0
$\begingroup$

I have tried to prove it in the following manner-
$||z|-1|=||z|-|1||\le|z-1|\ \forall z\in\Bbb{C}$ (by Triangle inequality)
Now, can I write $|z-1|\le|z+1|$ in $\Bbb{C}$? If yes then the proof is done.
But I can't get it. I don't know whether the statement is true. Can anybody solve it? Thanks for the assistance in advance.

$\endgroup$
4
$\begingroup$

$|z|=|z+1-1|\leq |z+1|+1$. Now just pull 1 on RHS to LHS.

$\endgroup$
  • $\begingroup$ Simple and concise(+1)! $\endgroup$ – Chinnapparaj R Apr 20 at 9:16
0
$\begingroup$

The triangle inequality doesn't say that $||z|-|1||\le|z-1|$. It says that $|a+b|\le|a|+|b|$ for all $a,b$.

What does it say if $a=z+1$ and $b=-1$?

$\endgroup$
0
$\begingroup$

Your approach with the reverse triangle inequality ($||z|-|w||\leq |z-w|$) works well with a slight adjustment.

Just note that

  • $a\leq |a|$ for all $a \in \mathbb{R}$ and
  • letting $w = -1$ you get immediately $$|z|-1 \leq ||z|-|-1||\leq |z -(-1)| = |z+1|$$
$\endgroup$

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.