Let \(z_0 = r_0e^{i\pi} \in (-\infty, 0) \subseteq \RR\), and show that \(z^{1\over 2}\) is not continuous along \((-\infty, 0)\) by computing \[ \lim_{z\in \gamma_1} f_1(z) &= i\sqrt{r_0} \\ \lim_{z\in \gamma_2} f_1(z) &= -i\sqrt{r_0} ,\]
where
- \(\gamma_1 = \ts{r_0 e^{it} \st t\in (0, \pi) }\),
- \(\gamma_2 = \ts{r_0 e^{-it} \st t\in (\pi, 0) }\)
Solution
\[ \lim _{(r, \theta) \rightarrow\left(r_{0}, \pi\right)} f_{1}\left(r e^{i \theta}\right) &=\lim _{(r, \theta) \rightarrow\left(r_{0}, \pi\right)} r^{\frac{1}{2}}\left(\cos \frac{\theta}{2}+i \sin \frac{\theta}{2}\right)=i r_{0}^{\frac{1}{2}}, \quad \text { and } \\ \lim _{(r, \theta) \rightarrow\left(r_{0},-\pi\right)} f_{1}\left(r e^{i \theta}\right) &=\lim _{(r, \theta) \rightarrow\left(r_{0},-\pi\right)} r^{\frac{1}{2}}\left(\cos \frac{\theta}{2}+i \sin \frac{\theta}{2}\right)=-i r_{0}^{\frac{1}{2}} . .\]