Continuity of conjugate of $z$: $f(z)=\bar z$ How to prove that the function $\;f(z)=\bar z\;$
is continuous on the whole plane?
Hint: $$ |z - z_0| = |\bar{z} - \bar{z_0}|. $$

Then use Weierstrass $\epsilon$-$\delta$ definition of continuity at $z_0$.


   We want to show that for every $\epsilon&gt;0$ there is $\delta&gt;0$ such that if $|z-z_0|&lt;\delta$, then $|f(z)-f(z_0)| = |\bar{z}-\bar{z_0}| = |z-z_0| &lt; \epsilon$. We can take $\delta$ equal to $\epsilon$ itself.
$$f(z+\delta)=\overline{z+\delta}=\overline z+\overline\delta=f(z)+\epsilon.$$
It suffices to set
$$\overline\delta=\epsilon.$$