1
$\begingroup$

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 :)

$\endgroup$
1
  • $\begingroup$ 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

7
$\begingroup$

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,$$|z+h|^2-|z|^2=2\operatorname{Re}\left(\overline z h\right)+|h|^2.$$

$\endgroup$
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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.