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Algebra Difficulty 5.7 AIME, harder Prove it

2. If the complex number zz satisfies ∣z2−1∣=∣z2∣\left|z^{2}-1\right|=\left|z^{2}\right|, then ∣z−1∣−∣z+1∣=|z-1|-|z+1|=

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Solution

2. ±2\pm \sqrt{2}.

From the condition, we have
(∣z−1∣−∣z+1∣)2=∣z−1∣2+∣z+1∣2−2∣z2−1∣=2∣z2∣+2−2∣z2−1∣=2, (|z-1|-|z+1|)^{2}=|z-1|^{2}+|z+1|^{2}-2\left|z^{2}-1\right|=2\left|z^{2}\right|+2-2\left|z^{2}-1\right|=2,

thus ∣z−1∣−∣z+1∣=±2|z-1|-|z+1|= \pm \sqrt{2}.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.