For example, $$\frac{3}{z^2-5z+4}$$ has a Laurent series expansion on the angular region $1<|z|<4$. Does the real function $$\dfrac{3}{x^2-5x+4}$$ have some sort of Laurent series on $(-4,-1)\cup (1,4)$?
I'm taking complex analysis, but I have almost no experience with series from calculus. It wasn't in the curriculum. I really need to fill in this gap in my math knowledge :(
The "real" Laurent series for $3/(x^2 - 5 x + 4)$ on $(-4,-1) \cup (1,4)$ is just the restriction to that domain of the Laurent series for $3/(z^2 - 5 z + 4)$ on the annulus $1 < |z| < 4$. But there's no advantage in looking at the real version: it's easier to understand the complex version of Laurent series than the real version.