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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!

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    $\begingroup$ If you approach along the $x$-axis, you get $\lim_{x\rightarrow 0}\frac{x}{x}=1$. If you approach along the $y$-axis, you get $\lim_{y\rightarrow 0}\frac{iy}{-iy}=-1$. The limit doesn't exists. $\endgroup$ – Jacky Chong Aug 5 '20 at 18:55
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    $\begingroup$ @JackyChong that's not using $\varepsilon-\delta$. $\endgroup$ – TSF Aug 5 '20 at 18:55
  • $\begingroup$ @JackyChong, I'm aware I can show the limit doesn't exist using different paths, but since it doesn't use $\epsilon - \delta$ I chose to avoid it. $\endgroup$ – Robert Lee Aug 5 '20 at 18:57
  • $\begingroup$ @RobertLee I am curious why you want to prove limit doesn't exist using $\varepsilon-\delta$. $\endgroup$ – Jacky Chong Aug 5 '20 at 18:58
  • $\begingroup$ "but since I don't know how to rigorously justify this I chose to avoid it." Let $A$ be the limit: $$\exists \epsilon = 1:\ \forall \delta > 0\ \exists x'=\delta, x''=\delta i: |x' - 0| \le \delta, |x'' - 0| \le \delta, \quad |f(x') - f(x'')| \ge 2 \Rightarrow \max(|f(x') - A|, |f(x'') - A|) \ge \epsilon $$ $\endgroup$ – Dmitry Aug 5 '20 at 18:59
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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 $|x| < \delta$, $|f(z) - L| < \epsilon$.

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  • $\begingroup$ I believe your proof is essentially the same as mine, with the difference that you take $L$ on the real axis but I don't simplify it. And I think that $|f(z) - L| = L+1$ is what I wrote as $|e^{i(2\theta)} - L| = r'+1$, or did I misunderstand what you meant? But as you point out, the idea that the difference can never be less than unity is exactly what I tried to use in my attempt. $\endgroup$ – Robert Lee Aug 5 '20 at 20:02
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    $\begingroup$ @RobertLee You are correct. Some simplification is possible because we do not need to generalize $L$ to be any argument due to radial symmetry. $\endgroup$ – heropup Aug 5 '20 at 20:08
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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

$$|f(z)-L|<\epsilon.$$

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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:

$$|z-0| = \frac\delta2 < \delta$$ $$|f(z)-L| = \left|\frac{i\frac\delta2}{-i\frac\delta2}-L\right|=|-1-L| = |L+1| > \epsilon$$

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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.

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