How to prove $f(z)=|z|$ is nowhere differentiable

Question

Hi so i am trying to prove that the function is not analytic but i am having trouble. I am supposed to use the definition $\frac{f(z+h)-f(z)}{h}$. I tried using the fact that $|z|=\sqrt{z\overline{z}}$ but I cannot see anything obvious. I know my aim is to show that if you approach $0$ at different angles you get different limits but to do so i have to somehow make the equation simpler.

Answer 1

For this, you'd probably be better served using polar form for the variable: $z=re^{i\theta}$ as then $|z|=r$. Applying the definition with this yields: $$\lim_{z\to z_0}\left[\frac{f(z)-f(z_0)}{z-z_0}\right]=\lim_{r\to r_0}\lim_{\theta\to\theta_0}\left[\frac{r-r_0}{re^{i\theta}-r_0e^{i\theta_0}}\right] $$

Then to take different paths, you fix one of the component variables, and then take the limit of the other, letting it vary. For instance, let $\theta\rightarrow\theta_0$, then you're just left with $$\lim_{r\to r_0}\left[\frac{r-r_0}{re^{i\theta_0}-r_0e^{i\theta_0}}\right]=\lim_{r\to r_0}\left[\frac{r-r_0}{e^{i\theta_0}(r-r_0)}\right]=\frac{1}{e^{i\theta_0}} $$

The idea would then be to hold $r$ constant and take the limit for $\theta$. Or to break it up into real and imaginary parts: $z = x+iy$, with the modulus ${\vert{z}\vert}={\sqrt{x^2 + y^2}}$. Then the definition is:$$\lim_{x\to x_0}\lim_{y\to y_0}\left[\frac{\sqrt{x^2 + y^2}-\sqrt{x_0^2+y_0^2}}{(x-x_0)+i(y-y_0)}\right]$$

In order for a function to be differentiable at a point, the limit must be the same from any path, so just repeat the same procedure as above, fixing one variable and letting the other vary until you end up getting two limits that'll be different for any point, at which juncture you'll of proved that the function is nowhere differentiable.

Answer 2

Try using the complex form of the Cauchy-Riemann equations: $\dfrac{\partial f}{\partial \bar{z}}=0$.

Answer 3

If one is required to apply the limit definition of the derivative, then one can proceed as follows.

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Analysis for $\displaystyle z\ne 0$

First, we assume that $z\ne 0$. Then, we can write

$$\begin{align} \frac{f(z+h)-f(z)}{h}&=\frac{|z+h|-|z|}{h}\\\\ &=\left(\frac{|z+h|-|z|}{h}\right)\left(\frac{|z+h|+|z|}{|z+h|+|z|}\right)\\\\ &=\frac{|z+h|^2-|z|^2}{h(|z+h|+|z|)}\\\\ &=\frac{2\text{Re}(\bar zh)+|h|^2}{h(|z+h|+|z|)} \end{align}$$

Next, we simply take the limits as $h$ approaches $0$ along the real and imaginary axes, respectively, and show that the limits are unequal.

Using $(1)$ it is easy to see that if $h\to 0$ along the real axis then

$$\lim_{h\to 0}=\text{Re}(z)/|z| \tag 2$$

while if $h\to 0$ along the imaginary axis then

$$\lim_{h\to 0}=\text{Im}(z)/|z| \tag 3$$

Since the limits in $(2)$ and $(3)$ are not equal, then $f'(z)$ fails to exist for $z\ne 0$.

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Analysis for $\displaystyle z= 0$ If $z=0$, then we have

$$\frac{f(h)-f(0)}{h}=\frac{|h|}{h}$$

which obviously fails to have a limit as $h\to 0$.

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Hence, $f'(z)$ fails to exist for all $z$.