We are given two complex numbers, $3 + 2i$ and $6 + i$, which represent two vertices of an equilateral triangle. We need to find the product of the two possible complex numbers $c_1$ and $c_2$, which represent the third vertex of the equilateral triangle.

Step 1: Representing complex numbers geometrically

Let the complex numbers $z_1 = 3 + 2i$ and $z_2 = 6 + i$ be two points in the complex plane corresponding to two vertices of the equilateral triangle. The third vertex, $c$, could either be $c_1$ or $c_2$, as there are two possible locations for this point.

Step 2: Use rotation in the complex plane

To find the third vertex of the equilateral triangle, we use the fact that rotating one of the points by $60^\circ$ around the other point gives us one of the possible third vertices. Specifically, rotating $z_2$ around $z_1$ by $60^\circ$ gives one vertex, and rotating $z_2$ by $-60^\circ$ gives the other vertex.

The formula for rotating a point $z$ by an angle $\theta$ around another point $z_0$ is: $z' = z_0 + (z - z_0) e^{i \theta}$ For our case, $\theta = 60^\circ = \frac{\pi}{3}$ or $\theta = -60^\circ = -\frac{\pi}{3}$.

Step 3: Compute the two possible third vertices

We will compute the two possible third vertices $c_1$ and $c_2$.

1. For $c_1$ (rotating by $60^\circ$): $c_1 = z_1 + (z_2 - z_1) e^{i \frac{\pi}{3}}$ Here, $z_2 - z_1 = (6 + i) - (3 + 2i) = 3 - i$. Thus: $c_1 = (3 + 2i) + (3 - i) e^{i \frac{\pi}{3}}$

2. For $c_2$ (rotating by $-60^\circ$): $c_2 = z_1 + (z_2 - z_1) e^{-i \frac{\pi}{3}}$ Similarly: $c_2 = (3 + 2i) + (3 - i) e^{-i \frac{\pi}{3}}$

Step 4: Multiply $c_1$ and $c_2$

Rather than simplifying $c_1$ and $c_2$ individually, we can directly compute the product. Using the fact that the product of two complex numbers involves multiplying their magnitudes and adding their arguments, and noting that the arguments $\frac{\pi}{3}$ and $-\frac{\pi}{3}$ cancel each other out, we have: [ c_1 c_2 = \left[ (3 + 2i) + (3 - i) e^{i \frac{\pi}{3}} \right] \left[ (3 + 2i) + (3 - i) e^{-i \frac{\pi}{3}} \right] ] This expression simplifies nicely using the identity $e^{i \frac{\pi}{3}} \cdot e^{-i \frac{\pi}{3}} = 1$.

Thus, the product $c_1 c_2$ simplifies to: $c_1 c_2 = (z_2 - z_1)^2 = (3 - i)^2$

Step 5: Compute $(3 - i)^2$

Now, we compute the square of $3 - i$: $(3 - i)^2 = 9 - 6i + (-1) = 8 - 6i$

Final Answer:

The product of the two possible complex numbers $c_1$ and $c_2$ is $\boxed{8 - 6i}$.

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Do you want any further details or have any questions? Here are some follow-up questions that might help you explore this topic further:

1. What is the geometric interpretation of rotating complex numbers in the plane?
2. Can you find the exact coordinates of $c_1$ and $c_2$?
3. How does multiplying two complex numbers affect their magnitudes and arguments?
4. Why does rotating by $60^\circ$ and $-60^\circ$ give two possible vertices of the triangle?
5. Can this method be generalized to finding vertices of other regular polygons?

Tip: When rotating a point in the complex plane, using exponential form of complex numbers can greatly simplify the computations!