Is inversion of bounded homeomorphisms continuous with the uniform metric? Consider the metric space $H$ of uniformly continuous homeomorphisms (with uniformly continuous inverses) of some open bounded subset $G$ of $\mathbb C$, with the uniform metric
$$d(f, g) = \sup_{z \in G} \left|f(z) - g(z)\right|.$$
It is clear that $(f, g) \mapsto f \circ g$ is a continuous map $H \times H \to H$ (and in fact this is due to the uniform metric, not the uniform continuity of any particular homeomorphism), but is inversion $f \mapsto f^{-1}$ a continuous map $H \to H$?
In other words, is it the case that whenever $(f_n) \to f$ uniformly on $G$, where $f$ and each $f_n$ is a uniformly continuous homeomorphism of $G$ with a uniformly continuous inverse, then $(f_n^{-1}) \to f^{-1}$ uniformly on $G$?
 A: Let $K$ be a compact metric space, $H_K$ is the group of self-homeomorphisms $K\to K$ equipped with the topology of uniform convergence. I will prove that the inversion is continuous on $H_K$. If this were not the case, there would exist an (equicontinuous) sequence of homeomorphisms $h_n\to h$ such that $(h_n^{-1})$ does not converge to $h^{-1}$.  By compactness, this means that (after passing to a subsequence in $(h_n)$) there exists $x\in K$, a sequence $(x_n)$ converging to $x$, while $h_n^{-1}(x_n)=y_n$ converges to $y\ne h^{-1}(x)$. Applying $h_n$ and $h$ and taking into account that $h_n\to h$, we obtain that $x_n=h_n(y_n)\to h(y)\ne x$. This contradicts the assumption that $x_n\to x$. qed
Now, apply this to the closure $K$ of your open and bounded subset $G\subset {\mathbb C}$. 
Edit: Suppose that $f_n: K\to K$ is a sequence of continuous maps. What does it mean that $f_n$ does not converge to $f: K\to K$ (a continuous map) in uniform toplogy? This means (just by negating the definition of uniform convergence) that there exist $r>0$ such that for arbitrarily large $n$, there exist $x_n\in K$ such that $d(f_n(x_n), f(x_n))> r$. Now, by compactness, after extraction, we obtain $x_n\to x$, $f_n(x_n)\to y$ and $y$ has to be different from $f(x)$ since $d(y, f(x))\ge r$. 
A: In fact $(H, \circ, d)$ will be a topological group whenever $(X, \rho)$
is a bounded metric space. (not necessarily totally bounded)
It is easy to see that 
$d(f \circ g, \operatorname{id}) \le d(f, \operatorname{id}) + d(g, \operatorname{id})$,
 hence composition is continuous at the identity.
Furthermore, it follows immediately from the surjectivity of $h$ that
$d(f \circ h, g \circ h) = d(f, g)$, hence $d$ is right-invariant, so
right translations are clearly continuous.
To show that $H$ is a topological group, it will suffice then to show that
inversion is continuous.
To do this we take arbitrary $f \in H$ and $\epsilon > 0$. By the
uniform continuity of $f^{-1}$ there is a $\delta > 0$ such that
$\rho(f^{-1}(x), f^{-1}(y)) < \epsilon$ whenever $\rho(x, y) < \delta$.
For any $g \in H$ such that $d(g, f) < \delta$, we have 
$$\rho(f^{-1}(g(x)), x) = \rho(f^{-1}(g(x)), f^{-1}(f(x))) < \epsilon/2$$ for all $x$, therefore $d(f^{-1} \circ g, \operatorname{id}) \le \epsilon/2 < \epsilon$.
From the right-invariance of $d$ it follows that $$
d(f^{-1} \circ g, \operatorname{id}) = 
d(f^{-1} \circ g, g^{-1} \circ g) = 
d(f^{-1}, g^{-1}) 
$$
so we have $d(f^{-1}, g^{-1}) < \epsilon$, which shows that $f\mapsto f^{-1}$ is continuous.

Additional note:
Without the assumption of uniform continuity, left translations may fail
to be continuous, in which case inversion can not be continuous either.
As an example take $X = (0,1)^2$ with $f(x,y) = (x^y, y)$ and
$$
g_t(x, y) = \cases{ (2^tx, y) &when $x \le 4^{-t}$ \\
                    ((x-1)/(2^{-t} +1) &otherwise.
}
$$
Here $d(g_t, \operatorname{id}) = 2^{-t} - 4^{-t} < 2^-t$, so $g_t \to \operatorname{id}$ as $t \to \infty$.
On the other hand, taking $x = 4^{-t}, y = 1/t$, we can see
that $d(f\circ g_t, f) \ge 2^{-1} - 4^{-1} \ge 1/4$,
so $f\circ g_t$ does not converge uniformly to $f$, which shows that the left translation by $f$ is not continuous.
