I was trying to proof $|z|$ was not differentiable at any $z\in \mathbb{C} $. I gave it couple of tries but couldn't simplify the following equation $ \frac{|z+h| - |z|}{h} $. I searched online for an answer and someone suggested that you can simplify the previous equation like the following: $$\frac{|z+h| - |z|}{h} = \frac{|z+h|^2 - |z|^2}{h(|z+h|+|z|)} = \frac{2\operatorname{Re}(\bar{z}h) + |h|^2}{h(|z+h|+|z|)}$$ I do get why the first equality holds however i don't see how the person came up with the second equality. If I assume the equality holds I understand how to prove my problem its just that I don't get the simplification.

Any help would be greatly appreciated :)

  • $\begingroup$ Have you tried using the definition of the absolute value of a complex number in the nominator? $\endgroup$
    – Carsten S
    Aug 8 at 6:57

1 Answer 1


If $a$ and $b$ are complex numbers, then$$|a+b|^2=|a|^2+2\operatorname{Re}\left(\overline a b\right)+|b|^2.$$Indeed, if $a=\alpha+\beta i$ and if $b=\gamma+\delta i$ (with $\alpha,\beta,\gamma,\delta\in\Bbb R$), then$$|a+b|^2=(\alpha+\gamma)^2+(\beta+\delta)^2$$and\begin{align}|a|^2+2\operatorname{Re}\left(\overline a b\right)+|b|^2&=\alpha^2+\beta^2+2(\alpha\gamma+\beta\delta)+\gamma^2+\delta^2\\&=(\alpha+\gamma)^2+(\beta+\delta)^2\\&=|a+b|^2.\end{align}So,$$|z+h|^2-|z|^2=2\operatorname{Re}\left(\overline z h\right)+|h|^2.$$

  • $\begingroup$ thank you very much for your help :)) $\endgroup$
    – luki luk
    Aug 8 at 9:08
  • 1
    $\begingroup$ I'm glad I could help. $\endgroup$ Aug 8 at 10:18

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