Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
up vote 1 down vote accepted

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 $r$.

share|cite|improve this answer
...which means that the limit is not well defined. – Fly by Night Aug 29 '12 at 18:29

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.