How do you prove that the complex inverse is continuous?

Question

I tried to show that the continuous at a point $\delta / \epsilon$ definition holds but failed. Now I'm thinking along the lines of multiplicative group: $C \rightarrow C, x \mapsto bx$ has inverse $x \mapsto b^{-1} x$ composing the maps is analogous to multiplication in $C$ and $\phi : C \times C \rightarrow C, (x,y) \mapsto xy $ is continuous so it's partial functions are continuous, then... and I'm stuck.

Answer 1

Let $a \neq 0$ be a complex number. We will prove that the map $f\colon \mathbb{C}^* \rightarrow \mathbb{C}^*$ defined by $f(x) = x^{-1}$ is continuous at $a$, where $\mathbb{C}^*$ denotes the multiplicatve group of $\mathbb{C}$.

We get $|f(x) - f(a)| = |x - a|/|x||a|$.

Let $\epsilon \gt 0$ be a positive number. Let $\delta =$ min$\{|a|/2, \epsilon |a|^2/2\}$. Suppose $|x - a| \lt \delta$. Since $|a| = |x - a - x| \le |x - a| + |x|$, $|a| - |x| \le |x - a| \lt \delta$ Hence $|a| - \delta \lt |x|$ Hence $|f(x) - f(a)| = |x - a|/|x||a| \lt \delta/(|a| - \delta)|a| \le 2\delta/|a|^2 \le \epsilon$. Hence $f(x)$ is continuous at $a$.

Answer 2

Compositions of continuous maps are continuous including compositions of the form $f(h(x), g(x))$. Conjugation is easily shown to be continuous and also the magnitude function $|x|$. And in $\mathbb{R}$ $1/f(x)$ is continuous at all $f(x) \neq 0$, since inverse is not defined at $0$ we don't need to worry. Since the inverse of a complex number $z$ equals $z^* / |z|^2$ and it's a composition of the continuous maps above it's also continuous!!!! :)

We should note that we transfered the common range of some of the above functions to $\mathbb{C}$ to make it work, that should be okay as the preimage $g^{-1}(U)$ for $U$ open in $\mathbb{C}$ is then $g^{-1}(U \cap \mathbb{R})$ for instance and $g^{-1}(\mathbb{R})$ is all of $\mathbb{C}$.