If $f$ is analytic on and outside the simple closed contour $C$ and has a zero of order 2 or more at $\infty$, then $\int_Cf(z)\,dz=0$. My Goal:
I want to prove that if $f$ is analytic on and outside the simple closed contour $C$ and has a zero of order 2 or more at $\infty$, then $\int_Cf(z)\,dz=0$.
Here is my attempt at my work and where I got stuck.
My Work:
Denote $w = \frac 1 z$. Since the zero is at $\infty$, then $f(\frac 1 w)$ has a zero of order $m \geq 2$ at $w = 0$. As such we can write $f(\frac 1 w) = w^m h(w), m \geq 2$ for some $h(w$) that is analytic on and outside the simple closed contour, and also analytic at $0$.
Now since $z = \frac 1 w$ then $\frac {dz}{dw} = -\frac 1 {w^2}$ and we have the integral
$$\int_Cf(z)\,dz = -\int_Cf(1/w)\frac{dw}{w^2} = -\int_C w^{m-2}h(w)\,dw.$$
I'd like to perhaps apply Cauchy's Integral Theorem, but I'm not sure how to do that, since it doesn't seem like the assumptions about simple connected domains and analyticity hold here - after all, we only knew that $f$ was analytic on and outside the contour, not inside. How can I finish up?
Note: This is a repeat of $f$ is analytic on and outside the simple closed contour $C$ and has a zero of order 2 or more at $\infty$. Show$\int_Cf(z)~dz=0$., which received no answers.
 A: Pick $z_0$ inside the domain enclosed by $C$, and take the Laurent series $$f(z)=\sum_{n=0}^\infty  \frac{a_n}{(z-z_0)^n}=\sum_{n=2}^\infty \frac{a_n}{(z-z_0)^n}$$
Then $$\int_{C} f(z)dz=\sum_{n=1}^\infty a_n\int_C \frac{1}{(z-z_0)^n}dz=0$$
since the residue of $\frac{1}{(z-z_0)^n}$ vanishes at $z=z_0$.
There are some technical details to be worried about, such as $C$ needs to be piecewise $C^1$ or at least rectifiable for the integral to be defined, but nothing too serious.
Here is another approach, using only the Cauchy integral formula. By the given condition, $g(w):=f(\frac{1}{w}+z_0)$ is analytic inside $C'=\{w=\frac{1}{z-z_0} : z\in C\}$ that encloses $0$, where $z_0$ is any point enclosed by $C$. Therefore by the Cauchy integral formula (or rather Cauchy's differentiation formula) $$g'(0)=\frac{1}{2\pi i}\int_{C'} \frac{g(w)}{w^2}dw=0$$ $$\Rightarrow\int_{C'}\frac{f(\frac{1}{w}+z_0)}{w^2}dw=0 \Rightarrow \int_{C} \frac{f(z)}{1/(z-z_0)^2}d(\frac{1}{z-z_0})=0\Rightarrow\int_C f(z)dz=0$$
