# what are the zeroes of $\sqrt{z}$?

well I need to know whether there are any zeroes of $\sqrt{z}$? so far I guess $0$ is a singular point of the function as it is not analytic at $z=0$, so It has no zeroes am I right? Is it an isolated singularity? please help.

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I wonder, does $(|z| = |w| \implies |f(z)| = |f(w)|) \implies$f is analytic? $\sqrt z$ falls under this case. – Thomas Eding Jun 10 '12 at 6:16
@trinithis That seems sort of unrelated, but consider $f(z)=\bar z$. – Dylan Moreland Jun 10 '12 at 6:18
Ah, right. Nevermind my previous post. – Thomas Eding Jun 10 '12 at 6:19
@DylanMoreland Does my question makes any sense? – Un Chien Andalou Jun 10 '12 at 6:22
Define $\sqrt{z}$... – Najib Idrissi Jun 10 '12 at 6:36

$\sqrt z=0\implies \sqrt z\sqrt z=z=0$
As pointed out in the comments, it depends on how you define $\sqrt{z}$. If one wants it to be analytic on its entire domain of definition, then take a branch of the logarithm and use that to define the square root off the branch cut as $\sqrt{z}=e^{\frac{1}{2}\log z}$ (or as -$e^{\frac{1}{2}\log z}$, if you prefer). Now, it isn't difficult to see that as $z\to 0$, then $\sqrt{z}\to 0$ in either case, so we can continuously extend our definition of $\sqrt{z}$ to say that $\sqrt{0}=0$, and in that case, can consider $0$ to be the sole zero of $\sqrt{z}$. We just can't analytically extend it that way, since $0$ is a branch point (of order $2$) of $\sqrt{z}$.