Let $f:\mathbb C \to \mathbb C$ be defined by $f(z)=\sqrt{z}=\sqrt{r}e^{i arg_\alpha(z)/2}$ where $arg_{\alpha}(z)$ is the angle in $(\alpha,\alpha+2\pi]$ and $A=\{z\in \mathbb C: z=te^{i\alpha}, t\ge 0\}$, how to show the discontinuity of $f$ at $z\in A$? Any help to understand the problem?
Is the set $A$ a bunch cut or a set of bunch points?
If you define $\sqrt {re^{it}}$ as $re^{it/2}$ when $0<t\leq 2\pi$ and take $\alpha =0$ then the statement is correct and it can be proved by showing that square root of $1+\frac i n$ tends to $1$ whereas square root of $1-\frac i n$ tends to $-1$. For other values of $\alpha$ you have to take $\alpha <t \leq 2\pi +\alpha $ instead of $0<\alpha \leq 2\pi$. In other words you haev to rotate $1\pm \frac i n$ by angle $\alpha$.