Csdrhrt

I'm asked to find the radius of convergence of $dfrac{z+1}{z-i}$ around $z_0 = 2+i$

The way I tried to go about it is by using the geometric series.

$$dfrac{z+1}{z-i}= dfrac{z}{z-i}+dfrac{1}{z-i} = dfrac{1}{Big(1-dfrac{i}{z}Big)} + dfrac{1}{z}.dfrac{1}{Big(1-dfrac{i}{z}Big)} = displaystylesumBig(dfrac{i}{z}Big)^k + dfrac{1}{z}∑Big(dfrac{i}{z}Big)^k .$$
we can then use that the geometric series $sum a^k$ converges to $dfrac{1}{1-a}$ for $|a| < 1$ with $a = dfrac{i}{z}$. So we say we must have $Big|dfrac{i}{z}Big|<1$ i.e. $|z|>1$, making the radius of convergence $R = 1$.

Now, I know the answer is $R = 2$, so I know this is wrong, I just don't know why or where. And I also don't understand the relevance of $z_0 = 2 + i$ and where to use that.

You should try to expand with respect to $w:=z-z_0=z-2-i$:
$$frac{z+1}{z-i}=frac{w+3+i}{w+2}=
frac{w/2}{1-(-w/2)}+frac{(3+i)/2}{1-(-w/2)}.$$
Note that the geometric series involved is convergent when $|-w/2|<1$, that is $|z-z_0|<R=2$.

Theorem. The radius of convergence of a power series development of an analytic function $f$ at a point $a$ is the radius of the largest disc centred at $a$ and contained in the domain of $f.$

Here the domain of $f$ is $mathbf{C}-{i}$ and the point $a = 2 + i,$ hence the largest disc has radius $|a-i|= 2.$ $square$