To prove {1+i*3^(1/2)}{1-i*3^1/2)}^2 = 4.

We know i = (-1)^(1/2) and therefore i^2 = -1.

The left side is in the form (a+b)(a-b) where a = 1 and b = i*3^(1/2)

We know (a+b)(a-b) = a^2 -ab+ab - b^2 = a^2-b^2.

Therefore (a+b)(a-b) = a^2-b^2....(1)

a^2 = 1^2 = 1.

b^2 ={ i*3^(1/2)}^2 = {(i)^2}{3^(1/2)}^2 = -1*3 = -3.

Therefore using (1) we get:

{1+ i*3^(1/2)}{1-i*3^(1/2)} = 1- (-3) = 4.

Therefore (1 + i*3^1/2 )^2 + (1 - i*3^1/2 )^2 = 4 . (But not equal to -4).

We'll expand the squares from the left sides, using the formula:

(a+b)^2 = a^2 + 2ab + b^2

We'll put a = 1 and b = i*sqrt3

(1 + i*sqrt3 )^2 = 1 + 2i*sqrt3 + 3i^2

But i^2 = -1

(1 + i*sqrt3 )^2 = 1 + 2i*sqrt3 - 3

We'll combine like terms:

(1 + i*sqrt3 )^2 = 2i*sqrt3 - 2 (1)

Now, we'll expand the square (1 - i*sqrt3 )^2:

(1 - i*sqrt3 )^2 = 1 - 2i*sqrt3 - 3

(1 - i*sqrt3 )^2 = -2i*sqrt3 - 2 (2)

We'll add (1) and (2):

2i*sqrt3 - 2 - 2i*sqrt3 - 2 = -4

We'll eliminate like terms:

**-4 = -4**

**The identity is verified!**