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Let $a,b\in\mathbb{C}$ be complex numbers. I am looking for some lower bound on $$||a-b|-|a+b||.$$ I've found an upper bound ( $2|b|$ ), but this is not quite what I need. I think there should be some lower bound as a function of $a,b$ (or their real/imaginary parts). Indeed, my intuition comes from the fact that if $a=a_1+\mathrm ia_2$ and $b=b_1+\mathrm i b_2,$ then $$|a-b|=|a+b|\Leftrightarrow(a_1+b_1)^2+(a_2+b_2)^2=(a_1-b_1)^2+(a_2-b_2)^2,$$ and this is equivalent to $$a_1b_1=-a_2b_2,$$ which is a tight condition, so that in most cases, $||a-b|-|a+b||>0.$

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  • $\begingroup$ do you want to know lower bound when $a_{1} b_{1}\neq -a_{2} b_{2}$ $\endgroup$
    – asimath
    Aug 13 '16 at 9:57
  • $\begingroup$ Yes, this would do the trick. $\endgroup$ Aug 13 '16 at 9:59
  • $\begingroup$ for $\epsilon >0$, let $a=(1+\epsilon, 1)$ and $b=(1,-1)$ then $||a-b|-|a+b||=|\sqrt{\epsilon ^2+4}-2+\epsilon | \to 0 $ as $\epsilon \to 0$. So I think lower bound should be zero even if $a_1b_1\neq a_2b_2$. $\endgroup$
    – asimath
    Aug 13 '16 at 10:29
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Does this help? If $a, b$ are not both zero then

$$ \bigl \lvert \lvert a-b \rvert - \lvert a+b \rvert \bigr \rvert = \frac{\bigl\lvert \lvert a-b \rvert^2 - \lvert a+b \rvert^2 \bigr \rvert}{\bigl\lvert \lvert a-b \rvert + \lvert a+b \rvert \bigr \rvert} \\ = \frac{4 \lvert \operatorname{Re} (a \overline b) \rvert}{\bigl\lvert \lvert a-b \rvert + \lvert a+b \rvert \bigr \rvert} \ge \frac{2 \lvert \operatorname{Re} (a \overline b) \rvert}{\lvert a \rvert + \lvert b \rvert} = \frac{2 \lvert a_1 b_1 + a_2 b_2\rvert}{\lvert a \rvert + \lvert b \rvert} $$

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  • $\begingroup$ Yes! I think this can help! Thank you! $\endgroup$ Aug 14 '16 at 7:28

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