square root commutes with multiplication for positive elements in a $C^*$ algebra? Let $A$ be a unital $C^*$ algebra.  If $z\in A$ is invertible, then so is $z^*$ and $z^*z$ and, furthermore, $z^*z$ is positive, so we can define using the functional calculus $|z|=\sqrt{z^*z}$.  My book then claims that $|z|$ is invertible with inverse $\sqrt{(z^*z)^{-1}}$.  Why is $|z|*|z|^{-1}=1$?  To me this looks like trying to say that $\sqrt{z^*z}*\sqrt{(z^*z)^{-1}}=\sqrt{(z^*z)(z^*z)^{-1}}=\sqrt1=1$, but I don't see why you can pull the product inside the square root like you can for reals (I don't know if this is actually how to prove the claim).
I don't see how some of the basic properties about continuous functions pass through the functional calculus and still hold inside of $A$.
 A: By definition, the (continuous) functional calculus associated to a normal operator $x$ is a continuous $*$-homomorphism $\varphi:C(\sigma(x))\to A$ sending the inclusion function $\sigma(x)\to\mathbb{C}$ to $x$, where $\sigma(x)$ is the spectrum of $x$.  In your case, $x=z^*z$, and $|z|$ is defined as $\varphi(f)$ where $f$ is the square root function.  Similarly, $|z|^{-1}$ is $\varphi(g)$, where $g$ is the function $g(t)=1/\sqrt{t}$ (which in $C(\sigma(x))$ since $0\not\in\sigma(x)$).  So $|z|\cdot|z|^{-1}=\varphi(f)\varphi(g)=\varphi(fg)$.  But the pointwise product of the functions $f$ and $g$ is just the constant function $1$, so $\varphi(fg)=\varphi(1)=1$.
More generally, the entire point of the functional calculus is that you can do everything "like you can for reals" (as long as you are only applying functions to a single normal operator, or more generally a commutative $*$-subalgebra), because addition and multiplication in $C(\sigma(x))$ are just pointwise addition and multiplication of ordinary numbers.
(If your definition of $|z|$ and/or $|z|^{-1}$ is not via the functional calculus on $x$, you can easily verify that the above functional calculus definitions meet the criteria of whatever other definition you have and thus coincide.)
A: Note that 
 $$\lvert z \rvert ^{-2} = ( \lvert z \rvert ^2)^{-1} = (z^*z)^{-2}
$$
Taking square roots gives the result.
A: Let $a=z^*z$ and $b=(z^*z)^{-1}=z^{-1}(z^*)^{-1}=z^{-1}(z^{-1})^*$. Then $a$ and $b$ are positive commuting elements for which $ab=1=ba$. Let $\sqrt{a}$ and $\sqrt{b}$ be the positive square roots of $a$ and $b$. Then $\sqrt{a}\sqrt{b}=\sqrt{b}\sqrt{a}$ because $\sqrt{a}$ and $\sqrt{b}$ are limits of polynomials in $a$ and $b$, respectively. Hence, $\sqrt{a}\sqrt{b}=\sqrt{b}\sqrt{a}$ is positive and
$$
            (\sqrt{a}\sqrt{b})^2 =ab=1.
$$
Therefore
$$
              (\sqrt{a}\sqrt{b}-1)(\sqrt{a}\sqrt{b}+1)=0.
$$
Because $\sqrt{a}\sqrt{b}$ is positive, then $\sqrt{a}\sqrt{b}+1$ is invertible. Hence $\sqrt{a}\sqrt{b}-1=0$, which is what you wanted to show:
$$
         \sqrt{a}^{-1}=\sqrt{b} \\
      \sqrt{z^*z}^{-1} = \sqrt{(z^*z)^{-1}}.
$$
