Question 1165360
This problem requires using the properties of complex numbers in both rectangular and polar forms, along with the geometric interpretation of multiplication.

Let $z_1 = x_1 + i y_1$ and $z_2 = x_2 + i y_2$.

### 1. Analyze the Given Information

* **$z_1$ in Rectangular Form:** We are given $\operatorname{Im}(z_1) = \sqrt{3}$.
    $$z_1 = x_1 + i\sqrt{3}$$
* **$z_2$ in Polar Form:** We are given $|z_2| = 2$.
    Let $z_2 = 2(\cos\theta_2 + i\sin\theta_2)$.
* **Product ($z_p$):** $z_p = z_1 z_2 = 4\left(\cos\left(\frac{5\pi}{6}\right) + i\sin\left(\frac{5\pi}{6}\right)\right)$.

### 2. Convert the Product to Rectangular Form

First, find the exact value of the product $z_p$.
$$\cos\left(\frac{5\pi}{6}\right) = -\cos\left(\frac{\pi}{6}\right) = -\frac{\sqrt{3}}{2}$$
$$\sin\left(\frac{5\pi}{6}\right) = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$$

$$z_p = 4\left(-\frac{\sqrt{3}}{2} + i\frac{1}{2}\right)$$
$$z_p = -2\sqrt{3} + 2i$$

### 3. Use the Modulus Property of the Product

The modulus of the product is the product of the moduli: $|z_1 z_2| = |z_1| |z_2|$.
From the product's polar form, $|z_1 z_2| = 4$.
We are given $|z_2| = 2$.
$$4 = |z_1| \cdot 2$$
$$|z_1| = 2$$

### 4. Determine $z_1$

We have $z_1 = x_1 + i\sqrt{3}$ and $|z_1| = 2$.
The modulus is calculated as $|z_1| = \sqrt{x_1^2 + y_1^2}$:
$$2 = \sqrt{x_1^2 + (\sqrt{3})^2}$$
$$4 = x_1^2 + 3$$
$$x_1^2 = 1$$
$$x_1 = \pm 1$$

Therefore, there are two possible solutions for $z_1$:
$$\mathbf{z_{1a} = 1 + i\sqrt{3}} \quad \text{or} \quad \mathbf{z_{1b} = -1 + i\sqrt{3}}$$

### 5. Determine $z_2$ (using $z_2 = \frac{z_p}{z_1}$)

We use the relationship $z_2 = \frac{z_p}{z_1}$, where $z_p = -2\sqrt{3} + 2i$.

#### Case A: Using $z_{1a} = 1 + i\sqrt{3}$

To divide, it's simplest to use the polar form for $z_{1a}$.
$|z_{1a}| = 2$.
$$\arg(z_{1a}) = \arctan\left(\frac{\sqrt{3}}{1}\right) = \frac{\pi}{3}$$
$$z_{1a} = 2\left(\cos\left(\frac{\pi}{3}\right) + i\sin\left(\frac{\pi}{3}\right)\right)$$

The argument of the quotient is the difference of the arguments: $\arg(z_2) = \arg(z_p) - \arg(z_{1a})$.
From $z_p = 4\left(\cos\left(\frac{5\pi}{6}\right) + i\sin\left(\frac{5\pi}{6}\right)\right)$, we have $\arg(z_p) = \frac{5\pi}{6}$.

$$\theta_{2a} = \frac{5\pi}{6} - \frac{\pi}{3} = \frac{5\pi}{6} - \frac{2\pi}{6} = \frac{3\pi}{6} = \frac{\pi}{2}$$

Since $|z_2| = 2$:
$$z_{2a} = 2\left(\cos\left(\frac{\pi}{2}\right) + i\sin\left(\frac{\pi}{2}\right)\right) = 2(0 + i\cdot 1)$$
$$\mathbf{z_{2a} = 2i}$$

#### Case B: Using $z_{1b} = -1 + i\sqrt{3}$

Polar form for $z_{1b}$:
$|z_{1b}| = 2$.
$$\arg(z_{1b}) = \pi - \arctan\left(\frac{\sqrt{3}}{1}\right) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$$
$$z_{1b} = 2\left(\cos\left(\frac{2\pi}{3}\right) + i\sin\left(\frac{2\pi}{3}\right)\right)$$

$$\theta_{2b} = \arg(z_p) - \arg(z_{1b})$$
$$\theta_{2b} = \frac{5\pi}{6} - \frac{2\pi}{3} = \frac{5\pi}{6} - \frac{4\pi}{6} = \frac{\pi}{6}$$

Since $|z_2| = 2$:
$$z_{2b} = 2\left(\cos\left(\frac{\pi}{6}\right) + i\sin\left(\frac{\pi}{6}\right)\right) = 2\left(\frac{\sqrt{3}}{2} + i\frac{1}{2}\right)$$
$$\mathbf{z_{2b} = \sqrt{3} + i}$$

### Final Solution

There are two pairs of complex numbers $(z_1, z_2)$ that satisfy the given conditions:

$$\text{Pair 1: } \mathbf{z_1 = 1 + i\sqrt{3}} \quad \text{and} \quad \mathbf{z_2 = 2i}$$
$$\text{Pair 2: } \mathbf{z_1 = -1 + i\sqrt{3}} \quad \text{and} \quad \mathbf{z_2 = \sqrt{3} + i}$$