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How to prove that the function $\;f(z)=\bar z\;$ is continuous on the whole plane?

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  • $\begingroup$ Sorry...I'm trying to put a short bar above z... $\endgroup$ – leave2014 Sep 5 '14 at 13:03
  • $\begingroup$ Cool.. Thanks... $\endgroup$ – leave2014 Sep 5 '14 at 13:04
  • $\begingroup$ Now that the notation is settled, can you tell where you're stuck and what you have attempted? $\endgroup$ – egreg Sep 5 '14 at 13:05
  • $\begingroup$ I tried the Cauchy Riemann Equation... Which tell me it's not analytic everywhere .. And by definition of limit .. I cannot get it be zero as the small change goes to0 $\endgroup$ – leave2014 Sep 5 '14 at 13:10
  • $\begingroup$ @leave2014 use Weierstrass $\epsilon$-$\delta$ definition of continuity at $z_0$. $\endgroup$ – ir7 Sep 5 '14 at 13:17
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$$f(z+\delta)=\overline{z+\delta}=\overline z+\overline\delta=f(z)+\epsilon.$$ It suffices to set $$\overline\delta=\epsilon.$$

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  • $\begingroup$ Is it able to prove $\lim {\frac{f(z+h)-f(z)}{h}} =\lim \frac{\bar h}{h} =0$ $\endgroup$ – leave2014 Sep 5 '14 at 13:17
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    $\begingroup$ Continuity still uses real $\epsilon-\delta$, of course... $\endgroup$ – Thomas Andrews Sep 5 '14 at 13:17
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    $\begingroup$ @leave2014 What do you mean "prove" that? The limit does not exist. But you only asked for continuity, not differentiability. $\endgroup$ – Thomas Andrews Sep 5 '14 at 13:18
  • $\begingroup$ Oh no... Am I confusing the definition of continuity and differentiable... $\endgroup$ – leave2014 Sep 5 '14 at 13:21
  • $\begingroup$ It appears you are, @leave2014 . $\endgroup$ – Thomas Andrews Sep 5 '14 at 14:40
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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 $\epsilon>0$ there is $\delta>0$ such that if $|z-z_0|<\delta$, then $|f(z)-f(z_0)| = |\bar{z}-\bar{z_0}| = |z-z_0| < \epsilon$. We can take $\delta$ equal to $\epsilon$ itself.

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As a function from $R^2$ to $R^2$, $f(x,y)=(x,-y)$. Both components are polynomial, so they are continuous, hence $f$ is.

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