is it true that A multiplied by its conjugate A* is always greater than zero?
and if so how do i prove/show this?
That's a strange sort of conjugation. I would have taken the conjugate of $\displaystyle \sqrt{2} - 2$ to be $\displaystyle -\sqrt{2} - 2$, in which case AA* is positive.
If A is meant to be a matrix and A* is its hermitian adjoint then AA* is always positive. Reason: $\displaystyle \langle AA^*x,x\rangle = \langle A^*x,A^*x\rangle = \|A^*x\|^2\geqslant0$.
another example is through the usage of complex numbers:
A=(a+j), A* = (a-j)
A.A*=(a^2 - aj + aj- j^2)
= a^2-(-1^(1/2))^2
=a^2 - (-1)
= a^2 + 1 which is always larger than 0
Complex Numbers
The word conjugate is one of those words in mathematics which has many different meanings. There is a standard algebraic/number theoretic definition. Let $\displaystyle F$ be an extension field over a smaller field $\displaystyle K$. Given two algebraic elements $\displaystyle \alpha,\beta\in F$ we say $\displaystyle \alpha,\beta$ are conjugates iff they has the same minimal polynomial. So $\displaystyle a+bi$ and $\displaystyle a-bi$ ($\displaystyle b\not = 0$) are conjugates because they have the same minimal polynomials.
Though that definition above is applied to fields we can extend it a little and apply it to $\displaystyle \mathbb{Z}$ (which is not a field) as well. Given two algebraic number $\displaystyle \alpha,\beta$ we say $\displaystyle \alpha$ and $\displaystyle \beta$ are conjugates if they have the same minimal polynomial over $\displaystyle \mathbb{Z}$. In your case $\displaystyle \alpha = \sqrt{2} - 2$ and $\displaystyle \beta = -\sqrt{2} - 2$ have $\displaystyle x^2+4x+2$ as their minimal polynomial.