I have a function $$f(z)=\frac{z^2-2z+5}{z-1}$$ and i'm supposed to use Laurent series to expand it at the singularity $z=1$
I tried to use geometric series, didn't work then i thought i could turn it into
$$f(z)=\frac{z(z-1)+5}{z-1}=z+\frac{5}{z-1}=z+\frac{1}{i}\frac{1}{1+\frac{z-1-i}{i}}=z-\frac{1}{i}\sum (\frac{z-1-i}{i})^n$$
But that's the most unfortunate thing i've seen in my life...So....
I'm also supposed to determine the residue and use it to calculate the integral along the unit circle around $z=1$ so i guess i can't use what i did to do that...
I guess i used the wrong series manipulation to do this...appreciate any help.
First, note that $z^2-2z+5=(z-1)^2+4\ne z(z-1)+5$. Then, we can write
$$\frac{z^2-2z+5}{z-1}=(z-1)+\frac4{z-1}$$
which constitutes the Laurent series around $z=1$. And it is easy to see that the residue is $4$.