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Prove the $\lim _{z \rightarrow 1}\frac{\overline{z}-1}{z-1}$ doesn't exist.

We see that $\frac{\overline{z}-1}{z-1}=\frac{x-iy-1}{x+iy-1}$, but when we take $z=1+iy$ we have $\lim _{z \rightarrow 1} \frac{1-iy-1}{1+iy-1}=-1$. But I don't see other points such that the limit is diffferent to $-1$. Could you help me with that? or can I take $z=1+x+i0$ such that $x \rightarrow 0$?

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  • $\begingroup$ If you take z to be real, then the limit is 1. $\endgroup$ Nov 12, 2020 at 15:41

3 Answers 3

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Consider $z = 1 + \varepsilon i$, where $\varepsilon > 0$. Then $\frac{\overline z - 1}{z - 1} = -1.$

Now consider $z = 1 + \varepsilon$, where $\varepsilon > 0$. Then $\frac{\overline z - 1}{z-1} = 1$.

Clearly, $1 \neq -1$.

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If $z=1+h$ with $h\in(0,\infty)$, then$$\frac{\overline z-1}{z-1}=\frac hh=1.$$This, together what you did, shows that the limit doesn't exist.

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How about, approach along a diagonal? $$ \lim_{x \rightarrow 0} \frac{(1+x) - \mathrm{i}x - 1}{(1+x) + \mathrm{i}x - 1} = -\mathrm{i} \text{.} $$

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