已知 $z_1 = -\frac{\text{i}}{2}$ 是方程 $2z^3 + (\text{i}-8)z^2 + az + 13\text{i} = 0$ 的一个根，求复数 $a$ 并解该方程，给出笛卡尔形式 $x + \text{i}y$ 的答案。 由此，以笛卡尔形式求方程
$2w^3 + (1+8\text{i})w^2 - aw - 13 = 0.$
的根。 解方程 $z^6 + 729 = 0$，将你的答案表示为 $r\text{e}^{\text{i}\theta}$ 的形式，其中 $r > 0$ 且 $-\pi < \theta \le \pi$。 已知 $z_1$ 是上述方程的一个根，且 $0 < \arg z_1 < \frac{\pi}{2}$。 如果 $\frac{\bar{z}_1^n}{z_1}$ 是一个正实数，求最小正整数 $n$。

1. Substitute $z = -\frac{\text{i}}{2}$ into the polynomial to form an equation in $a$.

Since $z = -\frac{\text{i}}{2}$ is a root, we substitute it into the given equation $P(z)=0$:
$2\left(-\frac{\text{i}}{2}\right)^3 + (\text{i}-8)\left(-\frac{\text{i}}{2}\right)^2 + a\left(-\frac{\text{i}}{2}\right) + 13\text{i} = 0$
$2\left(\frac{\text{i}}{8}\right) + (\text{i}-8)\left(-\frac{1}{4}\right) - \frac{a\text{i}}{2} + 13\text{i} = 0$
$\frac{\text{i}}{4} - \frac{\text{i}}{4} + 2 - \frac{a\text{i}}{2} + 13\text{i} = 0$
$2 + 13\text{i} - \frac{a\text{i}}{2} = 0$

2. Solve for the complex number $a$.

Isolating $a$:
$\frac{a\text{i}}{2} = 2 + 13\text{i}$
$a\text{i} = 4 + 26\text{i}$
$a = \frac{4 + 26\text{i}}{\text{i}}$
Multiplying numerator and denominator by $-\text{i}$:
$a = (4 + 26\text{i})(-\text{i}) = -4\text{i} - 26\text{i}^2 = 26 - 4\text{i}$

3. Factorise the cubic polynomial to find the remaining quadratic factor.

Since $z = -\frac{\text{i}}{2}$ is a root, $\left(z + \frac{\text{i}}{2}\right)$ is a factor, so $(2z + \text{i})$ is a factor of the cubic polynomial.
Let $2z^3 + (\text{i}-8)z^2 + (26-4\text{i})z + 13\text{i} = (2z + \text{i})(z^2 + Bz + C)$.
Comparing the constant terms:
$\text{i}C = 13\text{i} \implies C = 13$
Comparing the coefficients of $z^2$:
$2B + \text{i} = \text{i} - 8 \implies 2B = -8 \implies B = -4$
Thus, the remaining quadratic factor is $z^2 - 4z + 13$.

4. Solve the quadratic equation to find the other two roots.

Solve $z^2 - 4z + 13 = 0$ to find the remaining roots.
Using completing the square:
$(z - 2)^2 - 4 + 13 = 0$
$(z - 2)^2 = -9$
$z - 2 = \pm \sqrt{-9} = \pm 3\text{i}$
$z = 2 \pm 3\text{i}$
The roots of the first equation are $z = -\frac{\text{i}}{2},\; 2 + 3\text{i},\text{ and } 2 - 3\text{i}$.

5. Relate the new equation to the original cubic equation.

To relate the new equation $2w^3 + (1+8\text{i})w^2 - aw - 13 = 0$ to $P(z) = 0$, notice the constant terms $-13$ and $13\text{i}$. Multiplying the new equation by $-\text{i}$ yields:
$-2\text{i}w^3 - \text{i}(1+8\text{i})w^2 + a\text{i}w + 13\text{i} = 0$
$-2\text{i}w^3 + (8-\text{i})w^2 + a\text{i}w + 13\text{i} = 0$
Rewrite this to match the terms of $P(z)$:
$2(\text{i}w)^3 + (\text{i}-8)(\text{i}w)^2 + a(\text{i}w) + 13\text{i} = 0$
This is exactly the equation $P(\text{i}w) = 0$.

6. Use the relationship to find the roots of the new equation.

Since $P(\text{i}w) = 0$, the roots satisfy $\text{i}w = z \implies w = \frac{z}{\text{i}} = -\text{i}z$.
Using the roots $z$ found earlier:
For $z_1 = -\frac{\text{i}}{2}$, $w_1 = -\text{i}\left(-\frac{\text{i}}{2}\right) = -\frac{1}{2}$
For $z_2 = 2 + 3\text{i}$, $w_2 = -\text{i}(2 + 3\text{i}) = -2\text{i} - 3\text{i}^2 = 3 - 2\text{i}$
For $z_3 = 2 - 3\text{i}$, $w_3 = -\text{i}(2 - 3\text{i}) = -2\text{i} + 3\text{i}^2 = -3 - 2\text{i}$
The roots of the new equation are $w = -\frac{1}{2},\; 3 - 2\text{i},\text{ and } -3 - 2\text{i}$.

1. Set up the equation in exponential form to find the roots.

$z^6 = -729 = 729\text{e}^{\text{i}(\pi + 2k\pi)} \quad \text{for } k \in \mathbb{Z}$
$z = \sqrt[6]{729}\text{e}^{\text{i}\frac{\pi + 2k\pi}{6}} = 3\text{e}^{\text{i}\frac{(2k+1)\pi}{6}}$

2. Find all six roots of the equation in the specified range $-\pi < \theta \le \pi$.

Substituting $k = -3, -2, -1, 0, 1, 2$ to keep the argument within the given range:
For $k=0$, $z = 3\text{e}^{\text{i}\frac{\pi}{6}}$
For $k=1$, $z = 3\text{e}^{\text{i}\frac{\pi}{2}}$
For $k=2$, $z = 3\text{e}^{\text{i}\frac{5\pi}{6}}$
For $k=-1$, $z = 3\text{e}^{-\text{i}\frac{\pi}{6}}$
For $k=-2$, $z = 3\text{e}^{-\text{i}\frac{\pi}{2}}$
For $k=-3$, $z = 3\text{e}^{-\text{i}\frac{5\pi}{6}}$

The roots are $3\text{e}^{\text{i}\frac{\pi}{6}}, 3\text{e}^{\text{i}\frac{\pi}{2}}, 3\text{e}^{\text{i}\frac{5\pi}{6}}, 3\text{e}^{-\text{i}\frac{\pi}{6}}, 3\text{e}^{-\text{i}\frac{\pi}{2}}, 3\text{e}^{-\text{i}\frac{5\pi}{6}}$.

3. Identify the root $z_1$ that satisfies the condition $0 < \arg z_1 < \frac{\pi}{2}$.

Checking the arguments of the roots, the only one between $0$ and $\frac{\pi}{2}$ is $\frac{\pi}{6}$.
Thus, $z_1 = 3\text{e}^{\text{i}\frac{\pi}{6}}$.

4. Simplify the expression $\frac{\bar{z}_1^n}{z_1}$ in exponential form.

The complex conjugate is $\bar{z}_1 = 3\text{e}^{-\text{i}\frac{\pi}{6}}$.
Substituting into the expression:
$\frac{\bar{z}_1^n}{z_1} = \frac{(3\text{e}^{-\text{i}\frac{\pi}{6}})^n}{3\text{e}^{\text{i}\frac{\pi}{6}}} = \frac{3^n \text{e}^{-\text{i}\frac{n\pi}{6}}}{3\text{e}^{\text{i}\frac{\pi}{6}}}$
$= 3^{n-1} \text{e}^{-\text{i}\frac{(n+1)\pi}{6}}$

5. Determine the smallest positive integer $n$ for the expression to be a positive real number.

For $3^{n-1} \text{e}^{-\text{i}\frac{(n+1)\pi}{6}}$ to be a positive real number, its argument must be a multiple of $2\pi$.
$-\frac{(n+1)\pi}{6} = 2k\pi \quad \text{for some integer } k$
$n + 1 = -12k$
Since $n$ must be a positive integer, we test values to find the smallest $n$. The smallest valid positive value occurs when $k = -1$:
$n + 1 = 12 \implies n = 11$