# Is $f(z)=\bar{z}$ continuous?

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.

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

$f(z)$ 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

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

• That was a very clear and clean explanation, thank you! – Akaichan Feb 21 '14 at 17:43

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.

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.

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.