Find a power series expansion for $\frac{1}{z}$ around $z = 1 + i.$ Find a power series expansion for $\frac{1}{z}$ around $z = 1 + i.$
My Solution
For any complex $\alpha$,$\frac{1}{z}=\frac{1}{\alpha+z-\alpha}=\frac{1}{\alpha[1+\frac{z-\alpha}{\alpha}]}$
$=\frac{1}{\alpha}[1-\frac{z-\alpha}{\alpha}$$+\frac{(z-\alpha)^{2}}{\alpha^{2}}]-+...]$
$\displaystyle\sum_{k=0}^\infty \frac{(-1)^{k}(z-\alpha)^{k}}{\alpha^{k+1}}$ and then $\alpha=1+i$
Does anyone can help me improve it?
 A: When one defines a power series, it is also necessary to define what the domain of convergence is. It's basically similar as to when we define a function: we have to define a domain. You want to express the function
$$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|<a}$$
Note that it can't be the case that $a=0$.
Moving on to your problem. We have that, 
$$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 $z=i+1$.
