Prove that if $f$ has a pole of order $m$ in $z_0$, then there exist positive constants $r, C_1$ and $C_2$ such that $$C_1|z-z_0|^{-m}\leq |f(z)|\leq C_2|z-z_0|^{-m}$$ for all $z\in D'(z_0;r)$.
What I did was first write $f$ as $$f(z)=(z-z_0)^{-m}g(z)$$ where $f\in H(D(z_0;R))$ for some $R>0$ and $g(z_0)\neq 0$. Now, by the identity theorem, we can find sufficiently small $r<R$ such that $g(z)\neq 0$ for al $z\in D(z_0;r)$.
Now, $g$ is holomorphic on $D(z_0;r)$ and continuous on its boundary (since $g$ is also continuous on $D(z_0;R)$), so by the Maximum Modulus atheorem there is some $a\in \partial D(z_0;r)$ such that $|g(z)|\leq |g(a)|$ for all $z\in D(z_0;r)$. Letting $C_2=|g(a)|$ $$|f(z)|=|z-z_0|^{-m}|g(z)|\leq C_2|z-z_0|^{-m}$$
On the other hand, since $g(z)\neq 0$, $\frac{1}{g}$ is holomorphic in $D(z_0;r)$, somagain by the Maximum Modulus Theorem, there is some $b\in \partial D(z_0;r)$ such that $$ \frac{1}{|g(z)|}\leq \frac{1}{|g(b)|}$$ that is, $|g(b)|\leq |g(z)|$ for all $z\in D(z_0;r)$. Letting $C_1=|g(b)|$, we get the result.
Is this proof okay? I think I misused one of the theorems invoked above, but I'm not sure where or how. If not, how should I prove the result? Thanks in advance!
The proof is fine, just let me make some remarks:
you have $g(z) \ne 0$ on $D(z_0, r)$ not by the identity theorem (which is like using a cannon to shoot birds), but by the continuity of $g$ on $D(z_0, R)$ and that fact that $g(z_0) \ne 0$; so, it's continuity, not analyticity, that you use here;
a typo: in the sentence beginning with "On the other hand", it is $g(z_0) \ne 0$, not $g(z) \ne 0$;