Let $f$ be a holomorphic function on an open, connected set $\Omega$ in $\mathbb{C}$ and define $g$ by $g(z) = 1/f(z)$ if $f(z) \neq 0$. Recall that $f$ has a zero of order $k$ at $z \in$ $\Omega$ if $f(z) = f'(z) = f''(z) = ... = f^{(k-1)}(z) = 0 $ and $f^{(k)}(z)\neq 0$. Prove that:
(a) If $f$ has a zero of order $k$ at $z \in$ $\Omega$ then $g$ has a pole of order $k$ at $z$.
(b) If $g$ has a pole of order $k$ at $z \in$ $\Omega$ then $f$ has a zero of order $k$ at $z$.
For Part (a):
Let $f$ be a holomorphic function on an open, connected set $\Omega$ in $\mathbb{C}$.
Let $f(z)$ has a zero of order $k$ at $z \in$ $\Omega$
Then $f(z)=(z-a)^k \phi (z)$ where $\phi$ $(z)$ is an analytic function in $\Omega$ and $z = a$ is a zero of order $k$.
Now: $$g(z) = \frac{1}{f(z)}$$ $$g(z) = \frac{1}{(z-a)^k \phi (z)}$$ $$g(z) = \frac{\phi_1 (z)}{(z-a)^k}$$ (where $\phi_1 (z) = \frac{1}{\phi (z)}$)
Then: $z=a$ is a pole of order $k$ of $g$.
For Part (b):
Let $g$ has a pole of order $k$ at $z \in$ $\Omega$
Then $$g(z) = \frac{h(z)}{(z-a)^k}$$ where $h(z)$ is an analytic function in $\Omega$.
Now:$$f(z) = \frac{1}{g(z)} = \frac{1}{\frac{h(z)}{(z-a)^k}} = \frac{(z-a)^k}{h(z)}$$ $$f(z) = (z-a)^k h_1(z)$$ where $h_1(z) = \frac{1}{h(z)}$ since $h(z) \neq 0 ,\forall z \in$ $\Omega$
Then: $z = a$ is a pole of order $k$ of $f$.
Can someone please check this method of my proof?
The idea is good, but the proof have some important details left.
For example let's show (b).
Suppose that $g$ has a pole of order $k$ at $a\in\Omega.$ Then there exists an function $h$ analytic in a neighborhood $U$ of $a$ such that $$ g(z)=\frac{h(z)}{(z-a)^k}$$ in such neighborhood, and $h(a)\neq 0.$ Since $h$ is analytic in $U,$ is in particular continuous, hence $h(z)\neq 0$ for all $z\in V$ where $V$ is some neighborhood of $a$ contained in $U.$ Then $$ f(z)=\frac{(z-a)^k}{h(z)}$$ for all $z\in V,$ and $1/h$ is analytic in $V$ since $h$ is analytic and have no zeros in $V.$ Moreover, since $h(z)\neq 0$ for all $z\in V,$ then $f$ has a zero of order $k$ in $a.$
Part (a) is similar, you should try to fill the details left in your proof.