Show continuity of the following functions: a) $f: \mathbb C \to \mathbb C, f(z) = \vert z \vert$. b) $g: \mathbb C \to \mathbb C, g(z) = \ \bar z$

Question

Show continuity of the following functions:

a) $f: \mathbb C \to \mathbb C, f(z) = \vert z \vert$.

b) $g: \mathbb C \to \mathbb C, g(z) = \ \bar z$.

a) A function $f$ is continuous if $\lim_{z \to z_0} f(z) = f(z_0)$. So $\forall \epsilon \gt 0, \exists \delta \gt 0$ such as $\vert f(z) - \vert z \vert \vert \lt \epsilon \to \vert z - z \vert \lt \delta$. Choosing $\epsilon = \delta$ we finish?

b)$\vert g(z) - \bar z \vert \lt \epsilon \to \vert z - \bar z \vert = \vert 2bi\vert\lt \delta ?$

Is this correct?

Is there any way to do it without using $\epsilon$ and $\delta$? expanding z to $a + bi$?

Grateful for the attention.

Answer 1

b) We know that the function $z\to z$ is continuous everywhere, so since $$|\bar z-\bar z_0|=|\overline{z-z_0}|=|z-z_0|$$ just choose $\epsilon=\delta$.

a)Note that $$||z|-|z_0||\leq|z-z_0|$$ also with $\epsilon=\delta$ we have the continuity

Answer 2

Let us prove the first claim first.

According to the reverse triangle inequality, suppose that $|z - a| < \delta_{\varepsilon}$. Then we we arrive at: \begin{align*} ||z| - |a|| \leq |z - a| < \delta_{\varepsilon} \Rightarrow ||z| - |a|| < \delta_{\varepsilon} := \varepsilon \end{align*}

Thus for every $\varepsilon > 0$, there corresponds $\delta_{\varepsilon} = \varepsilon$ s.t. for every $z\in\mathbb{C}$ the limit definition is satisfied.

We may now prove the second claim.

According to the complex number's properties, if we assume that $|z - a| < \delta_{\varepsilon}$, it results that \begin{align*} |z - a| < \delta_{\varepsilon} \Rightarrow \overline{|z - a|} < \delta_{\varepsilon} \Rightarrow |\overline{z} - \overline{a}| < \delta_{\varepsilon} := \varepsilon \end{align*}

Thus for every $\varepsilon > 0$, there corresponds $\delta_{\varepsilon} = \varepsilon$ s.t. for every $z\in\mathbb{C}$ the limit definition is satisfied.

Hopefully this helps!