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So I want to show that $f(z)=|z|$ where $z$ is a complex number is not differentiable anywhere, and that $g(z)=|z|^2$ is differentiable at $z=0$ only. Now with some computation, I got:

$\lim_{h\rightarrow 0}\frac{|z+h|-|z|}{h}=\frac{\partial}{\partial x}\sqrt{x^2+y^2}=\frac{x}{\sqrt{x^2+y^2}}$ and $\lim_{ih\rightarrow 0}\frac{|z+ih|-|z|}{ih}=\frac{1}{i}\frac{\partial}{\partial y}\sqrt{x^2+y^2}=\frac{y}{i\sqrt{x^2+y^2}}$. Thus they are not equal for $z\ne x+iy.$

Now what I do not get it is, this result would imply what I have to prove in the question, but why would this mean that this $f(z)$ is not differentiable anywhere?

Similary, for $g(z)$, the computation I get for $\lim_{h\rightarrow 0}\frac{|z+h|^{2}-|z|^2}{h}=2x$ and $\lim_{ih\rightarrow 0}\frac{|z+ih|^{2}-|z|^2}{ih}=\frac{2y}{i}$, this is not equal unless $z=0$. The conclusion once again is $g$ is not differentiable unless at $z=0.$

Questions

1) Why is $f$ not differentiable anywhere? Is it because the only possible place where it is differentiable is if $z=0?$, but that would imply the denominator is $0 $ and hence undefined?

2) Why do we have to have the condition in which the partials have to be equal in order for the function to be differentiable at some given domain?

I would appreciate some clear explanation. Thank You.

José Carlos Santos
  • 427,504
Aurora Borealis
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  • If $|z|$ is differentiable, then so is $|z|^2 = z\bar{z}$. However, that function can't be differentiable outside of any neighbourhood of $0$, since assuming otherwise, we could divide by $z$ and get a differentiable function, but the result, $\bar{z}$ is not differentiable, see https://math.stackexchange.com/q/2199702/94959. – Eman Yalpsid Apr 07 '18 at 11:15

1 Answers1

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I don't know why you are using partial derivatives here. It's more basic than that. The function $f$ is not differentiable at $0$ because the limit$$\lim_{h\to0}\frac{|h|}h$$doesn't exist. And it doesn't exist because if you take $h\in(0,+\infty)$ then the limit is $1$, and if you take $h\in(-\infty,0)$, the limit is $-1$.

This doesn't prove that $f$ isn't differentiable anywhere. It only proves that it is not differentiable at $0$.

Now, the function $g$ is differentiable at $0$, because$$\lim_{h\to0}\frac{g(h)}h=\lim_{h\to0}\frac{h\overline h}h=\lim_{h\to0}\overline h=0.$$But it's differentiable only there. You can prove it using the Cauchy-Riemann equations: $g(x+yi)=u(x,y)+v(w,y)i$ with $u(x,y)=x^2+y^2$ and $v(x,y)=0$. And you only have $u_x=v_y(=0)$ and $u_y=-v_x(=0)$ when $x+yi=0$.

And if $f$ was differentiable at a point $z\neq0$, so would be $f^2$, which is $g$.

José Carlos Santos
  • 427,504
  • Thank you for this explanation and as much as I get this, I want to try and understand if my working in the above is correct. And if so, why the different partials imply the non-differentiability of the complex function. – Aurora Borealis Apr 07 '18 at 10:39