Prove $\lim_{z \to 0} \frac{z}{\overline{z}}$ doesn't exist using $\varepsilon-\delta$. I'm trying to prove that the limit
$$
\lim_{z \to 0} \frac{z}{\overline{z}} \quad \qquad z \neq 0
$$
doesn't exist. Up to this point, the only definition of a limit for complex functions I know is as that $\lim_{z \to w} f(z) = L$ if and only if

$$
\forall \varepsilon >0, \ \exists \delta >0 \text{ such that if }\lvert z-w \rvert < \delta \implies \lvert f(z)- L\rvert< \varepsilon
$$

So I wanted to solve my problem using only this. I know that I could use paths and show that approaching $0$ in different ways
gives different limits, but since I don't know how to rigorously justify this I chose to avoid it.

My idea was to argue by contradiction. So I would assume that the limit existed and that it was equal to some complex number $L$, and then I would show that this assumption would lead to problems.
My attempt
The first thing I notice is that I can simplify the function as follows
$$
\lim_{z \to 0} \frac{z}{\overline{z}} = \lim_{z \to 0} \frac{z^2}{|z|^2} = \lim_{z \to 0} \frac{\left(re^{i\theta}\right)^2}{r^2}= \lim_{z \to 0} e^{i(2\theta)}
$$
where $\theta = \arg(z)$ is a function of $z$.
Now, since we assume that the limit does exist and that it's equal to $L \in \mathbb{C}$, we can write $L$ as
$$
L = r' e^{i \theta'}
$$
where $r'\ge 0$ (i.e. $r' \nless 0$) and $\theta'$ are some fixed real numbers.
Since we're assuming that the limit exists, if I choose the value $\varepsilon =1 $ I know there exists a $\delta$ such that $\lvert z-0 \rvert < \delta \implies \lvert e^{i(2\theta)}- L\rvert< \varepsilon$.
If I then choose to analyze the complex number $ z =  \frac{\delta}{2} e^{i\left(\frac{\theta' + \pi }{2}\right)}$ I see that
$$
\lvert z -0 \rvert = \Biggl\lvert\frac{\delta}{2} e^{i\left(\frac{\theta' + \pi }{2}\right)} -0 \Biggr\rvert =  \Bigl\lvert\frac{\delta}{2} \Bigr\rvert \cdot \Biggl\lvert e^{i\left(\frac{\theta' + \pi }{2}\right)}\Biggl\lvert = \frac{\delta}{2} < \delta
$$
which means that for $\theta = \arg\left( \frac{\delta}{2} e^{i\left(\frac{\theta' + \pi }{2}\right)}\right)$ it should be the case that $\lvert e^{i(2\theta)}- L\rvert< \varepsilon$, but here we see that
\begin{align}
\Bigl\lvert e^{i(2\theta)} - L\Bigr\rvert &=  \Bigl\lvert  e^{i\left(2\frac{\theta' + \pi }{2}\right)} - r' e^{i\theta}\Bigr\rvert = \Bigl\lvert e^{i\theta'}\left( e^{i\pi}  - r'\right) \Bigl\lvert \\
&= \bigl\lvert e^{i\theta'}\bigl\lvert \cdot \bigl\lvert-\left( 1  + r'\right)\bigl\lvert = 1 + r' \nless 1  = \varepsilon 
\end{align}
where we get the contradiction we wanted.

The idea of my attempt was that I noticed that the function always outputted numbers on the unit circle, which meant that even though I could find a $z$ really close to $0$, the output couldn't get as close to some limit $L$ as it wanted since it had to be on the unit circle.
I'm not sure if my proof used the contradiction correctly, more specifically, I don't know if my final equation implies that my original assumption was wrong or if I can conclude anything from it at all. I'm also unsure if there's a problem with me choosing a specific $z$ which depends on $\delta$.
Could anyone tell me if my attempt is correct? And if it isn't, could someone tell me how I could make a correct proof? Thank you very much!
 A: The epsilon-delta argument can be made very simply, once you know that the limiting value is path-dependent.  Let $$f(z) = z/\bar z = e^{2i\arg(z)}.$$  Then suppose there exists an $L \in \mathbb C$ satisfying the definition; then the claim the limit exists is equivalent to stipulating that $|e^{2i \arg(z)} - L|$ can be made arbitrarily small when $z$ is in a neighborhood of $0$.  But you can see right away where this will not work:  the magnitude of $e^{2i \arg (z)}$ is always unity irrespective of the size of the neighborhood, but the argument is $2 \arg (z)$; thus if you choose any fixed $L$, the supremum of the modulus of the difference is never less than unity.  Geometrically, this is equivalent to saying that for any choice of a point in the plane, the maximum distance of that point to any point on a unit circle is never less than $1$.  This furnishes the intuition for proceeding with a more formal argument, the outline of which is as follows:
We may assume without loss of generality that $\Im(L) = 0$ and $\Re(L) \ge 0$.  Then we compute for such an $L$ the maximum value of $|f(z) - L|$, which occurs for $\arg(z) = \pm \pi/2$; hence $|f(z) - L| = L+1$, and it follows that for any choice of $\epsilon < 1$, it is impossible to choose $\delta > 0$ such that whenever $|x| < \delta$, $|f(z) - L| < \epsilon$.
A: With $z=e^{i\theta}$ we have
$$\frac z{\bar z}=e^{2i\theta}=\cos2\theta+i\sin2\theta,$$ independently of $r$.
Then as
$$\left|\cos2\cdot0-\cos2\frac\pi2\right|=2,$$ for $\epsilon<1$, no $\delta$ can satisfy the condition
$$|f(z)-L|<\epsilon.$$
A: Either $L=-1$ or $L\neq-1$.  First, let $L=-1$.  Let $\epsilon=1$, $z=\frac\delta2$.  Then:
$$|z-0| = \frac\delta2 < \delta$$
$$|f(z)-L| = |f(z)+1| = \left|\frac{\frac\delta2}{\frac\delta2}+1\right| = 2 > \epsilon$$
So now let $L\neq-1$.  Let $\epsilon=\frac{|L+1|}{2}$, $z=i\frac\delta2$.  Then:
$$|z-0| = \frac\delta2 < \delta$$
$$|f(z)-L| = \left|\frac{i\frac\delta2}{-i\frac\delta2}-L\right|=|-1-L| = |L+1| > \epsilon$$
A: We prove by contradiction that the limit $\;\lim_\limits{z \to 0} \frac{z}{\overline{z}}\;$ does not exist.
If, by absurdum, the limit $\;\lim_\limits{z \to 0} \frac{z}{\overline{z}}$ existed, since $\left|\frac{z}{\overline{z}}\right|=1$ for all $z\in\mathbb{C}\setminus\{0\}$, the limit would be finite, so it would exist $L\in\mathbb{C}$ such that $\;\lim_\limits{z \to 0} \frac{z}{\overline{z}}=L$.
And, by the definition of limit, we get that
for $\;\epsilon=1>0\;,\;\exists\delta>0$ such that for all $z\in\mathbb{C}\setminus\{0\}\land|z|<\delta\;$ it results that $\;\left|\frac{z}{\overline{z}}-L\right|<1$.
Since $\;z_1=\frac{1}{2}\delta\;$ and $\;z_2=\frac{1}{2}\delta i\;$ satisfy the condition
$“\;z\in\mathbb{C}\setminus\{0\}\land|z|<\delta\;”\;,\;\;$ it follows that
$\left|\frac{z_1}{\overline{z_1}}-L\right|=\left|1-L\right|<1\;\;$ and
$\left|\frac{z_2}{\overline{z_2}}-L\right|=\left|-1-L\right|=\left|1+L\right|<1\;.$
So we get that
$|1+1|=|1-L+1+L|\le|1-L|+|1+L|<1+1=2$,
that is $\;|1+1|<2\;$ which is a contradiction.
Hence the limit $\;\lim_\limits{z \to 0} \frac{z}{\overline{z}}\;$ does not exist.
