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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?

2 Answers2

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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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    @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
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$\lim_{x\to 0+}(-1+xi)^{1/2} =i$ and $\lim_{x\to 0+}(-1-xi)^{1/2}\to -i$.

Thomas Andrews
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