# Complex conjugation continuous function

How can I show that complex conjugation is a continuous function? I tried looking at open sets $U$ and then the preimage. Can assume preimage is not empty so that if $z$ is in preimage then $f(z) \in U$. How to go from here?

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You can see it as a linear function on $\mathbb{C}$ identified with $\mathbb{R}^2$. Its matrix is, indeed, diagonal, with $1$ and $-1$. As a linear transformation from $\mathbb{R}^2$ onto itself, it is continuous. – matgaio May 1 '12 at 6:22
Picture. We are reflecting in the $x$-axis. Algebra. $|z-w|=|\overline{z}-\overline{w}|$. – André Nicolas May 1 '12 at 6:30
how to make rigorous? – user23086 May 1 '12 at 6:31
Lets show this is continuous on $z=a+ib$. Lets call the conjugation by $T$. So, $T(a+ib)=a-ib$. Take $\epsilon>0$ and verify that the ball $B(a+ib,\epsilon)$ is taken to the ball $B(a-ib,\epsilon)$, calculating the norm os the complex numbers inside these balls. – matgaio May 1 '12 at 6:34

1) Complex conjugation as a function $c:\Bbb C\rightarrow\Bbb C$ is an involution, i.e. $c^2=\text{id}_{\Bbb C}$. Thus, it coincides with its inverse. Thus the pre-image of an open set $U\subset\Bbb C$ is the same as its image. Now: what happens when you take the complex conjugates of a ball centered at $z_0$ of radius $r$?
2) Under the Argand-Gauss identification $\Bbb C=\Bbb R^2$ the standard topology in $\Bbb C$ is the product of the standard topologies in $\Bbb R$. Now, if you have a function $$f:S\longrightarrow X\times Y$$ between topological spaces, you know that $f$ is continuous if and only if $\text{pr}_X\circ f$ and $\text{pr}_Y\circ f$ are continous, where $\text{pr}_X$ and $\text{pr}_Y$ denote the projection onto $X$ and $Y$ respectively. What you have if you compose complex conjugation with the two projections on the real and imaginary axes?
Let $z_0$ be a complex number. To make $|\overline{z}-\overline{z_0}|<\epsilon$, it is enough to make $|z-z_0|<\epsilon$, since $|\overline{z}-\overline{z_0}|=|z-z_0|$. So in the definition of continuity (for metric spaces) we are choosing $\delta=\epsilon$.