Representation of unitary matrix. If $S$ is a skew-Hermitian matrix, then it is known that  $A= (I+S)(I-S)^{-1}$ is a unitary matrix.
Now how to show that 
 every unitary matrix can be expressed in the above form provided $-1$ is not an Eigen value of $A$?
 A: We note to begin that, for invertible matrices $B$,
$BB^{-1} = I, \tag 1$
whence
$(B^{-1})^\dagger B^\dagger = (BB^{-1})^\dagger = I^\dagger = I, \tag 2$
whence
$(B^{-1})^\dagger = (B^\dagger)^{-1}, \tag 3$
which will prove its utility in what follows.
We validate the assertion that $S$ skew-Hermitian implies $A$ unitary, i.e., that
$S^\dagger = -S \Longrightarrow A^\dagger A = I \; \text{or} \; A^\dagger = A^{-1}; \tag 4$
with
$A = (I + S)(I - S)^{-1}, \tag 5$
we have
$A^\dagger = (  (I + S)(I - S)^{-1} )^\dagger = ((I - S)^{-1})^\dagger (I + S)^\dagger$
$=  ((I - S)^{-1})^\dagger (I + S^\dagger) = ((I - S)^{-1})^\dagger (I - S); \tag 6$
we apply (3) to (6):
$A^\dagger = ((I - S)^{-1})^\dagger (I - S) = ((I - S)^\dagger)^{-1} (I - S)$ $= (I - S^\dagger)^{-1} (I - S) = (I + S)^{-1} (I - S); \tag 7$
then
$A^\dagger A =  ((I + S)^{-1} (I - S))((I + S)(I - S)^{-1})$
$= ((I + S)^{-1}(I + S))( (I - S)(I - S)^{-1}) = II = I; \tag 8$
also,
$AA^\dagger = ((I + S)(I - S)^{-1})((I + S)^{-1} (I - S))$
$= ((I + S)(I + S)^{-1})( (I - S)^{-1}(I - S)) = II =I; \tag 9$
(8) and (9) show, jointly and separately, that $A$ is unitary.
We can also solve for $S$; from (5), it follows that $AS = SA$, and that
$A(I - S) = I + S; \tag{10}$
$A - AS = I + S; \tag{11}$
$S + SA = S + AS = A - I; \tag{12}$
$S(I + A) = A - I; \tag{13}$
given that $-1$ is not an eigenvalue of $A$, we obtain 
$\det(A + I) = \det(A - (-1)I) \ne 0, \tag{14}$
and thus $I + A$ is invertible, allowing us to write
$S = (A - I)(I + A)^{-1} = (I + A)^{-1}(A - I); \tag{15}$
with $A$ unitary,
$A^\dagger = A^{-1}, \tag{16}$
we have
$S^\dagger = ((A - I)(I + A)^{-1})^\dagger = ((I + A)^{-1})^\dagger (A - I)^\dagger = ((I + A)^{-1})^\dagger (A^\dagger - I); \tag{17}$
thus, again invoking (3),
$S^\dagger = ((I + A)^{-1})^\dagger (A^\dagger - I) = (I + A^\dagger)^{-1}( A^\dagger - I) = (I + A^{-1})^{-1}(A^{-1} - I)$
$= (A^{-1}(A + I))^{-1}(A^{-1} - I) = (A + I)^{-1}(A^{-1})^{-1}(A^{-1} - I) = (A + I)^{-1}A(A^{-1} - I)$
$= (A + I)^{-1}(I - A) = -(A + I)^{-1}(A - I) = -S, \tag{18}$
and we have shown $S$ to be skew-Hermitian.
In closing, I add a few remarks on the ivertibility of $I \pm S$, which has been assumed at several points in the above argument; we note that $S^\dagger = -S$ implies $(iS)^\dagger = -iS^\dagger = -i(-S) = iS$, that is, $iS$ is a Hermitian matrix; as such, all of its eigenvalues are real, and thus the non-zero eigenvalues of $S$ itself are purely imaginary; the eigenvalues of $I \pm S$ are precisely those complex numbers of the form $\pm \mu + 1$, $\mu$ being an eigenvalue of $S$; therefore the eigenvalues of $I \pm S$, each having real part $1$, are all non-zero, and this is sufficient for the ivertibility of $I \pm S$, since it forces $\det(I \pm S) \ne 0$.
