Exercise · ZW-COMPLEX-CONTINUITY-004 · ★☆☆ Basic
The function \(f(z)=\bar z\) is:
Choose one option.
Show answer
- B. continuous on all of \(\mathbb{C}\), since \(|f(z)-f(z_0)|=|z-z_0|\)
Exercise · ZW-COMPLEX-CONTINUITY-004 · ★☆☆ Basic
Choose one option.
We have \(|\bar z-\bar z_0|=|\overline{z-z_0}|=|z-z_0|\), so \(\delta=\varepsilon\) works. Continuity does not require holomorphy.
A function \(f\) is continuous at a point \(z_0\) of its domain if \(f(z)\) tends to the value \(f(z_0)\) as \(z\) tends to \(z_0\); it is continuous if it is so at every point of its domain.