Show that the map is onto 
Suppose $R=\{A\in GL(n,\mathbb C):A^{-1}=\bar A\}$, where $\bar A$ is the conjugate matrix of $A$. Show that $$b:GL(n,\mathbb C)\to R,\quad A\mapsto A\bar{A}^{-1}$$ is onto.

I don't know where to start. Is it possible to construct directly for any $M\in R$ a matrix $A$ such that $b(A)=M$? 
(The problem comes from a description of totally real subspaces of $\mathbb C^n$.)
 A: cf. https://mathoverflow.net/questions/221340/parameterize-unitary-without-transpose/222121#222121
It is shown that, if $A\in R$, then there is a real matrix $T$ s.t. $A=e^{iT}$. Thus $b(e^{iT/2})=e^{iT/2}e^{iT/2}=A$.
A: Here is an elementary proof. It suffices to show that:

If $C$ is a complex square matrix such that $C\bar{C}=I$, then $C=A\bar{A}^{-1}$ for some matrix $A$.

As all eigenvalues of $C\bar{C}$ are nonnegative, $C$ admits a Youla normal form $C=U^TLU$, where $U$ is unitary and $L$ is triangular (cf. corollary 3 of D.C. Youla, A normal form for a matrix under the unitary congruence group, Canad. J. Math. 13, 1961, pp.694-704).
So, from $C\bar{C}=I$, we get $L\bar{L}=I$. Yet, $L$ is triangular. Hence all its eigenvalues have unit moduli. Therefore, by absorbing an appropriate unit diagonal matrix into $U$, we may further assume that all eigenvalues of $L$ are equal to $1$. It follows that we may construct a matrix square root of $L$ using a real Hermite interpolating polynomial. In other words, we may pick $L^{1/2}=p(L)$ for some polynomial $p$ with real coefficients.
Therefore, $\overline{L^{1/2}} = \overline{p(L)} = p(\bar{L}) = p(L^{-1})$. Since $L^{-1}$ is a polynomial in $L$ and in turn a polynomial in $L^{1/2}$, we see that $\overline{L^{1/2}}$ commutes with $L^{1/2}$. So, from $L\bar{L}=I$, we get $\left(L^{1/2}\overline{L^{1/2}}\right)^2=I$. Yet, $L^{1/2}\overline{L^{1/2}}$ is a triangular matrix with all diagonal entries equal to $1$. So we must have $L^{1/2}\overline{L^{1/2}}=I$ and in turn $\overline{L^{1/2}}^{-1}=L^{1/2}$. Now define $A=U^TL^{1/2}$. Then
$$
A\bar{A}^{-1}=U^TL^{1/2}\overline{L^{1/2}}^{-1}\overline{U^T}^{-1}=U^TLU=C.
$$
