C* Algebra, f(x,z)

Question

Let $A$ be a $C^*$ algebra, $x\in A$ and $||x|| < 1$.

Let $f(x,z) = (1-x x^*)^{-\frac{1}{2}}(1+zx)$, $|z|=1, z\in \mathbb{C}$, $\mathbb{C}$ is the complex field.

How to prove:

$$ f(x,z)^* f(x,z) +1 = (1-xx^*)^{-1} + (1- x^* x)^{-1} + (1-xx^*)^{-1}zx + (1- x^*x)^{-1} \overline{z}x^*, $$

and

$$ x (1-x^*x)^{\frac{1}{2}} = (1-x x^*)^{\frac{1}{2}} x$$

Answer 1

For the second calculation, show first that for any polynomial p(t) with p(0) = 0, you get that $$x p(x^*x) = p(xx^*)x.$$ Then the fact that $$x(1 - x^*x)^{-\frac{1}{2}} = (1 - xx^*)^{-\frac{1}{2}}x$$ follows from the functional calculus applied to $x^*x$ and $xx^*$.

The first should be just a straightforward calculation after observing easy things like the fact that $(1 - xx^*)$ is self-adjoint, hence $(1 - xx^*)^{-1}$ is also self-adjoint.