The complex numbers $z$ and $w$ are such that $z = 1 + i a,\quad w = -b - i$ where $a$ and $b$ are real and positive. Given that $zw = 6 - 11i$, find the exact values of $a$ and $b. The conjugate of a complex number $u$ has modulus 2 and argument $\frac{2}{3}\pi$ and another complex number $v$ has modulus 5 and argument $\frac{3}{4}\pi$. Find the exact values of the modulus and argument of $\dfrac{v}{u^2}$. Hence, find the smallest positive integer $n$ such that the point representing $\left(\dfrac{v}{u^2}\right)^n$ lies on the negative imaginary axis.

1. Set up the equation and expand the product $zw$.

$\begin{aligned} zw &= (1 + i a)(-b - i) \\ &= -b - i - i ab - i^2 a \\ &= -b - i(1 + ab) - (-1)a \\ &= (a - b) - i(1 + ab) \end{aligned}$

2. Equate the real and imaginary parts to those of $6 - 11i$.

$(a - b) - i(1 + ab) = 6 - 11i$

Equating real parts:
$a - b = 6 \implies a = b + 6 \quad \text{--- (1)}$

Equating imaginary parts:
$-(1 + ab) = -11 \implies ab = 10 \quad \text{--- (2)}$

3. Substitute to form a quadratic equation in $b$.

Substitute (1) into (2):
$(b + 6)b = 10$
$b^2 + 6b - 10 = 0$

4. Solve the quadratic equation and select the valid roots for $a$ and $b$.

Using the quadratic formula:
$b = \frac{-6 \pm \sqrt{6^2 - 4(1)(-10)}}{2}$
$b = \frac{-6 \pm \sqrt{76}}{2}$
$b = \frac{-6 \pm 2\sqrt{19}}{2}$
$b = -3 \pm \sqrt{19}$

Since we are given that $b > 0$, we reject $-3 - \sqrt{19}$. Thus,
$b = -3 + \sqrt{19}$

Substitute back to find $a$:
$a = (-3 + \sqrt{19}) + 6 = 3 + \sqrt{19}$
(Note that $a > 0$ is also satisfied since $\sqrt{19} > 4$)

1. Determine the modulus and argument of $u$ from its conjugate $u^*$.

$|u| = |u^*| = 2$
$\arg(u) = -\arg(u^*) = -\frac{2}{3}\pi$

2. Calculate the exact modulus of $\frac{v}{u^2}$.

$\left|\frac{v}{u^2}\right| = \frac{|v|}{|u|^2}$
$= \frac{5}{2^2} = \frac{5}{4}$

3. Calculate the exact principal argument of $\frac{v}{u^2}$.

$\arg\left(\frac{v}{u^2}\right) = \arg(v) - 2\arg(u)$
$= \frac{3}{4}\pi - 2\left(-\frac{2}{3}\pi\right)$
$= \frac{3}{4}\pi + \frac{4}{3}\pi = \frac{25}{12}\pi$
To find the principal argument, we subtract $2\pi$ to bring it within the range $(-\pi, \pi]$:
$\frac{25}{12}\pi - 2\pi = \frac{1}{12}\pi$

1. Calculate the principal argument of $\frac{v}{u^2}$ using the given arguments.

Given $\arg(u^*) = \frac{2}{3}\pi$, we have $\arg(u) = -\frac{2}{3}\pi$.
$\arg\left(\frac{v}{u^2}\right) = \arg(v) - 2\arg(u) = \frac{3}{4}\pi - 2\left(-\frac{2}{3}\pi\right) = \frac{25}{12}\pi$
The principal argument is $\frac{25}{12}\pi - 2\pi = \frac{1}{12}\pi$.

2. Express the argument of the $n$-th power.

Using the properties of arguments:
$\arg\left[\left(\frac{v}{u^2}\right)^n\right] = n \arg\left(\frac{v}{u^2}\right) = \frac{n\pi}{12}$

3. Set up the condition for the point to lie on the negative imaginary axis and solve for $n$.

For the point to lie on the negative imaginary axis, its argument must be $\frac{3\pi}{2} + 2k\pi$ (or equivalently $-\frac{\pi}{2} + 2k\pi$) for some integer $k$.
$\frac{n\pi}{12} = \frac{3\pi}{2} + 2k\pi$
Dividing by $\pi$ and multiplying by $12$:
$n = 18 + 24k$
To find the smallest positive integer $n$, we take $k = 0$:
$n = 18$