设复数 $z$ 为 $-1+i\sqrt{3}$。求： $\arg(z)$， 实数 $a$，使得 $\arg\big(z(z+a)\big)=\frac{5\pi}{6}$。 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 $-\pi<\theta\le\pi$。 以类似的形式推导方程 $-w^{3}=3-3i$ 的根。

1. Calculate the argument of $z$ by finding its basic angle and identifying its quadrant

$z = -1 + i\sqrt{3}$
Since the real part is negative and the imaginary part is positive, $z$ lies in the second quadrant.
The basic angle $\alpha = \tan^{-1}\left(\frac{\sqrt{3}}{1}\right) = \frac{\pi}{3}$.
$\arg(z) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$

1. Find the argument of $z$ (as required in the previous sub-part).

Given $z = -1 + i\sqrt{3}$, the point lies in the second quadrant.
$\arg(z) = \pi - \tan^{-1}\left(\frac{\sqrt{3}}{1}\right)$
$\arg(z) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$

2. Use the multiplicative property of arguments to find $\arg(z+a)$.

Using the property $\arg(w_1 w_2) = \arg(w_1) + \arg(w_2) + 2k\pi$:
$\arg\big(z(z+a)\big) = \arg(z) + \arg(z+a)$
$\frac{2\pi}{3} + \arg(z+a) = \frac{5\pi}{6}$
$\arg(z+a) = \frac{5\pi}{6} - \frac{4\pi}{6} = \frac{\pi}{6}$

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

$z+a = (-1 + a) + i\sqrt{3} = (a - 1) + i\sqrt{3}$
Since $\arg(z+a) = \frac{\pi}{6}$ is in the first quadrant, the real part $a - 1 > 0$ and we can set up the equation using the tangent function:
$\tan\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{a - 1}$

4. Solve the equation for $a$.

$\frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{a - 1}$
Cross-multiplying gives:
$a - 1 = (\sqrt{3})(\sqrt{3})$
$a - 1 = 3$
$a = 4$

1. Isolate $z^3$ and express it in Cartesian form.

$i z^3 = 3 - 3i$
$z^3 = \frac{3 - 3i}{i}$
$z^3 = (3 - 3i)(-i) = -3i + 3i^2 = -3 - 3i$

2. Convert $-3 - 3i$ to exponential form.

Modulus:
$r = |-3 - 3i| = \sqrt{(-3)^2 + (-3)^2} = \sqrt{18} = 3\sqrt{2}$

Argument:
Since $-3 - 3i$ is in the third quadrant,
$\theta = -\pi + \tan^{-1}\left(\frac{3}{3}\right) = -\pi + \frac{\pi}{4} = -\frac{3\pi}{4}$

$z^3 = 3\sqrt{2} e^{-i\frac{3\pi}{4}}$

3. Set up the general formula for the cube roots.

$z = (3\sqrt{2})^{\frac{1}{3}} e^{i\left(\frac{-\frac{3\pi}{4} + 2k\pi}{3}\right)}$
where $k \in \mathbb{Z}$. Note that $(3\sqrt{2})^{\frac{1}{3}} = (18^{\frac{1}{2}})^{\frac{1}{3}} = 18^{\frac{1}{6}}$.

4. Evaluate the roots for $-\pi < \theta \le \pi$.

For $k = 0$:
$z_1 = 18^{\frac{1}{6}} e^{-i\frac{\pi}{4}}$

For $k = 1$:
$z_2 = 18^{\frac{1}{6}} e^{i\left(\frac{-\frac{3\pi}{4} + 2\pi}{3}\right)} = 18^{\frac{1}{6}} e^{i\frac{5\pi}{12}}$

For $k = -1$:
$z_3 = 18^{\frac{1}{6}} e^{i\left(\frac{-\frac{3\pi}{4} - 2\pi}{3}\right)} = 18^{\frac{1}{6}} e^{-i\frac{11\pi}{12}}$

1. Find the roots of $iz^3 = 3-3i$ from part (b)(i) to use in the deduction.

$iz^3 = 3 - 3i \implies z^3 = \frac{3-3i}{i} = -3 - 3i$

Calculate the modulus and argument:
$|-3 - 3i| = \sqrt{(-3)^2 + (-3)^2} = \sqrt{18}$
$\arg(-3 - 3i) = -\frac{3\pi}{4}$

So, $z^3 = \sqrt{18} e^{i\left(-\frac{3\pi}{4} + 2k\pi\right)}$. Taking the cube root gives:
$z = 18^{\frac{1}{6}} e^{i\left(-\frac{\pi}{4} + \frac{2k\pi}{3}\right)}$ for $k = -1, 0, 1$.

This yields the roots for $z$:
$z_1 = 18^{\frac{1}{6}} e^{-i\frac{\pi}{4}}$
$z_2 = 18^{\frac{1}{6}} e^{i\frac{5\pi}{12}}$
$z_3 = 18^{\frac{1}{6}} e^{-i\frac{11\pi}{12}}$

2. Relate the new equation to the one in (b)(i) using complex conjugates.

Rearrange the given equation:
$-w^3 = 3 - 3i \implies w^3 = -3 + 3i$

Compare this with $z^3 = -3 - 3i$ from the previous part. We can see that $w^3$ is exactly the complex conjugate of $z^3$:
$w^3 = (-3 - 3i)^* = (z^3)^*$

Using the property of complex conjugates, $(z^3)^* = (z^*)^3$.
Therefore, $w^3 = (z^*)^3$, which implies the roots for $w$ are simply the complex conjugates of the roots for $z$.

3. Deduce the roots for $w$ by conjugating the roots of $z$.

The complex conjugate of $re^{i\theta}$ is $re^{-i\theta}$. To find the roots for $w$, negate the arguments of the $z$ roots:
$w_1 = z_1^* = 18^{\frac{1}{6}} e^{i\frac{\pi}{4}}$
$w_2 = z_2^* = 18^{\frac{1}{6}} e^{-i\frac{5\pi}{12}}$
$w_3 = z_3^* = 18^{\frac{1}{6}} e^{i\frac{11\pi}{12}}$

All arguments ($-\frac{5\pi}{12}, \frac{\pi}{4}, \frac{11\pi}{12}$) fall perfectly within the required principal range $-\pi < \theta \le \pi$.