Square root of a positive operator I was reading the proof of the existence of the unique positive square root operator of positive operator from the Reed & Simons Book. The author starts with the power series expansion of the function $\sqrt{(1-z)}=1 + \sum_{k=1}^\infty c_kz^k$ on $|z|\leq 1$. I have the following questions:
(1) Why the power series expansion is valid in $|z|\leq 1$? Why there is an equality? I feel to make this function analytic, we need to modify its domain by introducing a Branch cut $(1,\infty)$.
(2) Is not the variable $z$, when replaced by the operator $A$, define the square root of the operator $(I-A)$? To prove the existence of the square root operator, how can we use square root of another operator?
I do apologize if my question seems silly. I will be happy if the proof of Reed and Simons explained in details or any other proof of the existence of the square root operator is given (of course in the infinite dimensional setting).
 A: $(1)$ The cut $(1,\infty)$ does not intersect the closed disk $\bar D\subset \Bbb C$ centered in zero with radius one. Also, we can show the convergence of the corresponding power series for any $z\in \bar D$. Let us compute the coefficients $(c_k)$.
The Taylor series is the one for $(1-z)^{1/2}$, so the binomial coefficients show up:
$$
\begin{aligned}
(1+z)^{1/2} &=\sum_{k\ge 0}\binom{1/2}kz^k\ ,\\
(1-z)^{1/2} &=\sum_{k\ge 0}(-1)^k\binom{1/2}k z^k\ ,\\
(1-z)^{1/2} &=\sum_{k\ge 0}\underbrace{(-1)^k\binom{1/2}k}_{c_k} z^k\ ,\\
c_k &= (-1)^k\binom{1/2}k =
(-1)^k\cdot\frac 1{k!}\cdot\frac12\left(\frac12-1\right)\dots\left(\frac12-k+1\right)
\ ,
\\
&\qquad\text{ so for instance}\\
c_0 &= +1\ ,
\\
c_1 &= -\frac 12\ ,
\\
c_2 &= +\frac 12\left(-\frac 12\right)\cdot\frac 1{2!}=-\frac 18\ ,
\\
c_3 &= -\frac 12\left(-\frac 12\right)\left(-\frac 32\right)\cdot\frac 1{3!}
=-\frac 1{2\cdot 3}\cdot \frac{1\cdot 3}{2\cdot 4}=-\frac 1{16}\ ,
\\
c_4 &= +\frac 12
\left(-\frac 12\right)
\left(-\frac 32\right)
\left(-\frac 52\right)
\cdot\frac 1{4!}
=-\frac 1{2\cdot 4}\cdot \frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}
=-\frac 5{128}\ ,
\\
&\qquad\text{ and so on, so that we have in general}
\\
c_k &=-\frac 1{2k}\cdot
\frac
{1\cdot 3\cdot5\dots (2k-3)}
{2\cdot 4\cdot6\dots (2k-2)}\ .
\end{aligned}
$$
In order to show absolute convergence on $D$, it is enough to show convergence for $1+|c_1|+|c_2|+|c_3|+\dots$ which is computable, and equal to
$$
\begin{aligned}
&1+|c_1|+|c_2|+|c_3|+\dots
=1-c_1-c_2-c_3-\dots\\
&\qquad=2-(1+c_1\cdot 1+c_2\cdot 1^2+c_3\cdot 1^3+\dots)=2-(1-1)^{1/2}=2\ .
\end{aligned}
$$
We have thus a complex analytic function on the open disk $D$ centered in $0$ with radius one, with expansion $\sum c_kz^k$, which is continuous on the closed disk $\bar D$. We are in position to apply analytic functional calculus.

$(2)$ Now Theorem VI.9 from loc. cit. works as follows.
Start with $A\ge 0$. We may and do assume for our purposes that $A\le I$.
(I will write below $1$ instead of $I$.)
Else replace $A$ by $A/\|A\|$. From $0\le A\le 1$ we get
Now $0\le 1-A\le 1$. Denote by $Z$ this operator, $Z=1-A$. We plug it in into the series $\sum c_kz^k$, obtain the series of operators
$$
\sum c_kZ^k\ ,
$$
which is also "absolute convergent", and a Cauchy-like criterion for it shows that this limit of partial sums exists in our operator space, which is a Banach space. Explicitly, fix some $N$, consider $m,n$ with $N\le m< n$, then  use the estimation
$$
\left\| 
\sum_{N\le m\le k\le n } c_kZ^k
\right\|
\le 
\sum_{N\le m\le k\le n } 
|c_k|
\left\| 
Z^k
\right\|
\le 
\sum_{N\le m\le k\le n } 
|c_k|\;\|Z\|^k
\le 
\sum_{N\le m\le k\le n } 
|c_k|
\ ,
$$
and use the knowledge of the Cauchy property for the real series $\sum|c_k|$
to conclude. Let $B$ be the limit,
$$
B = \sum c_kZ^k\ .
$$
Then there is a similar argument to estimate and compute the product of series  as in real analysis, and we get
$$
\begin{aligned}
B^2&
=\left(\sum c_kZ^k\right)^2
=\left(\left(\sum c_k z^k\right)_{\text{in z=Z}}\right)^2
=\left(\left(\sum c_k z^k\right)^2\right)_{\text{in z=Z}}
\\
&=(1-z)_{\text{in z=Z}}=1-Z
=1-(1-A)=A\ .
\end{aligned}
$$
$\square$
