I know that we can express the function $f(x)=\frac{1}{1-x}$ as a power series given $|x|<1$:
$\frac{1}{1-x}=\sum_{n=0}^{\infty}x^{n}=1+x+x^2+...$
I am similarly trying to write out the function
$$\frac{1}{-xi+(\sqrt{5}+5i)y}$$
as a power series developed at $0$.
Here $x$ and $y$ are real variables. I am trying to find a formula for this online, but cannot seem to find any. My best idea is to fix either $x$ or $y$, but I feel like it would be better to treat $x$ and $y$ as separate variables.
Here's an approach that applies the usual geometric series $\frac{1}{1-x}=\sum_{j\ge0}x^j$ when $\left|x\right|\le 1.$ As mentioned by Thomas this series won't be centered about $\left(x=0,y=0\right)$.
If $y$ is nonzero, then
$$\frac{1}{-xi+(\sqrt{5}+5i)y}=\frac{1}{y\left(\sqrt5+5i\right)}\left(\frac{1}{1-x\frac{i}{y\left(\sqrt5 + 5i\right)}}\right)\\=\frac{1}{y\left(\sqrt5+5i\right)}\left(\sum_{j\ge 0}\left(\frac{xi}{y\left(\sqrt5+5i\right)}\right)^j\right)\\=\sum_{j\ge0}\frac{x^ji^j}{y^{j+1}\left(\sqrt{5}+5i\right)^{j+1}},$$ and the latest expression can be cleaned up a bit by using the fact that the powers of $i$ cycle from $i,-1,-i,1$. As long as $y\not=0$, this series is defined when $x=0$. Moreover, we can deal with $y$ separately by expanding $1/y$ as a series (but not centered at $y=0$ since $1/y$ is not defined there). E.g., we can write $$\frac{1}{y}=\frac{1}{1-(1-y)}=\sum_{j=0}^\infty \left(1-y\right)^j,$$ which is a series for $y$ centered at $1$. Combine this with your previous expression to obtain a power representation with respect to $x=0, y=1$. If you want the series centered at another pair $x,y$ then use some slightly different algebra than I did above.
Remark: If you want the series centered at $(0,1)=i$ in the complex plane, then the common term in the series should be of the form $(z-i)^j$ (from what I understand, that is a different question than what is currently being asked).
