Let $f$ be analytic on $B'(a;\delta)$. Then $f$ has a pole of order $m$ at $a$ iff $\frac {1} {f}$ has a zero of order $m$ at $a$.
My attempt $:$
Since $f$ has pole of order $m$ at $a$, so $m$ is the least positive integer for which $(z-a)^m f(z)$ has a removable singularity at $a$. Now let us consider a function $h$ on $B(a;\delta)$ defined by $h(z)=(z-a)^m f(z)$ for $z \in B'(a;\delta)$ and $h(a)=\lim_{z \rightarrow a} (z-a)^mf(z)$. Then $h$ is analytic on $B(a;\delta)$. Clearly $h(a) \neq 0$ for otherwise $(z-a)^kf(z)$ will have a removable singularity at $a$ for some $k < m$ which violates the minimality of such an $m$. Now since $f$ has a pole at $a$, $\frac {1} {f}$ has a removable singularity at $a$. We define $\frac {1} {f} (a) = \lim_{z \rightarrow a} \frac {1} {f(z)} = 0$. Then $\frac {1} {f}$ becomes analytic on $B(a;\delta)$. Also $\frac {1} {f} (z) = \frac {(z-a)^m} {h(z)}$ whenever $z \in B'(a;\delta)$ and $\frac {1} {f} (a) = 0 = \left [\frac {(z-a)^m} {h(z)} \right ]_{z=a}$. This shows that $\frac {1} {f} (z) = \frac {(z-a)^m} {h(z)}$, $\forall z \in B(a;\delta)$. Clearly $\frac {1} {f}$ has a zero of order $m$ at $a$.
Conversely if $\frac {1} {f}$ has a zero of order $m$ at $a$ then $\frac {1} {f} (z)=(z-a)^m f_1 (z)$ for all $z \in B(a;\delta)$, where $f_1$ is analytic in $B(a;\delta)$ and $f_1 (a) \neq 0$. Then $f(z) = \frac {1} {(z-a)^m f_1 (z)}$.Now $\lim_{z \rightarrow a} (z-a)^mf(z)$ is a non-zero complex number. So $(z-a)^mf(z)$ has a removable singularity at $a$ . Now for any $k<m$ it is easy to see that $\lim_{z \rightarrow a} |(z-a)^kf(z)|= \infty$ which proves the minimality of such an $m$. Hence $f$ has a pole of order $m$ at $a$.
Please check my answer whether it is correct or not.
Thank you in advance.