How to prove directly from the definition that the complex norm isn't differentiable at any point?

Question

How to prove directly from the definition that the complex norm isn't differentiable at any point?

I used both ways of the definition:

$$\lim_{z\to z_0}\frac{|z|-|z_0|}{z-z_0} \ \text{ and }\ \lim_{h\to 0}\frac{|z_0+h|-|z_0|}{h}$$

but I don't even know how to begin.

Answer 1

If $z_0 = 0$, then since $t \mapsto |t|$ is not differentiable at $t=0$ we see that $z \mapsto |z|$ is not differentiable at $z_0=0$ .

If $z_0 \neq 0$ then we have $\lim_{r \to0} {|z_0+rz_0| - |z_0| \over r z_0}= {|z_0| \over z_0}$, $\lim_{r \to0}{|z_0+riz_0| - |z_0| \over r iz_0}= 0$, hence ${|z_0+h| - |z_0| \over h}$ has no limit as $h \to 0$.

Aside: ${|z_0+riz_0| - |z_0| \over r iz_0} = {(|1+ri| - 1)|z_0| \over r iz_0} = {(|1+ri| - 1) \over r } \cdot {|z_0| \over iz_0} = {(\sqrt{1+r^2} - 1) \over r } \cdot {|z_0| \over iz_0} $. Note that ${(\sqrt{1+r^2} - 1) \over r } \to 0$.

Answer 2

Consider, for $z\ne0$ and $h\ne0$, $$ \frac{|z+h|-|z|}{h}=\frac{|z|}{z}\frac{|1+(h/z)|-1}{h/z} $$ so it's not restrictive to show that $$ \lim_{h\to0}\frac{|1+h|-1}{h} $$ doesn't exist. If it exists, then it's the same along every path. For $h$ real we have $$ \lim_{h\to0}\frac{|1+h|-1}{h}=1 $$ For $h=it$ with $t$ real, $$ \lim_{t\to0}\frac{|1+it|-1}{it}= \lim_{t\to0}\frac{\sqrt{1+t^2}-1}{it}=0 $$ Do similarly at $z=0$.

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More generally, if $f$ is defined on a region in $\mathbb{C}$ and only assumes real values, it can only be differentiable at points where $$ \lim_{h\to0}\frac{f(z+h)-f(z)}{h}=0 $$ because the limit must be both real ($h$ approaching $0$ on the real axis) and purely imaginary ($h$ approaching $0$ on the imaginary axis).