(a) $a_{-1} \ne 0$, $a_{-n} = 0 \quad$ for all $n \ge 2$
(b) $a_{-N} \ne 0$ for some $N \ge 1$ and $a_{-n} = 0$ for all $n > N$
(c) $a_{-n} = 0$ for all $n \ge 1$
(d) $a_{-n} \ne 0$ for all $n \ge 1$
Explanation: (b) is correct.
Given that $\lim_{z \to 0} |f(z)| = \infty$, i.e. $z=0$ is a pole of $f(z)$ (But we don't have information about its order).
If $a_{-1} \neq 0, a_{-n}=0$ for $n \ge 2$
$ \implies f(z)$ has a term $\dfrac{a_{-1}}{z}$ as the only negative power of $(z-0)$
$ \implies z=0$ is a pole of order 1. But we don't have information about the order
$\implies$ (a) is false.
If $a_{-n}=0$ for $n \ge 1$ then $ f(z)$ has no negative term in the Laurent series.
$\implies z=0$ is a removable singularity, but given $z=0$ is a pole
$\implies$ (c) is false.
If $a_{-n} \neq 0$ for $n \ge 1$, then $f(z)$ has infinite negative powers of $(z-0) $
$\implies z=0$ is an essential singularity. Again, incorrect.
$\implies$ Option (b) is true, which says $f(z)$ at $z=0$ has a pole of some order $N$.