TITLE: Is $f(z)=\bar{z}$ continuous?
QUESTION [7 upvotes]: I have $z\in \mathbb{C}$, is $f(z)=\bar{z}$ continuous on the whole complex plane?
Note that $\bar{z}$ is the conjugate of $z\in \mathbb{C}$ 
I was thinking that if $z$ is on the real line, then $f(z)=z$, but if $z$ is on the imaginary line, then $f(z)=-z$, so I am still questioning about that. 
I'm sorry if I am asking something so trivial, but lots of things do not seem that straight forward to me. Thank you for input.

REPLY [10 votes]: $f(z)$ is continuous:
$$\lim_{z\to w}|f(z)-f(w)|=\lim_{z\to w}|\bar z-\bar w|=\lim_{z\to w}|\overline{z-w}|=\lim_{z\to w}|z-w|=0$$
To convince yourself of this fact, I recommend you to regard $\Bbb C$ as the plane $\Bbb R^2$ with coordinates $x,y$. Then, complex conjugation is the obviously continuous map
$$(x,y)\mapsto (x,-y)$$

REPLY [2 votes]: Let $\phi(z) = \bar{z}$. Since $|\bar{z}| = |z|$, we have $|\phi(z_1)-\phi(z_2)| = |z_1-z_2|$, and so $\phi$ is Lipschitz continuous with rank 1.