I have the function $$ f(z)={\frac{z}{z^2-2z+5}}, z_0=1 $$ i.e. I need to have it in form $f(z)={\sum{C_k(z-z_0)^n}}={\sum{C_k(z-1)^n}}$
I do $$ z-1=t, z=t+1, $$ $$ {\phi}(t)=f(t+1)={\frac{t+1}{(t+1)^2-2(t+1)+5}}={\frac{t+1}{t^2+2t+1-2t-2+5}} ={\frac{t+1}{t^2+4}}$$ I don't get how to continue from here. Usually I get something in the form $1+{\frac{C}{g(t)}}$ and try to use it, but here I don't see it.
Hint: One way is writing your result as geometric series $$\dfrac{1}{4+t^2}=\dfrac{1/4}{1+(\frac{t}{2})^2}=\dfrac14\left(1-\dfrac{t^2}{4}+\dfrac{t^4}{16}-\dfrac{t^6}{64}+\cdots\right)$$ where $|t|<2$.
Edit: $$\dfrac{t+1}{4+t^2}=\dfrac14\left[\left(1-\dfrac{t^2}{4}+\dfrac{t^4}{16}-\dfrac{t^6}{64}+\cdots\right)+t\left(1-\dfrac{t^2}{4}+\dfrac{t^4}{16}-\dfrac{t^6}{64}+\cdots\right)\right]$$ $$=\dfrac14\left[1-\dfrac{t^2}{4}+\dfrac{t^4}{16}-\dfrac{t^6}{64}+\cdots+t-\dfrac{t^3}{4}+\dfrac{t^5}{16}-\dfrac{t^7}{64}+\cdots\right]$$ $$=\dfrac14+\dfrac{t}{4}-\dfrac{t^2}{16}-\dfrac{t^3}{16}+\dfrac{t^4}{64}+\dfrac{t^5}{64}\pm\cdots$$