Does $\lim_{x \to 0}({z^2\over \overline z})$ exist? $(z\in \mathbb{C})$

Question

I am trying to figure out if $\lim_{x \to 0}({z^2\over \overline z})$ exists or not. This is a way I though to show that this does not exist but I am not entirely sure.

Let $a_n={1\over n}$ and $b_n={i\over n}$. $f(z)={z^2\over \overline z}={z^2\overline{z^2}\over \overline z^3}$=${|z|^2\over \overline z^3}$.

Thus, $f(a_n)=n$ and $f(b_n)=-{n\over i}$

$\lim_{n \to \infty}f(a_n) \rightarrow +\infty$ and

$\lim_{n \to \infty}f(b_n) \rightarrow -\infty$. Thus the limit cannot exist. Is this correct? If someone has a better answer or any other method of solving this please do share. Thanks

Answer 1

you're mistaken. actually $z \bar{z} = |z|^2$ and not $|z|$ which is what you're using in your argument. The limit actually exists and is equal to $0$

oh, and if you're looking for a precise argument you can try the following: let $z_n \rightarrow 0$. Write $z_n = r_n e^{i \alpha_n}$ - in trigonometric form. You now know that $r_n \rightarrow 0$ and $\bar{z} = r_n e^{-i \alpha_n}$.