# What is the limit of function $\frac{z}{\bar{z}-z}$ at $z=0$?

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 exist – M Turgeon Aug 29 '12 at 14:21
Consequently, it cannot be analytic in the unit circle. – Old John Aug 29 '12 at 14:23
@MTurgeon: How to show that it is undefined on the whole real line? Just hint. – alvoutila Aug 29 '12 at 14:34
@alvoutila what is $\bar z - z$ when $z$ is real? – axblount Aug 29 '12 at 14:37
@alvoutila You cannot divide by zero; your function is simply undefined on the real line (in particular, it is not equal to $\infty$). – M Turgeon Aug 29 '12 at 15:09

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 $r$.
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.