2
$\begingroup$

Without using Cauchy-Riemann.

I know the trick is to make $z_0 + \Delta z \rightarrow z_0 $ along a circle so that $ \lvert z_0 + \Delta z\rvert $ stays constant- small changes in the domain do not lead to small changes in the range. But I'm not sure how to prove it.

This is not the same as showing that $g(z) =\bar{z}$ is not differentiable everywhere.

  • 0
    You can show that it is not a conformal map. – 2017-01-26
  • 0
    better http://math.stackexchange.com/questions/1232372/show-that-gz-barz-is-continuous-at-every-point-of-mathbbc-and-that – 2017-01-26
  • 2
    Possible duplicate of [Show that $g(z) =\bar{z}$ is continuous at every point of $\mathbb{C}$ and that it's not differentiable at any point.](http://math.stackexchange.com/questions/1232372/show-that-gz-barz-is-continuous-at-every-point-of-mathbbc-and-that) – 2017-01-26
  • 0
    @MyGlasses The OP is requesting a proof without the CREs. The possible duplicate uses the CREs. – 2017-01-26
  • 0
    @Dr.MV Yes. Second link prepares this. – 2017-01-26
  • 0
    Well, I would not call that a duplicate. – 2017-01-26

1 Answers 1

5

Let $f(z)=|z|$. Then, $f$ is differentiable then the limit $\lim_{\Delta z\to 0}\frac{f(z+\Delta z)-f(z)}{\Delta z}$ exists and is independent of the manner in which $\Delta z$ approaches $0$.

Suppose $\Delta z=\Delta x$. Then,

$$\begin{align} \lim_{\Delta x\to 0}\frac{|z+\Delta x|-|z|}{\Delta x}&=\lim_{\Delta x\to 0}\frac{|z+\Delta x|^2-|z|^2}{\Delta x(|z+\Delta x|+|z|)}\\\\ &=\lim_{\Delta x\to 0}\frac{2x+|\Delta x|}{(|z+\Delta x|+|z|)}\\\\ &=\frac{x}{\sqrt{x^2+y^2}}\tag 1 \end{align}$$

Instead, suppose that $\Delta z=iy$. Then,

$$\begin{align} \lim_{\Delta y\to 0}\frac{|z+i\Delta y|-|z|}{i\Delta y}&=\lim_{\Delta x\to 0}\frac{|z+i\Delta y|^2-|z|^2}{i\Delta y(|z+i\Delta y|+|z|)}\\\\ &=\lim_{\Delta x\to 0}\frac{2y+|\Delta y|}{i(|z+\Delta y|+|z|)}\\\\ &=\frac{-iy}{\sqrt{x^2+y^2}}\tag 2 \end{align}$$

Since $(1)$ and $(2)$ are not equal, the limit of interest fails to exist and hence $f$ is nowhere differentiable.