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Just want to check. What is the limit of function $\frac{z}{\bar{z}-z}$ at $z=0$? I got $\lim_{\substack{z \to 0 \\ z \in \mathbb{R}}} \frac{z}{\overline{z}-z} =-\infty$

and $\lim_{\substack{z \to 0 \\ z \in i\mathbb{R}}} \frac{z}{\overline{z}-z} =-\frac{1}{2}$, so $f$ is not defined at $z=0$? Byt the way does this have any singularities? And finally is this analytic in unit circle?

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    This function is undefined on the whole real line. Hence, the limit does not exist2012-08-29
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    Consequently, it cannot be analytic in the unit circle.2012-08-29
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    @MTurgeon: How to show that it is undefined on the whole real line? Just hint.2012-08-29
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    @alvoutila what is $\bar z - z$ when $z$ is real?2012-08-29
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    @MTurgeon: Now I got it. $\bar{z}-z=x-iy-x-iy=-i2y=0$, when $z$ is real. So regardless of $z$ $\frac{z}{\bar{z}-z} = \infty$?2012-08-29
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    @alvoutila You cannot divide by zero; your function is simply undefined on the real line (in particular, it is not equal to $\infty$).2012-08-29
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    So I guess because limit does not exist, $f$ is not continuous at $z=0$2012-08-29

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It is easy to see that the limit of $f(z) = \frac{z}{\bar z - z}$ as $z\to0$ depends on the direction of approach. For that we write $z = r e^{i\theta}$. Then $$f(z) = \frac{r e^{i\theta}}{r e^{-i\theta} -r e^{i \theta}} = \frac{1}{e^{-2i\theta} -1}$$ and the value of $f(z)$ depends on the angle $\theta$ but not on $r$.

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    ...which means that the limit is not well defined.2012-08-29
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Change to polar coordinates $z=r {\rm e}^{i \theta} $ and notice that your function depends only on $\theta.$ This tells you that the limit does not exist.