To solve $(\sqrt{3} + i)^2 (1 - i)^2$ and express it in the form $x + iy$, let's go through the calculations step-by-step.

Step 1: Expand $(\sqrt{3} + i)^2$

Using the formula $(a + b)^2 = a^2 + 2ab + b^2$:

1. $a = \sqrt{3}$ and $b = i$,
2. $a^2 = (\sqrt{3})^2 = 3$,
3. $b^2 = i^2 = -1$,
4. $2ab = 2 \cdot \sqrt{3} \cdot i = 2\sqrt{3} \, i$.

Therefore, $(\sqrt{3} + i)^2 = 3 + 2\sqrt{3} \, i - 1 = 2 + 2\sqrt{3} \, i.$

Step 2: Expand $(1 - i)^2$

Using the same formula:

1. $a = 1$ and $b = -i$,
2. $a^2 = 1$,
3. $b^2 = (-i)^2 = -1$,
4. $2ab = 2 \cdot 1 \cdot (-i) = -2i$.

So, $(1 - i)^2 = 1 - 2i - 1 = -2i.$

Step 3: Multiply the Results

Now, we have: $(\sqrt{3} + i)^2 (1 - i)^2 = (2 + 2\sqrt{3} \, i)(-2i).$

Distribute $-2i$ across each term in $2 + 2\sqrt{3} \, i$:

1. $2 \cdot (-2i) = -4i$,
2. $2\sqrt{3} \, i \cdot (-2i) = -4\sqrt{3} \, i^2 = -4\sqrt{3} \cdot (-1) = 4\sqrt{3}$.

Combining these, we get: $(\sqrt{3} + i)^2 (1 - i)^2 = 4\sqrt{3} - 4i.$

Final Answer

The expression in $x + iy$ form is: $4\sqrt{3} - 4i.$

Would you like further details or have any questions?

Here are five related questions to deepen your understanding:

1. How would the result change if we computed $(\sqrt{3} - i)^2(1 + i)^2$ instead?
2. What is the modulus of $4\sqrt{3} - 4i$?
3. Can you express $4\sqrt{3} - 4i$ in polar form?
4. How do we square complex numbers in polar form?
5. What is the significance of the imaginary unit $i$ in complex number multiplication?

Tip: When dealing with complex numbers in the form $a + bi$, always separate real and imaginary parts to simplify calculations and keep track of $i^2 = -1$.