How to handle an identical zero pole when expanding to a Laurent Series?

Question

I have a function given as $f(z)=\frac{2(z-3)}{z^2-8z+15},$ which is clearly the same as $f(z)=\frac{2(z-3)}{(z-3)(z-5)}$ when $z\neq3$. Typically I would break these into two separate fractions using $f(z) = \frac{A}{z-3} + \frac{B}{z-5},$ but because of the apparent reducibility I get the obvious $f(z) = \frac{2}{z-5}$. But it doesn't seem this is correct. How would I go about doing this properly?

Answer 1

Yes, $f(z)=\dfrac2{z-5}$, for each $z\in\mathbb{C}\setminus\{3,5\}$. Note, however, that the domain of $f$ doesn't change. You can extend $f$ to a (continuous) functions from $\mathbb{C}\setminus\{3\}$, but that's an extension, not the original function. However, as far as the computation of the Laurent series centered at $3$, this is irrelevant; just work with $\dfrac2{z-5}$.