometric equation for $\alpha$.

$2\alpha = n\pi, \quad n \in \mathbb{Z}$
$\alpha = \frac{n\pi}{2}, \quad n \in \mathbb{Z}$

1. Separate the modulus and use $|w|=1$ to rewrite the constant $1$ in the denominator.

$\left|\frac{3(w - z)}{1 - \overline{z}w}\right| = \frac{3|w - z|}{|1 - \overline{z}w|}$
Given $|w| = 1$, we have $|w|^2 = 1$.
Using the property $|w|^2 = w\overline{w}$, we obtain $w\overline{w} = 1$.
Substitute this into the denominator's expression:
$1 - \overline{z}w = w\overline{w} - \overline{z}w$

2. Factorise the denominator and apply conjugate properties.

Factorise $w$ from the expression:
$w\overline{w} - \overline{z}w = w(\overline{w} - \overline{z})$
Using the property of complex conjugates that the difference of conjugates is the conjugate of the difference ($\overline{w} - \overline{z} = \overline{w - z}$):
$1 - \overline{z}w = w(\overline{w - z})$

3. Find the modulus of the simplified denominator.

Take the modulus of both sides:
$|1 - \overline{z}w| = |w(\overline{w - z})|$
Using the multiplicative property of the modulus ($|ab| = |a||b|$):
$|1 - \overline{z}w| = |w||\overline{w - z}|$
Since $|w| = 1$ and the modulus of a conjugate is equal to the modulus of the original complex number ($|\overline{w - z}| = |w - z|$):
$|1 - \overline{z}w| = 1 \cdot |w - z| = |w - z|$

4. Evaluate the original fraction.

Substitute the result back into the separated modulus fraction:
$\left|\frac{3(w - z)}{1 - \overline{z}w}\right| = \frac{3|w - z|}{|w - z|}$
We are given that $z$ and $w$ are different non-zero complex numbers, meaning $w - z \neq 0$ and therefore $|w - z| > 0$. We can safely cancel the common factor:
$\frac{3|w - z|}{|w - z|} = 3$