Geometrical proof of: | x + y |^2 = | x |^2 + | y |^2 + x y* + x* y, where x, y ∈C?

You can interpret this equation as the law of cosines.

Introduce z = -y, and write the numbers in the polar form

x = a exp(i φ), z = b exp(i ψ).

These numbers corespond to vectors in the Re-Im plane. The lengths of the vectors are a and b, and they form angles φ and ψ with the real axis.

|x-z| it the length of the third side of the triangle built on vectors x and y. According to the law of cosines

|x-z|^2 = a^2 + b^2 - 2 a b cos( φ - ψ) =

a^2 + b^2 - a b exp[ i (φ - ψ) + exp(-i (φ - ψ))] =

a^2 + b^2 - a exp(i φ) b exp(- i ψ) - a exp(-i φ) b exp(i ψ) =

|x|^2 + |z|^2 - x z* - x* z.

Substitute here z = - y to get your equality.