Assuming you've shown
f is a homomorphism then take two distinct elements in your subring which is
M2(R) with some extra conditions on the elements of the matrix.
A general element of your subring looks like
(a−bba).
Consider where they are mapped to under
f is it possible for the two elements to get mapped to the same element in
C and still be distinct? ... Conclude
f is injective.
I think this a more "direct approach" alternatively note that
f is injective if and only if
ker(f) is trivial. So you could show that the only element that gets mapped to the zero in the complex numbers under f is in fact the zero of your subring.
Pick an arbitrary element in
C can you find a matrix in your subring that gets maps to your arbitrary complex number under
f? ... Conclude
f is surjective.
Hence
f is a homomorphism and bijective so
f is indeed a isomorphism.