2
$\begingroup$

I know that $\bar{z} = \frac{1}{z}=e^{-i\theta}=\cos \theta - i \sin \theta$

We also know that $z+dz = (x+\delta x) + i(y + \delta y)$, therefore $\overline{z+dz} = 1/ ((x+\delta x) + i(y + \delta y))$

and therefore all we have to do is to show that two paths lead to a different result.

Using the definition of the derivative we have that:

$$f'(z)=\lim_{\delta z \rightarrow 0}\frac{\overline{z+\delta z}-\bar{z}}{\delta x + i \delta y}$$

Am I on the right track? Because if all of the above is correct, then I am having trouble simplifying this.

$\endgroup$
3
  • $\begingroup$ I don't really understand your notation. I'll use a notation I know. In order to prove that $f$ isn't differentiable in the complex plane, it's enough to prove it isn't differentiable somewhere. I'll choose the origin. You want to look at $\lim \limits_{z\to 0}\left(\dfrac{\overline z}{z}\right)$, or, with $x:=\Re(z)$ and $y:=\Im(z)$ (for all $z\in \mathbb C$), $\lim\limits_{(x,y)\to (0,0)}\left(\dfrac{x-iy}{x+iy}\right)$. Now you just have two find two paths that lead to different results. Hint: Consider the limits along the real axis and along the imaginary axis. $\endgroup$
    – Git Gud
    Feb 21, 2016 at 10:53
  • $\begingroup$ thanks. Why are you looking at: $\frac{\bar{z}}{z}$ and not $\bar{z}$? $\endgroup$
    – Naz
    Feb 21, 2016 at 10:54
  • $\begingroup$ The definition of derivative of $f$ at $0$ I'm using is $\lim \limits_{z\to 0}\left(\dfrac{f(z)-0}{z-0}\right)$, when this limit exists. $\endgroup$
    – Git Gud
    Feb 21, 2016 at 10:55

3 Answers 3

Reset to default
4
$\begingroup$

Not like that. Write $z=z_0+h$ with $h=re^{i\theta}$ and note that the limit $$ \lim_{z\to z_0}\frac{\overline z-\overline z_0}{z-z_0}=\lim_{h\to0}\frac{\overline h}h=\lim_{h\to0}e^{-2i\theta} $$ does not exist, since it would give different values for different values of $\theta$.

$\endgroup$
7
  • $\begingroup$ thanks. Is this true? $\bar{z} = \overline{z_0 + h} = \bar{z_0} + \bar{h}$? $\endgroup$
    – Naz
    Feb 21, 2016 at 10:58
  • 1
    $\begingroup$ Yes, that's true. $\endgroup$
    – John B
    Feb 21, 2016 at 11:01
  • $\begingroup$ hmm, but then if that was true, then I would have that $\overline{z+dz} = \bar{z} + \bar{dz}$ which I do not think is correct. Because then I would have $\frac{\bar{dz}}{\bar{dz}}$ which would lead to the result of 1. But in my notes I have that that derivative simplifies to: $\frac{\delta x - i\delta y}{\delta x + i \delta y}$ $\endgroup$
    – Naz
    Feb 21, 2016 at 11:11
  • $\begingroup$ I regret that I don't understand your notation. Never seen it anyways. $\endgroup$
    – John B
    Feb 21, 2016 at 11:17
  • $\begingroup$ which part of it? I can explian $\endgroup$
    – Naz
    Feb 21, 2016 at 11:18
1
$\begingroup$

use the cauchy riemann equtaion in polar form $$\frac {\partial f}{\partial \bar z}=0$$ what you get is $$1=0$$ and that is never true or absurd hence the function is not differentiable in the entire complex plane

$\endgroup$
1
$\begingroup$

For my own reference:

$\overline{z+dz}$ is conjugate of $z+dz$, which is $(x+\delta x) + i(y+\delta y)$. Therefore,$\overline{z+dz}= (x+\delta x) - i(y+\delta y)$.

Hence: $$\overline{z+dz}-\bar{z} = (x+\delta x) - i(y+\delta y) - x - iy = \delta x - i\delta y$$

Thus we have

$$\lim_{\delta z \rightarrow 0} \frac{\delta x - i \delta y}{\delta x + i \delta y}$$

When we make $\delta x \rightarrow 0$ we get $-1$.\ When we make $\delta y \rightarrow 0$ we get $1$.

Thus, $\bar{z}$ is not differentiable everywhere.

$\endgroup$
1
  • $\begingroup$ In this case, I hope you mean 'as $\delta y \to 0$' $\endgroup$
    – João Víctor Melo
    Oct 16, 2022 at 7:30

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .