设 $z$ 为复数 $-1+i\sqrt{3}$。求： $\arg(z)$ 使得 $\arg\big(z(z+a)\big)=\frac{5\pi}{6}$ 的实数 $a$。 解方程 $i z^{3}=3-3i$，将你的答案写成 $re^{i\theta}$ 的形式，其中 $r>0$ 且 $-\pi<\theta\le\pi$。 用类似的形式推导方程 $-w^{3}=3-3i$ 的根。

1. Determine the quadrant and the basic reference angle

The complex number $z = -1 + i\sqrt{3}$ has a negative real part and a positive imaginary part, so it lies in the second quadrant.
The reference angle $\alpha$ is given by:
$\alpha = \tan^{-1}\left|\frac{\sqrt{3}}{-1}\right| = \frac{\pi}{3}$

2. Calculate the principal argument

$\arg(z) = \pi - \alpha = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$

1. Determine $\arg(z+a)$ using the properties of arguments.

From part (a)(i), we have $\arg(z) = \frac{2\pi}{3}$.
Using the property $\arg(zw) = \arg(z) + \arg(w)$:
$\arg\big(z(z+a)\big) = \arg(z) + \arg(z+a)$
$\frac{2\pi}{3} + \arg(z+a) = \frac{5\pi}{6}$
$\arg(z+a) = \frac{5\pi}{6} - \frac{2\pi}{3} = \frac{\pi}{6}$

2. Express $z+a$ in Cartesian form and relate its real and imaginary parts to its argument.

$z+a = -1 + i\sqrt{3} + a = (a-1) + i\sqrt{3}$
Since $\arg(z+a) = \frac{\pi}{6}$, we can equate the tangent of the argument to the ratio of the imaginary and real parts:
$\tan\big(\arg(z+a)\big) = \frac{\operatorname{Im}(z+a)}{\operatorname{Re}(z+a)}$
$\tan\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{a-1}$
$\frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{a-1}$

3. Solve the equation for the real number $a$.

$a - 1 = (\sqrt{3})(\sqrt{3})$
$a - 1 = 3$
$a = 4$
(Checking the domains, for $a=4$, $\operatorname{Re}(z+a) = 3 > 0$ and $\operatorname{Im}(z+a) = \sqrt{3} > 0$, matching the first-quadrant angle $\frac{\pi}{6}$.)

1. Concisely state the roots of $iz^3 = 3-3i$ from the preceding part.

Since the target part depends on the roots of $iz^3 = 3-3i$ from part (b), we first concisely calculate them:
$z^3 = \frac{3-3i}{i} = -3-3i = \sqrt{18}e^{-i\frac{3\pi}{4}}$
Taking the cube roots, we add $2k\pi$ to the argument:
$z = 18^{\frac{1}{6}} e^{i\left(-\frac{\pi}{4} + \frac{2k\pi}{3}\right)}, \quad k = -1, 0, 1$
This yields the roots:
$z_1 = 18^{\frac{1}{6}}e^{-i\frac{\pi}{4}}, \quad z_2 = 18^{\frac{1}{6}}e^{i\frac{5\pi}{12}}, \quad z_3 = 18^{\frac{1}{6}}e^{-i\frac{11\pi}{12}}$

2. Equate the equations to deduce the transformation from $z$ to $w$.

Substitute $iz^3 = 3-3i$ into the new equation $-w^3 = 3-3i$:
$-w^3 = iz^3$
$w^3 = -iz^3$
Using the property that $i^3 = i^2 \cdot i = -i$, we can rewrite this as:
$w^3 = i^3 z^3 = (iz)^3$
This implies the set of roots for $w$ can be obtained by multiplying each root of $z$ by $i$, where $i = e^{i\frac{\pi}{2}}$.

3. Apply the transformation to deduce the roots of $w$.

Multiply each root of $z$ by $e^{i\frac{\pi}{2}}$ to deduce the roots of $w$:
$w_1 = z_1 \cdot e^{i\frac{\pi}{2}} = 18^{\frac{1}{6}}e^{-i\frac{\pi}{4}} \cdot e^{i\frac{\pi}{2}} = 18^{\frac{1}{6}}e^{i\frac{\pi}{4}}$
$w_2 = z_2 \cdot e^{i\frac{\pi}{2}} = 18^{\frac{1}{6}}e^{i\frac{5\pi}{12}} \cdot e^{i\frac{\pi}{2}} = 18^{\frac{1}{6}}e^{i\frac{11\pi}{12}}$
$w_3 = z_3 \cdot e^{i\frac{\pi}{2}} = 18^{\frac{1}{6}}e^{-i\frac{11\pi}{12}} \cdot e^{i\frac{\pi}{2}} = 18^{\frac{1}{6}}e^{-i\frac{5\pi}{12}}$
All arguments are within $(-\pi, \pi]$, satisfying the required form.