How do you calculate $\lim_{z \to 0} (\frac{z}{\bar z})^\frac{1}{4}$ I obtained $(\frac{x+iy}{x-iy})^\frac{1}{4}$
Let x -> 0, I obtained that the limit is (-1)$^\frac{1}{4}$ which is not real
Let y -> 0, I obtained that the limit is (1)$^\frac{1}{4}$ = 1
Does this necessarily meant that I just proved that the limit of this does not exist since one is real and one is not real.
Thanks a lot!
 A: Arguing with polar coordinates is also very easy:
let $z=r e^{i\varphi}$, then (if the limit would exist) $$\lim_{z\rightarrow 0}\left({\frac{z}{\bar{z}}}\right)^{1/4}=\lim_{r\rightarrow 0}\left({\frac{r e^{i\varphi}}{r e^{-i\varphi}}}\right)^{1/4}=e^{i\varphi/2}.$$
But any two choices of $\varphi$ will lead to different results. Thus the limit cannot exist.
A: Yes, then the limit wouldn't exist. You can also try the following approach: since $\;z\overline z =|z|\;$ ,then
$$\frac z{\overline z}=\frac{z^2}{|z|^2}=\left(\frac z{|z|}\right)^2$$
and then it is easy to check that over the real axis, when $\;z\to0\;$ we get $\;1$, and over the imaginary axis, when $\;\text{Im}\,z\to0^+\;$, we get
$$\lim_{y\to0^+}\left(\frac{iy}{|iy|}\right)^2=\lim_{y\to0^+}-\frac{y^2}{y^2}=-1$$
A: *

*$$\lim_{\text{z}\to0}\left(\frac{\text{z}}{\overline{\text{z}}}\right)^\text{n}=\exp\left[n\lim_{\text{z}\to0}\ln\left(\frac{\text{z}}{\overline{\text{z}}}\right)\right]$$

*When $\text{z}\in\mathbb{C}$:
$$\frac{\text{z}}{\overline{\text{z}}}=\frac{\text{z}^2}{|\text{z}|^2}=\frac{\left(\Re[\text{z}]+\Im[\text{z}]i\right)^2}{|\Re[\text{z}]+\Im[\text{z}]i|^2}=\frac{\Re^2[\text{z}]-\Im^2[\text{z}]+2\Re[\text{z}]\Im[\text{z}]i}{\Re^2[\text{z}]+\Im^2[\text{z}]}$$

*When $\text{z}\in\mathbb{C}$:
$$\ln(\text{z})=\ln\left(|\text{z}|e^{\left(\arg(\text{z})+2\pi k\right)i}\right)=\ln|\text{z}|+\ln\left(e^{\left(\arg(\text{z})+2\pi k\right)i}\right)$$
Where $k\in\mathbb{Z}$



Using (2):


*

*When $\Re[\text{z}]=0$:
$$\frac{\text{z}}{\overline{\text{z}}}=\frac{0^2-\Im^2[\text{z}]+2\cdot0\cdot\Im[\text{z}]i}{0^2+\Im^2[\text{z}]}=\frac{-\Im^2[\text{z}]}{\Im^2[\text{z}]}=-1\to\ln(-1)=\pi i$$

*When $\Im[\text{z}]=0$:
$$\frac{\text{z}}{\overline{\text{z}}}=\frac{\Re^2[\text{z}]-0^2+2\Re[\text{z}]\cdot0i}{\Re^2[\text{z}]+0^2}=\frac{\Re^2[\text{z}]}{\Re^2[\text{z}]}=1\to\ln(1)=0$$


So, the limit does not exist.
