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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,

$|(x,y) - (x_0,y_0)|^2 = |(x-x_0, y-y_0)|^2 = |x-x_0|^2 + |y-y_0|^2 = (x-x_0)(\bar{x} - \bar{x_0}) + \dots = |x|^2 + |y|^2 - 2Re(xx_0) - 2Re(yy_0)$

and

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

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

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  • $\begingroup$ Does writing $|xy-x_0y_0|=|x(y-y_0)+(x-x_0)y_0|$ helps ? $\endgroup$ – Bertrand R Aug 13 '13 at 23:31
  • $\begingroup$ Oh, the old add and subtract the same thing trick. Lemme see... $\endgroup$ – AbstractAlgebraLearner Aug 13 '13 at 23:38
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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 $|x-x_0|<\frac{\epsilon}{2|y|}$ and $|y-y_0|<\frac{\epsilon}{2|x_0|}$. For $x_0$ or $y_0=0$ this will need just a slight modification.

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  • $\begingroup$ You need to choose $y_{max} = \sup \{|y| : |y-y_0| < \frac{\epsilon}{2|x_0|}\}$, which will be bounded. $\endgroup$ – AbstractAlgebraLearner Aug 14 '13 at 0:40
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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.

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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 $\Re(p(x,y))=x_1y_1-x_2y_2 = \pi_1(\pi_x(x,y))\pi_1(\pi_y(x,y))-,\cdots$.

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.

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  • $\begingroup$ I will look at this next, but my method almost works, so still looking at it. $\endgroup$ – AbstractAlgebraLearner Aug 13 '13 at 23:46

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