Does the identity $\det(I+g^{-1})\det(I+g)=|\det(g-I)|^2$ hold for $g \in U(n)$?

Question

In a paper (corollary 1, p.14) the following identity is used:

Let g be a unitary matrix. Then:

$$\det(I+g^{-1})\det(I+g)=|\det(g-I)|^2 \text{ for }g \in U(n)$$

Now my question is why this holds

I calculated:

$$\det(I+g^{-1})\det(I+g)=\overline{\det(I+g^t)}\det(I+g)=\overline{\det(I+g)}\det(I+g)=|\det(I+g)|^2$$

Where the second equality holds as $I$ has only entries in the diagonal ($I$ is of course the unit matrix). But this is not the same as on the right side.

(I also thought that maybe there was a typo on the left side where should be minus-signs. However in the paper itself it is needed that there are plus-signs.)

Thanks for any hints.

Edit: This equality was in the scope of an integral: $$\int_{U(n)}\prod_{l=1}^{k}det(I+g^{-1})\prod_{l=1}^{k}\det(I+g)dg=\int_{U(n)}|\det(g-I)|^{2k}dg$$

With a change of variable it was solved with my calculation done above. See Giuseppe's answer.

Answer 1

Dear AndreasS I have given a look at the paper.

I think that the calculation $$\det(I+g^{-1})\det(I+g)=|\det(I+g)|^2 \text{ for }g \in U(n)$$ is correct.
But in order to obtain the result stated in Corollary 1, you just need the change of variable $g\mapsto -g$ in the integral over $U(n)$, so that $$\int_{U(n)}|\det(g-I)|^{2k}dg=\int_{U(n)}|\det(I+g)|^{2k}dg.$$ Then in the paper you find how factorize $|\det(I+g)|^{2k}.$

I hope that it helps.