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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.

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  • $\begingroup$ It doesn't make sense to write $z<-1$ is $z$ is a complex number $\endgroup$ – Tony S.F. Nov 12 '18 at 8:37
  • $\begingroup$ You are correct, sorry about that. I have fixed this. $\endgroup$ – user557493 Nov 12 '18 at 8:38
  • $\begingroup$ Bell, if you wanted to say that $f$ is analytic on $\mathbb C\setminus (-\infty,-1]$, that's definitely not the same as $\operatorname{Re}z\leq 1$. I would return to what you wrote originally. As Tony said, it doesn't make sense to write $z\leq -1$ if $z$ is complex number, so one should automatically assume that $z$ is real seeing that. $\endgroup$ – Ennar Nov 12 '18 at 9:37
  • $\begingroup$ I do not understand your comment. If $f$ is analytic on $\mathbb{C}$\ $(-\infty,-1]$, is this not the same as saying $f$ is not analytic for $x=\Re(z)\leq -1$? $\endgroup$ – user557493 Nov 12 '18 at 10:05
  • $\begingroup$ Is $z = -2+3i$ in both sets or just one of them? $\endgroup$ – Ennar Nov 12 '18 at 10:14
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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 $-1$.

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  • $\begingroup$ If you choose the principal branch of the log, then you will have $f(z)$ continuous at -1 because the log will tend to $-\infty$. $\endgroup$ – Swann Nov 12 '18 at 8:53

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