# Where is $\sqrt{z+1}$ analytic and continuous?

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.

• It doesn't make sense to write $z<-1$ is $z$ is a complex number – Tony S.F. Nov 12 '18 at 8:37
• You are correct, sorry about that. I have fixed this. – user557493 Nov 12 '18 at 8:38
• 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. – Ennar Nov 12 '18 at 9:37
• 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 $x=\Re(z)\leq -1$? – user557493 Nov 12 '18 at 10:05
• Is $z = -2+3i$ in both sets or just one of them? – Ennar Nov 12 '18 at 10:14

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