We are given the complex numbers $z_1 = -1 + i\sqrt{3}$ and $z_2 = 1 - i\sqrt{3}$. We need to compute $z_1 + z_2$, $z_1 - z_2$, and $z_1 \times z_2$. Then we need to write the complex numbers $z_1 - z_2$ and $z_1 \times z_2$ in trigonometric form.

1. Problem Description

We are given the complex numbers $z_1 = -1 + i\sqrt{3}$ and $z_2 = 1 - i\sqrt{3}$. We need to compute $z_1 + z_2$, $z_1 - z_2$, and $z_1 \times z_2$. Then we need to write the complex numbers $z_1 - z_2$ and $z_1 \times z_2$ in trigonometric form.

2. Solution Steps

First, let's compute $z_1 + z_2$:

$z_1 + z_2 = (-1 + i\sqrt{3}) + (1 - i\sqrt{3}) = -1 + 1 + i\sqrt{3} - i\sqrt{3} = 0$

Next, let's compute $z_1 - z_2$:

$z_1 - z_2 = (-1 + i\sqrt{3}) - (1 - i\sqrt{3}) = -1 - 1 + i\sqrt{3} + i\sqrt{3} = -2 + 2i\sqrt{3}$

Now, let's compute $z_1 \times z_2$:

$z_1 \times z_2 = (-1 + i\sqrt{3})(1 - i\sqrt{3}) = -1 + i\sqrt{3} + i\sqrt{3} - i^2(3) = -1 + 2i\sqrt{3} + 3 = 2 + 2i\sqrt{3}$

Now we need to write $z_1 - z_2$ and $z_1 \times z_2$ in trigonometric form.

For $z_1 - z_2 = -2 + 2i\sqrt{3}$, the modulus is

$r = \sqrt{(-2)^2 + (2\sqrt{3})^2} = \sqrt{4 + 12} = \sqrt{16} = 4$.

The argument $\theta$ satisfies $\tan(\theta) = \frac{2\sqrt{3}}{-2} = -\sqrt{3}$. Since the complex number is in the second quadrant, $\theta = \frac{2\pi}{3}$.

So $z_1 - z_2 = 4(\cos(\frac{2\pi}{3}) + i\sin(\frac{2\pi}{3}))$.

For $z_1 \times z_2 = 2 + 2i\sqrt{3}$, the modulus is

$r = \sqrt{2^2 + (2\sqrt{3})^2} = \sqrt{4 + 12} = \sqrt{16} = 4$.

The argument $\theta$ satisfies $\tan(\theta) = \frac{2\sqrt{3}}{2} = \sqrt{3}$. Since the complex number is in the first quadrant, $\theta = \frac{\pi}{3}$.

So $z_1 \times z_2 = 4(\cos(\frac{\pi}{3}) + i\sin(\frac{\pi}{3}))$.

3. Final Answer

$z_1 + z_2 = 0$

$z_1 - z_2 = -2 + 2i\sqrt{3} = 4(\cos(\frac{2\pi}{3}) + i\sin(\frac{2\pi}{3}))$

$z_1 \times z_2 = 2 + 2i\sqrt{3} = 4(\cos(\frac{\pi}{3}) + i\sin(\frac{\pi}{3}))$