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. Multiply $z$ and $w$ and group the real and imaginary parts.

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

2. Equate the real and imaginary parts to $6 - 11i$ to form simultaneous equations.

Given $zw = 6 - 11i$, we can equate the real and imaginary parts:
Real part: $a - b = 6 \implies a = b + 6$
Imaginary part: $-(1 + ab) = -11 \implies ab = 10$

3. Solve the simultaneous equations to obtain a quadratic equation in $b$.

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

4. Solve the quadratic equation for $b$, reject the negative root, and find $a$.

Using the quadratic formula:
$b = \frac{-6 \pm \sqrt{6^2 - 4(1)(-10)}}{2}$
$b = \frac{-6 \pm \sqrt{76}}{2}$
$b = \frac{-6 \pm 2\sqrt{19}}{2} = -3 \pm \sqrt{19}$
Since $b$ is positive and $\sqrt{19} > 4$, we reject the negative root $b = -3 - \sqrt{19}$.
Therefore, $b = -3 + \sqrt{19}$.

Substitute $b$ back to find $a$:
$a = (-3 + \sqrt{19}) + 6$
$a = 3 + \sqrt{19}$

1. Find the modulus and argument of $u$ from its conjugate.

Given that the conjugate $u^*$ has $|u^*| = 2$ and $\arg(u^*) = \frac{2}{3}\pi$, we have:
$|u| = |u^*| = 2$
$\arg(u) = -\arg(u^*) = -\frac{2}{3}\pi$

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

Using the multiplicative properties of the modulus:
$\left|\frac{v}{u^2}\right| = \frac{|v|}{|u|^2}$
$\left|\frac{v}{u^2}\right| = \frac{5}{2^2} = \frac{5}{4}$

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

Using the properties of the argument:
$\arg\left(\frac{v}{u^2}\right) = \arg(v) - \arg(u^2)$
$\arg\left(\frac{v}{u^2}\right) = \arg(v) - 2\arg(u)$
$= \frac{3}{4}\pi - 2\left(-\frac{2}{3}\pi\right)$
$= \frac{3}{4}\pi + \frac{4}{3}\pi$
$= \frac{9\pi + 16\pi}{12} = \frac{25}{12}\pi$

To find the principal argument, we adjust the value to lie within $(-\pi, \pi]$ by subtracting $2\pi$:
$\frac{25}{12}\pi - 2\pi = \frac{1}{12}\pi$

1. Find the argument of $\frac{v}{u^2}$ using the properties of arguments.

$\arg(u) = -\arg(u^*) = -\frac{2\pi}{3}$
$\arg\left(\frac{v}{u^2}\right) = \arg(v) - 2\arg(u)$
$\arg\left(\frac{v}{u^2}\right) = \frac{3\pi}{4} - 2\left(-\frac{2\pi}{3}\right) = \frac{25\pi}{12}$
$\text{Principal argument } \equiv \frac{\pi}{12}$

2. Set up an equation for $n$ using the condition for the negative imaginary axis.

$\arg\left(\left(\frac{v}{u^2}\right)^n\right) = n\left(\frac{\pi}{12}\right)$
For the point to lie on the negative imaginary axis, its argument must be $-\frac{\pi}{2} + 2k\pi$ where $k \in \mathbb{Z}$.
Since $n$ must be a positive integer, we seek the smallest positive argument corresponding to this axis, which is $\frac{3\pi}{2}$ (when $k=1$):
$\frac{n\pi}{12} = \frac{3\pi}{2}$

3. Solve for the smallest positive integer $n$.

$n = \frac{3\pi}{2} \times \frac{12}{\pi}$
$n = 18$