Do not use a calculator in answering this question.

1. Solve the quadratic equation to find the complex roots in Cartesian form.

$z = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(36)}}{2}$
$z = \frac{6 \pm \sqrt{36 - 144}}{2}$
$z = \frac{6 \pm \sqrt{-108}}{2}$
$z = \frac{6 \pm 6\sqrt{3}i}{2}$
$z = 3 \pm 3\sqrt{3}i$

2. Calculate the modulus of the roots.

For both roots, the modulus $r$ is:
$r = \sqrt{3^2 + (\pm 3\sqrt{3})^2}$
$r = \sqrt{9 + 27}$
$r = \sqrt{36} = 6$

3. Find the arguments, distinguish $z_1$ and $z_2$, and express them in $re^{i\theta}$ form.

For $3 + 3\sqrt{3}i$ (first quadrant):
$\arg(z) = \tan^{-1}\left(\frac{3\sqrt{3}}{3}\right) = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3}$
For $3 - 3\sqrt{3}i$ (fourth quadrant):
$\arg(z) = -\tan^{-1}\left(\frac{3\sqrt{3}}{3}\right) = -\frac{\pi}{3}$

Given $\arg(z_1) > \arg(z_2)$, we have $\arg(z_1) = \frac{\pi}{3}$ and $\arg(z_2) = -\frac{\pi}{3}$.
Therefore, in the exact form $re^{i\theta}$:
$z_1 = 6e^{i\frac{\pi}{3}}$
$z_2 = 6e^{-i\frac{\pi}{3}}$

1. Concisely solve part (i) to find the exponential forms of $z_1$ and $z_2$ (which are required for this part).

$z^2 - 6z + 36 = 0 \implies z = \dfrac{6 \pm \sqrt{36 - 144}}{2} = 3 \pm 3\sqrt{3}i$

Modulus $r = \sqrt{3^2 + (3\sqrt{3})^2} = \sqrt{9 + 27} = 6$.
Argument $\theta = \pm\tan^{-1}\left(\dfrac{3\sqrt{3}}{3}\right) = \pm\dfrac{\pi}{3}$.

Given $\arg(z_1) > \arg(z_2)$, we have $z_1 = 6e^{i\frac{\pi}{3}}$ and $z_2 = 6e^{-i\frac{\pi}{3}}$.

2. Compute the modulus of the required expression.

$\left| \dfrac{z_1^4}{i z_2} \right| = \dfrac{|z_1|^4}{|i| |z_2|} = \dfrac{6^4}{(1)(6)} = 216$

3. Compute the argument of the required expression.

$\arg\left( \dfrac{z_1^4}{i z_2} \right) = 4\arg(z_1) - \arg(i) - \arg(z_2)$
$= 4\left(\dfrac{\pi}{3}\right) - \dfrac{\pi}{2} - \left(-\dfrac{\pi}{3}\right)$
$= \dfrac{4\pi}{3} - \dfrac{\pi}{2} + \dfrac{\pi}{3} = \dfrac{7\pi}{6}$

4. Adjust the argument to the principal range and express the result in exact polar form.

The principal argument must lie in $(-\pi, \pi]$.
$\theta = \dfrac{7\pi}{6} - 2\pi = -\dfrac{5\pi}{6}$

In exact polar form, the complex number is:
$216\left(\cos\left(-\dfrac{5\pi}{6}\right) + i\sin\left(-\dfrac{5\pi}{6}\right)\right)$

1. Express $z_2^n$ in exponential form using the result from part (i)

From part (i), $z_2 = 6e^{-\mathrm{i}\frac{\pi}{3}}$.

Applying index laws (De Moivre's theorem implicitly):
$z_2^n = \left(6e^{-\mathrm{i}\frac{\pi}{3}}\right)^n = 6^n e^{-\mathrm{i}\frac{n\pi}{3}}$

2. Determine the condition for $z_2^n$ to be a positive real number and find $n$

For $z_2^n$ to be a strictly positive real number, its argument must be an integer multiple of $2\pi$ (since an argument of an odd multiple of $\pi$ would yield a negative real number).
$-\frac{n\pi}{3} = 2k\pi, \quad k \in \mathbb{Z}$
$n = -6k$
To find the smallest positive integer $n$, we take $k = -1$:
$n = -6(-1) = 6$