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Let z and omega be two complex numberssuch that |z|le 1, |omega| le 1 and |z+ iomega| = |z_(1)-z_(2)|is equal to |
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Answer» `(2)/(3)` `therefore |z|+|omega|le2""...(1)` But it is given that `|z|le1 and |omega|le1`. `rArr |z|+|omega|le2""...(2)` From (1) and (2), `|z|=|omega|=1` Also, `|z+iomega|=|z-(ibaromega)|` `rArr |z-(-iomega)|=|z-(ibaromega)|` This means that z lies on perpendicularbisector of theline segment JOINING `(-iomega) and (ibaromega)`, which is real axis, as `(-iomega) and (ibaromega)` are CONJUGATE to each other. For`z, Im(z) = 0` If `z =x, " then " |z|le1` `rArr x^(2)le1` `rArr -1le x le 1` Therefore, (3) is the correct option. |
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