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I'm making an exercise about the derivative. I needed to prove that $f(x)=|x|$ isn't differentiable at zero. Now I was wondering if we had a function $f:\mathbb{C} \to \mathbb{C} :z \to |z|$ so a complex function, in which points is this $f$ differentiable.

If we say $z=a+bi$, for $a=b=0$ it's not differentiable and for $a$ not equal to zero and $b=0$ it's differentiable (I think) but are there other problematic points for this function?

Bernard
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questmath
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1 Answers1

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HINT

Let $g(z)=f^2(z)=f^2(x+yi)=x^2+y^2\in\mathbb{R}$. Then $g$ is differentiable $\iff$ $f$ is constant by Cauchy-Riemann equations.

  • what do you mean by: f is constant by cauchy riemann equations? – questmath Aug 05 '20 at 17:28
  • https://mathworld.wolfram.com/Cauchy-RiemannEquations.html –  Aug 05 '20 at 17:32
  • isn't it $f^{2}(z)$=$|x^{2}-y^{2}+2xyi|$ EDIT: i'm sorry i confussed it with absolut value – questmath Aug 05 '20 at 17:34
  • if $f(z)=f(x+yi)=\sqrt{x^2+y^2}$ then $f^2(x+yi)=(\sqrt{x^2+y^2})^2=x^2+y^2$. What you were doing is $f(z^2)$. –  Aug 05 '20 at 17:37
  • correct i'm sorry – questmath Aug 05 '20 at 17:37
  • but i'm not sure how i can prove in which points f is differentiable – questmath Aug 05 '20 at 17:38
  • but i'm not sure how i can prove in which points f is differentiable. Because it's not because g is differentiable in a point that f is differentiable in that point. The other implicitation is valid but not this one. – questmath Aug 05 '20 at 17:44
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    Mathmath, what you're missing here is the standard exercise that a real-valued complex-differentiable function must be constant. Write $g = u+iv$ and note that $v=0$, so $\partial u/\partial x = \partial u/\partial y = 0$. – Ted Shifrin Aug 05 '20 at 18:50