Limit of a complex number over its conjugate, as z approach the infinity. this question must be pretty easy but i´m just taking my firs course in complex variables, i need to find the limit of Z over it conjugate as z reach the infinity. I need to know how to do it in a rigorous way and informal way, so i need your help. By formal, i mean with neighborhood and epsilon definition of limit.
I have tried with trig form of Z, but i don´t reach any answer.
Thanks
 A: We have $z\bar{z}=|z|^2$ so that $z/\bar{z}=z^2/|z|^2=\big(z/|z|\,\big)^2$, which very much depends on $\arg z$.
Indeed, picking any $M>0$ and $s$ with $|s|=1$ we can select a $z$ with $|z|=M$ and $(z/|z|)^2=s$; can you see why this should be true algebraically and geometrically? This means every value on the unit circle is obtained somewhere in every neighborhood of $\infty$; can all of the distances between points on the unit circle be made as small as we wish?
A: Hint: The trigonometric form is a good idea. What is $\dfrac{r(\cos\theta+i\sin\theta)}{r(\cos\theta-i\sin\theta)}$ when $r$ is large?  Does it depend on $\theta$?
A: Perhaps even easier than the trig form is the exponential form. If $z=re^{i\theta}$, then $\bar z=re^{-i\theta}$, so you’re looking at the limit of $$\frac{re^{i\theta}}{re^{-i\theta}}=e^{2i\theta}$$ as $r$ gets large.
A: If $z=r e^{i \theta}$, then $\frac{z}{\overline{z}}  = e^{i 2 \theta}$.
If I choose $r_n = n$, and $\theta_n = n \frac{\pi}{2}$, then with $z_n=r_n e^{i \theta_n}$, we have $z_n \to \infty$ and $\frac{z_n}{\overline{z_n}} = (-1)^n$, consequently the ratio has no limit.
