Isn't this condition for $f$ to have a pole of order $k$ redundant?

Question

From Complex Analysis by Bak and Newman (p. 118):

If $f$ is analytic in a deleted neighborhood of $z_0$ and if there exists a positive integer $k$ such that $$\lim_{z→z_0} (z − z_0)^k f (z) \neq 0 \hspace{1em} \text{but} \hspace{1em} \lim_{z→z_0}(z − z_0)^{k+1} f(z) = 0,$$ then $f$ has a pole of order $k$ at $z_0$.

Why do we need this fragment

$\text{but} \hspace{1em} \lim_{z→z_0}(z − z_0)^{k+1} f(z) = 0$

isn't it redundant?

Answer 1

That requirement is to guarantee that the pole is of exactly order $k$, if the pole if of higher order we would have that $(z-z_0)^kf(z)$ would have a pole as well.

One could alternatively have written the extra requirement as $\lim (z-z_0)^kf(z) \ne \infty$ instead.

A simple example is $f(z) = (z-z_0)^{-k-1}$ which means that $(z-z_0)^kf(z) = (z-z_0)^{-1}$ and we would have that $f(z)$ has a pole of order $k+1$ there, but then $\lim (z-z_0)^kf(z) = \infty$ and $\lim (z-z_0)^{k+1}f(z) = \lim 1=1\ne0$.

As a note one can know that $f$ has a pole at all and it's not a essential singularity since the limit exists (due to Picard's theorem).

Answer 2

No, because if $\hspace{1em} \lim_{z→z_0}(z − z_0)^{k+1} f(z) = a \neq 0$

Then the Laurent expansion would have a non-zero factor on the $(z-z_0)^{-(k+1)}$ parameter and then the pole will be of order $k+1$ at least