Help proving that complex multiplication is continuous.

Question

Let $(\cdot) : C\times C \rightarrow C, (x,y) \mapsto x\cdot y = xy$, where $C$ is the field of complex numbers. Let $|\cdot|$ denote the standard norm on $C$ and on $C^2$ let the norm be the most common which is given by $|(z,z')|^2 = |z|^2 + |z'|^2$. To show that complex multiplication is continuous under these norms it's enough to show that it's continuous at each point using the open-ball definition: Let $\epsilon, \delta \in \mathbb{R}$. If for any $(x_0,y_0) \in C^2$, for all $\epsilon \gt 0$ there exists $\delta \gt 0$ such that $$0 \lt |(x,y) - (x_0,y_0)| \lt \delta \implies |xy - x_0y_0| \lt \epsilon$$

then complex multiplication is continuous.

So my approach goes like proving that $(\mathbb{R},+)$ is a topological group, by expanding the norms and finding an appropriate $\delta$ which in that example turned out to be $\sqrt{\epsilon}$. So,

\begin{align} |(x,y) - (x_0,y_0)|^2 &= |(x-x_0, y-y_0)|^2 = |x-x_0|^2 + |y-y_0|^2 \\ &= (x-x_0)(\bar{x} - \bar{x_0}) + \dots \\&= |x|^2 + |y|^2 - 2\text{Re}(xx_0) - 2\text{Re}(yy_0) \end{align}

and

$$|xy - x_0y_0|^2 = (xy - x_0y_0)(\bar{xy} - \bar{x_0y_0}) = |xy|^2 + |x_0y_0|^2 -2\text{Re}(xy\bar{(x_0y_0)})$$

Then I'm thinking of squaring the first one. But that brings the expressions to another level of complexity. What would you do here?

Answer 1

This is similar in spirit to your method: $$xy-x_0y_0=xy-x_0y+x_0y-x_0y_0=y(x-x_0)+x_0(y-y_0),$$ so just pick $x,y$ such that $|x-x_0|<\dfrac{\epsilon}{2|y|}$ and $|y-y_0|<\dfrac{\epsilon}{2|x_0|}$.

For $x_0$ or $y_0=0$ this will need just a slight modification.

Answer 2

You should know that the inner product map $\langle-,-\rangle:\mathbb{C}\times\mathbb{C}\to\mathbb{C}$ is continuous (it's used to define the topology!).

Note then that the multiplication map $m(x,y)=xy$ is just $\langle -,c(-)\rangle$ where $c$ is the conjugation map, which is evidently continuous.

Answer 3

Rather than using a norm, you can view the complex numbers as a topological product space, $\Bbb C = \Bbb R\times \Bbb R$.

$p(x,y)=(x_1,x_2)\cdot(y_1,y_2) = (x_1y_1-x_2y_2,x_1y_2+x_2y_1)$.

A function into a product space is continuous iff its composition with each projection is continuous. The first projection gets you

$$\eqalign{\Re(p(x,y)) &= x_1y_1-x_2y_2 \\ &= \pi_1(\pi_1(x,y))\pi_1(\pi_2(x,y)) \\ &\qquad - \pi_2(\pi_1(x,y))\pi_2(\pi_2(x,y))}$$

Compositions of continuous functions are continuous, and products and sums of continuous real-valued functions are continuous, so this is continuous. The same holds for the second projection.