4
$\begingroup$

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.

$\endgroup$
  • 3
    $\begingroup$ A complex-valued function is continuous if and only if both, its real part and its imaginary part are continuous. $\endgroup$ – Daniel Fischer Feb 21 '14 at 17:14
  • 1
    $\begingroup$ Both the functions $x \rightarrow x $ and $y \rightarrow -y$ are continuous. $\endgroup$ – user99680 Feb 21 '14 at 17:15
  • $\begingroup$ It may help to think of the complex plane as $\mathbb{R}^2$, and conjugation is the operator $\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}$. $\endgroup$ – copper.hat Feb 21 '14 at 17:24
  • $\begingroup$ Yes, it's continuous, as others have said. It's also nowhere differentiable! $\endgroup$ – Dave L. Renfro Feb 21 '14 at 17:46
7
$\begingroup$

$f(z)$ 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

$$(x,y)\mapsto (x,-y)$$

$\endgroup$
  • 1
    $\begingroup$ That was a very clear and clean explanation, thank you! $\endgroup$ – Akaichan Feb 21 '14 at 17:43
2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

Take $z=x+iy$, so $\overline{z}=x-iy$. Since the monomials $x,y$ are continuous, so is the conjugate.

Try to check that the conjugate is not holomorphic. To this end check the Cauchy-Riemann equations.

$\endgroup$
0
$\begingroup$

Yes, it is continuous; both of the functions $$x \rightarrow x $$ and $$ y \rightarrow -y$$ are continuous. For more rigor, you can use this to do a $\epsilon -\delta$ argument.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.