Why does it seem like we are selectively cancelling $(z-z_0)$ terms in the Laurent series in the limit $z\rightarrow z_0$?

Question

(Disclaimer, I am a Physics student not a mathematician).

I am looking at this nice video about the Laurent series. At 4:00 we are talking about a specific example of a complex function $f(z)$ with a simple pole at $z_0$. In the penultimate line on this slide we multiply both sides by $(z-z_0)$ which is fine to get

$$(z-z_0)f(z)=\sum_{k=0}^\infty a_k(z-z_0)^{k+1}+b_1, \tag{1}$$

then we claim that as $z\rightarrow z_0$ the $a_k$ terms cancel, but we don't say the same for the term on the left hand side? Is the assumption that the polynomial terms approach zero more quickly? Or is there something else I am missing?

Answer 1

Any power series defines a function, say $g$, that in particular is continuous on its region of convergence. So you can simple take $z=z_0$, that is, $$\lim_{z\to z_0}g(z)=g(z_0).$$ In your case this gives $$ \lim_{z\to z_0}\sum_{k=0}^\infty a_k(z-z_0)^{k+1}=\sum_{k=0}^\infty a_k(z-z_0)^{k+1}\Big|_{z=z_0}=0. $$

Answer 2

If you are given that $f(z)$ has a simple pole at $z_0$, then the left side approaches a nonzero number as $z \to z_0$. All the terms in the sum go to zero as $z \to z_0$, so $b_1$ must be the limit of the left side. Was $b_1$ the term in an expansion in powers of $z-z_0$ of something?. If $b_0\neq 0$ the right side will blow up and all the other $b$ terms go to zero because they have a higher power of $z-z_0$.