My complex analysis textbook claims that $f(z) = z^{1/2} = r^{1/2}cos(\frac{\theta}{2}) + ir^{1/2}sin(\frac{\theta}{2})$ , where $ r>0$, and $-\pi<\theta \leq \pi$, and this means that $f$ is analytic at all points other than the origin and the negative real axis. The reasoning is that it is not continuous on the negative real axis. why is this?
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Related. – Cameron Buie Dec 17 '15 at 03:45
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$\lim_{\theta\to\pi} f(e^{i\theta}) = \lim_{\theta\to\pi} e^{i\theta/2} = e^{i\pi/2} = i$
$\lim_{\theta\to-\pi} f(e^{i\theta}) = \lim_{\theta\to-\pi} e^{i\theta/2} = e^{-i\pi/2}=-i$

Andrew Dudzik
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Thanks! at the origin though, $\theta$ can take any value , so why is it not continuous there? – Dr. John A Zoidberg Dec 17 '15 at 02:55
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1@user3677952 It's continuous at the origin, but it's not analytic there. If it were analytic at the origin, it would have to be continuous in some neighborhood of the origin. – Andrew Dudzik Dec 17 '15 at 03:06