- #1
Silviu
- 624
- 11
<Moderator note: Merger of two threads on the topic.>
Hello! I am reading some basic stuff on complex integrals using branch cuts and i found the problem in the attachment. I am not sure I understand why the branch cut is along ##R^+##. I thought that branch cut is, loosely speaking, a line where the function is not continuous (and thus not holomorphic). But in the presented problem, the function is continuous on ##R^+## as ##lim_{\theta \to 0} = \sqrt{r}## and ##lim_{\theta \to 2\pi} = -\sqrt{r}##. The limits are not equal, but they don't have to be, as the funtion is not defined for ##\theta = 2\pi##. However, the function is not continuous for ##\theta = \pi##, as, coming from above and below x-axis, gives different values for ##sin(\theta)##. So, isn't the branch cut on ##R^-##, or did I get something wrong about the definition of branch cut? Thank you!
Hello! I am reading some basic stuff on complex integrals using branch cuts and i found the problem in the attachment. I am not sure I understand why the branch cut is along ##R^+##. I thought that branch cut is, loosely speaking, a line where the function is not continuous (and thus not holomorphic). But in the presented problem, the function is continuous on ##R^+## as ##lim_{\theta \to 0} = \sqrt{r}## and ##lim_{\theta \to 2\pi} = -\sqrt{r}##. The limits are not equal, but they don't have to be, as the funtion is not defined for ##\theta = 2\pi##. However, the function is not continuous for ##\theta = \pi##, as, coming from above and below x-axis, gives different values for ##sin(\theta)##. So, isn't the branch cut on ##R^-##, or did I get something wrong about the definition of branch cut? Thank you!
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