Given $\alpha \in \mathbb{C}$ and $F$ a subfield of $\mathbb{C}$, define $F(\alpha) = \{a_n\alpha^n + \cdots + a_1\alpha + a_0 \mid a_i \in F\}$. How do we show $F(\alpha) = F(\alpha^{-1})$ if $\alpha$ is algebraic over $F$?
I have proven that $\alpha^{-1}$ is also algebraic over $F$ and want to use this to prove this. I know if $f(\alpha) = a_n\alpha^n + \cdots + a_1\alpha + a_0 =0$ then I can have
$$g(\alpha^{-1}) = a_n + a_{n-1}\alpha^{-1} + \cdots + a_1\alpha^{-(n-1)} + a_0\alpha^{-n} = \frac{1}{\alpha^n}f(\alpha) = 0$$
But if I am given $f(\alpha) = y$ for some arbitrary $y \in \mathbb{C}$ I couldn't seem to find the $g$ that $g(\alpha^{-1}) =y$ using similar method.
It is of course not restrictive to assume that $\alpha\notin F$.
Since $\alpha$ is algebraic, there exists a polynomial $f(x)=c_0+c_1x+\dots+c_mx^m$ with $c_0\ne0$ and $c_n\ne0$ such that $f(\alpha)=0$. The minimal polynomial would do, for instance. Note that $m>1$ as $\alpha\notin F$.
Then $$ -c_0\alpha^{-1}=c_1+c_2\alpha+\dots+c_n\alpha^{n-1} $$ easily translates to $$ \alpha^{-1}=b_0+b_1\alpha+\dots+b_{n-1}\alpha^{n-1} $$ with obvious meaning of $b_i$.
This proves that $\alpha^{-1}\in F(\alpha)$. Since you have certainly proved that $F(\alpha)$ is a ring, it follows that $F(\alpha^{-1})\subseteq F(\alpha)$.
We have not needed that $\alpha^{-1}$ is algebraic. Indeed, this is a consequence: any element $\beta\in F(\alpha)$ is algebraic over $F$, because $F(\beta)$ is finite dimensional over $F$, being a subspace of $F(\alpha)$, which is generated as a vector space by $\{1,\alpha,\dots,\alpha^{m-1}\}$ (because of polynomial division).
Now the converse inclusion follows by symmetry.
More standard notation for your $F(\alpha)$ is $F[\alpha]$, reserving the former for the quotient field of the latter. The two rings coincide if and only if $\alpha$ is algebraic over $F$.
Personally, I'd prefer a different path.
If $K$ is an extension field of $F$ and $\alpha\in K$, $\alpha\ne0$, define $$ F[\alpha]=\{f(\alpha):f(x)\in F[x]\} $$ This is easily seen to be a subring of $K$ containing both $F$ and $\alpha$. Indeed, it is the image of the ring homomorphism $$ v_\alpha\colon F[x]\to K,\qquad v_\alpha(f)=f(\alpha) $$ (evaluation of the polynomial $f$ at $\alpha$). This homomorphism is injective if and only if $\alpha$ is a root of no nonzero polynomial in $F[x]$, that is to say $\alpha$ is transcendental over $F$.
Otherwise the homomorphism has a nonzero kernel; by general facts on $F[x]$, the kernel is a principal ideal generated by a unique monic polynomial $g(x)$, called the minimal polynomial of $\alpha$ over $F$. Let's say that the degree of $g(x)=c_0+c_1x+\dots+c_mx^m$ is $m$; by construction, $g(\alpha)=0$. It's easy to prove that $c_0\ne0$.
In this case we can consider, for every polynomial $f(x)\in F[x]$, the division with remainder $f(x)=g(x)q(x)+r(x)$, where $r$ has degree less than $m$. In particular $f(\alpha)=g(\alpha)q(\alpha)+r(\alpha)=r(\alpha)$, so we see that every element of $F[\alpha]$ has a representation as $$ a_0+a_1\alpha+\dots+a_{m-1}\alpha^{m-1} $$ and therefore $F[\alpha]$ is a finite dimensional vector space over $F$.
Conversely, if $F[\alpha]$ is finite dimensional over $F$, $\alpha$ is algebraic. Indeed, if the dimension is $k$, the elements $1,\alpha,\dots,\alpha^k$ are linearly dependent over $F$, which implies $\alpha$ is algebraic.
Now we can prove that when $\alpha$ is algebraic, $F[\alpha]$ is a field. Take $\beta\in F[\alpha]$. Then $F[\beta]$ is a subspace of $F[\alpha]$, so it is finite dimensional and therefore $\beta$ is algebraic over $F$.
If $b_0+b_1x+\dots+b_kx^k$ is the minimal polynomial, then we have $$ -b_0\beta^{-1}=b_1+b_2\beta+\dots+b_k\beta^{k-1} $$ and, since $b_0\ne0$, we get that $\beta^{-1}\in F[\beta]\subseteq F[\alpha]$.
Therefore $F[\alpha]$ is a field. In particular $F[\alpha^{-1}]\subseteq F[\alpha]$ and so they are equal by symmetry.