When one defines a power series, it is also necessary to define what the domain of convergence is. It's basically similar as to when we define a function: we have to define a domain. You want to express the function
$f(z)=\frac 1 z $
as a power series around $z=i+1$. You cleverly now that the function
$ g(z)=\frac{1}{z-a}$
can be expanded as
$ g(z)=\frac{1}{z-a} =-\frac 1 a\frac{1}{1-z/a} =-\frac 1 a \sum_{n=0}^\infty \left(\frac z a \right)^n$
But when is this representation legitimate? If we plug any $z$ such that $|z|>a$, we'll find ourself with a divergent series. So, what we should really write is
$ g(z)=-\frac 1 a \sum_{n=0}^\infty \left(\frac z a \right)^n\text{ ; }\color{red}{ |z|
Note that it can't be the case that $a=0$.
Moving on to your problem. We have that,
$f(z)=\frac 1 z =\frac 1 {z-a+a}=-\frac 1 a \frac 1 {1-\frac {z+a}{a}}$ so again we write
$f(z)=-\frac 1 a \frac 1 {1-\frac {z+a}{a}}=-\frac 1 a \sum_{n=0}^\infty \left(\frac {z+a}{a} \right)^n$
Note that again it cant be the case that $a=0$. And similarily, we need that
$|z+a|<|a|$ for the series to converge. So our domain of convergence will be
$\Bbb D = \{ z \in \Bbb C : |z+a|<|a|\}$
In this case we want to expand around $z=i+1$, so we choose $a=-(i+1)$, which gives
$f(z)=\frac 1 {i+1} \sum_{n=0}^\infty (-1)^n \left(\frac {z-(i+1)}{i+1} \right)^n$
$\frac 1 {i+1}=\frac {1-i}{2}$
so we can write this as
$f(z)=\frac {1-i}{2} \sum_{n=0}^\infty (-1)^n \left(\frac {1-i}{2}\right)^n \left( z-(i+1) \right)^n$
As a final step, we find what $\Bbb D$ should be.
$|z-(i+1)|<|i+1|$
$|z-(i+1)|< \sqrt 2$
So the series can be used in $\Bbb D$, a disk of radius $\sqrt 2$ and center at $z=i+1$.