Show that $f(z) = |z|$ is not differentiable anywhere.

Question

Show that $f(z) = |z|$ is not differentiable anywhere.

I already know that $f(z) = |z|^2$ is differentiable at $0$, but I am unsure how to show that $f(z) = |z|$ is not. I have set up the definition of a derivative below, but I do not know how to approach solving this, particularly the part $|z+\lambda|$, since I know for complex numbers $|z| = \sqrt {x^2 + y^2}$.

$\lim _{\lambda\to \:0}\left(\frac{\left|z+\lambda\right|-\left|z\right|}{\lambda}\right)$

Answer 1

So you'll want to use the Cauchy-Riemann equations, that for $x + iy = f(a+bi)$, $x,y,a,b$ all real, $f(z)$ is differentiable iff:

$$\frac{\delta x}{\delta a} = \frac{\delta y}{\delta b}$$ $$\frac{\delta y}{\delta a} = -\frac{\delta x}{\delta b}$$

In this case, $\delta y/\delta a$ and $\delta y/\delta b$ are both identically zero, since the absolute value has no imaginary part. The partials of the real part, which is $\sqrt{a^2+b^2}$, are $\frac{a}{\sqrt{a^2+b^2}}$ and $\frac{b}{\sqrt{a^2+b^2}}$ respectively. The equations are only satisfied when these are both zero, but since each is zero only if $a$ or $b$ is zero, and neither exists if both are zero, the equations are never satisfied.

Answer 2

Take $z_0\in\Bbb C$. If $z_0=0$, then $f$ is not differentiable at $z_0$ because the limit$$\lim_{z\to z_0}\frac{f(z)-f(z_0)}{z-z_0}=\lim_{z\to0}\frac{|z|}z$$doesn't exist (if $z\in(0,\infty)$, then $\frac{|z|}z=1$ and if $z\in(-\infty,0)$, then $\frac{|z|}z=-1$). And if $z_0\ne0$, if $f$ was differentiable at $z_0$, then $f^2$ would also be differentiable at $z_0$. But you already know that $f^2$ is not differentiable at $z_0$ when $z_0\ne0$.