The polynomial $\text{P}(z)$ is given by $z^3 - 8z + a$, where $a$ is a real constant. Given that $\sqrt{3} + \text{i}$ is a root of the equation $\text{P}(z) = 0$, solve exactly the equation $\text{P}(z) = 0$. It is given that $w = -\sqrt{3} + \text{i}$. Write down the exact value of $\arg(w)$ and hence find the three smallest positive integer values of $n$ such that $w^n$ is purely imaginary.

1. State the complex conjugate root.

Since the polynomial $\text{P}(z)$ has a real constant $a$, all of its coefficients are real. Thus, the complex conjugate of the given root must also be a root:
$z = \sqrt{3} - \text{i}$

2. Find the quadratic factor containing the two complex roots.

Sum of the roots $= (\sqrt{3} + \text{i}) + (\sqrt{3} - \text{i}) = 2\sqrt{3}$
Product of the roots $= (\sqrt{3} + \text{i})(\sqrt{3} - \text{i}) = (\sqrt{3})^2 - (\text{i})^2 = 3 - (-1) = 4$

The quadratic factor is therefore:
$z^2 - (2\sqrt{3})z + 4$

3. Express the polynomial as a product of factors and find the real root.

Let the third real root be $\alpha$. We can express $\text{P}(z)$ as the product of the quadratic factor and a linear factor:
$z^3 - 8z + a = (z^2 - 2\sqrt{3}z + 4)(z - \alpha)$

Comparing the coefficients of $z^2$ on both sides:
$0 = -\alpha - 2\sqrt{3}$
$\alpha = -2\sqrt{3}$

4. State all the exact solutions to the equation.

The solutions to $\text{P}(z) = 0$ are the three roots we have identified:
$z = \sqrt{3} + \text{i}, \quad z = \sqrt{3} - \text{i}, \quad z = -2\sqrt{3}$

1. Calculate the exact value of $\arg(w)$.

Given $w = -\sqrt{3} + \text{i}$, the complex number has a negative real part and a positive imaginary part, so it lies in the second quadrant.
The basic angle is $\alpha = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}$.
Therefore, the principal argument is:
$\arg(w) = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$

2. Set up the condition for $w^n$ to be purely imaginary using its argument.

Using the properties of arguments, $\arg(w^n) = n\arg(w) = \frac{5n\pi}{6}$.
For $w^n$ to be purely imaginary, its argument must lie on the imaginary axis, which means it must be an odd multiple of $\frac{\pi}{2}$:
$\frac{5n\pi}{6} = \frac{\pi}{2} + k\pi$ for some integer $k$.
Dividing by $\pi$ and multiplying by $6$ gives:
$5n = 3 + 6k$

3. Solve for the three smallest positive integer values of $n$.

We require $5n$ to be $3$ more than a multiple of $6$, i.e., $5n \equiv 3 \pmod 6$.
Since $5 \equiv -1 \pmod 6$, we can write $-n \equiv 3 \pmod 6$, which implies $n \equiv 3 \pmod 6$.
Alternatively, testing integer values of $n > 0$:

• $n = 1 \implies 5(1) = 5$ (not of the form $3+6k$)
• $n = 2 \implies 5(2) = 10$ (not)
• $n = 3 \implies 5(3) = 15 = 3 + 6(2)$ (valid)
  Since $n \equiv 3 \pmod 6$, the sequence of valid positive integers is $n = 3, 9, 15, 21, \dots$
  The three smallest positive integer values are $3, 9,$ and $15$.