Is this Determinant and Trace identity equivalent to Unitary matrix? Thanks for any help in advance. 
I have this equality for a 2x2 invertible complex matrix:
$$\text{Tr}(AA^*)=2|\text{det}(A)|^2$$
where $*$ is complex conjugate transposition. 
Is this equality equivalent to saying that $A$ is a unitary matrix? One direction is easy. Perhaps I am missing something obvious for the other direction. I haven't really been able to come up with a counter example yet either.
If it is equivalent I would love just a hint for the proof please! :)
Thanks again! 
 A: Writing $A = \begin{bmatrix}
a & b \\
c & d
\end{bmatrix}$, the equality becomes 
$$|a|^2 + |b|^2 + |c|^2 + |d|^2 = 2(|a|^2|d|^2 - 2\text{Re}(\overline{ad}bc) + |b|^2|c|^2)$$
If $|a| = 1$, $d = 1/a$, $c = 1$, this turns into
$$|b|^2 + 3 = 2(1 - 2\text{Re}(b) + |b|^2) \iff |b|^2 - 4\text{Re}(b) = 1$$ A concrete example is $A = \begin{bmatrix}
i & 2+\sqrt{5}i \\
1 & -i
\end{bmatrix}$, which is not unitary as $|\det A| \ne 1$.
A: If we assume that $A$ is normal and non-singular (there are no singular, non-zero solutions), then this becomes an equation on eigenvalues:
$$
|\lambda_1|^2 + |\lambda_2|^2 = 2|\lambda_1|^2 |\lambda_2|^2 \implies\\
2|\lambda_1|^2 |\lambda_2|^2 - |\lambda_1|^2  - |\lambda_2|^2 = 0 \implies\\
\frac 12(2 |\lambda_1|^2 - 1)(2|\lambda_2|^2 - 1) - \frac 12 = 0\implies\\
(2 |\lambda_1|^2 - 1)(2|\lambda_2|^2 - 1) = 1
$$
So, one solution is when $|\lambda_1| = |\lambda_2| = 1$, but we could equally have $|\lambda_1|^2 = 4$ and $|\lambda_2|^2 = \frac 47$, as in the matrix
$$
A = \pmatrix{2 & 0\\0&2/\sqrt{7}}
$$
I am not sure if there are any non-normal solutions.
