I am trying to determine where $$f(z)=\sqrt{z+1}$$ is analytic, where the square root is the principal branch.
I know that $\sqrt{w}$ is analytic for $\mathbb{C}\setminus(-\infty,0]$. So, I think $f(z)$ is not analytic when $\Re(z)\leq-1$. Is this correct?
As for continuity, I would like to determine if $f(z)$ is continuous at $\Re(z)=-1$ by taking the limit from above and below. But I am unsure of how to do this.
Thank you very much.
Best Answer
For any $\theta\in \mathbb{R}$ there is an continuous determination of the argument defined on $\mathbb{C}\setminus e^{i\theta}\mathbb{R_-}$. (The formula is $\operatorname{Arg}(z)= 2\operatorname{Arctan}(\frac{\operatorname{Im}(e^{-i\theta}z)}{|z|+\operatorname{Re}(e^{-i\theta}z)})+\theta$). So there is an continuous log on the same open given by $\log(z)=\ln|z|+i\operatorname{Arg}(z)$, and which is then analytic. You can then define your function with the formula $f(z)=\exp(\log(z+1)/2)$ which is analytic on $\{z\in\mathbb{C} \mid \operatorname{Arg(z+1)}\neq \theta\}$. If $\theta\neq 0$, your function is continuous at $-1$.