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

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 $$x \rightarrow 0$$?

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

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