# 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!

• 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. – Jacky Chong Aug 5 '20 at 18:55
• @JackyChong that's not using $\varepsilon-\delta$. – TSF Aug 5 '20 at 18:55
• @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. – Robert Lee Aug 5 '20 at 18:57
• @RobertLee I am curious why you want to prove limit doesn't exist using $\varepsilon-\delta$. – Jacky Chong Aug 5 '20 at 18:58
• "but since I don't know how to rigorously justify this I chose to avoid it." Let $A$ be the limit: $$\exists \epsilon = 1:\ \forall \delta > 0\ \exists x'=\delta, x''=\delta i: |x' - 0| \le \delta, |x'' - 0| \le \delta, \quad |f(x') - f(x'')| \ge 2 \Rightarrow \max(|f(x') - A|, |f(x'') - A|) \ge \epsilon$$ – Dmitry Aug 5 '20 at 18:59

## 4 Answers

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

• 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. – Robert Lee Aug 5 '20 at 20:02
• @RobertLee You are correct. Some simplification is possible because we do not need to generalize $L$ to be any argument due to radial symmetry. – heropup Aug 5 '20 at 20:08

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

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

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:

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

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.