1. The problem involves analyzing inequalities with absolute values: (i) $|r + a - \sigma| > |r + a + \sigma|$ (ii) $|r + a - \sigma| > |r + a + \sigma|$ (iii) $|r + a - \sigma| > |r + a|$ 2. Notice that (i) and (ii) are identical, so we only need to analyze two distinct inequalities. 3. For (i) and (ii): $|r + a - \sigma| > |r + a + \sigma|$ - Let $x = r + a$. - The inequality becomes $|x - \sigma| > |x + \sigma|$. 4. To analyze $|x - \sigma| > |x + \sigma|$, consider the squares (since absolute values are non-negative): $$|x - \sigma|^2 > |x + \sigma|^2$$ $$ (x - \sigma)^2 > (x + \sigma)^2 $$ 5. Expand both sides: $$ x^2 - 2x\sigma + \sigma^2 > x^2 + 2x\sigma + \sigma^2 $$ 6. Simplify by canceling $x^2$ and $\sigma^2$: $$ -2x\sigma > 2x\sigma $$ 7. Combine terms: $$ -2x\sigma - 2x\sigma > 0 $$ $$ -4x\sigma > 0 $$ 8. Divide both sides by $-4$ (note inequality reverses because dividing by negative): $$ x\sigma < 0 $$ 9. So the inequality holds when $x\sigma < 0$, i.e., when $r + a$ and $\sigma$ have opposite signs. 10. For (iii): $|r + a - \sigma| > |r + a|$ - Again, let $x = r + a$. - The inequality is $|x - \sigma| > |x|$. 11. Square both sides: $$ (x - \sigma)^2 > x^2 $$ $$ x^2 - 2x\sigma + \sigma^2 > x^2 $$ 12. Simplify: $$ -2x\sigma + \sigma^2 > 0 $$ 13. Rearrange: $$ \sigma^2 > 2x\sigma $$ 14. This inequality depends on the signs of $x$ and $\sigma$: - If $\sigma > 0$, then $\sigma^2 > 2x\sigma$ implies $\sigma > 2x$. - If $\sigma < 0$, then $\sigma^2 > 2x\sigma$ implies $\sigma < 2x$. 15. "Strong rotor" likely refers to the condition where these inequalities hold, indicating a strong effect or dominance of $\sigma$ relative to $r + a$. Summary: - (i) and (ii) hold when $(r + a)\sigma < 0$. - (iii) holds when $\sigma^2 > 2(r + a)\sigma$. These conditions describe regions where the absolute value expressions are greater, possibly indicating a "strong rotor" effect.