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For $\mathrm{a} \in \mathrm{C}$, let $\mathrm{A}=\{\mathrm{z} \in \mathrm{C}: \operatorname{Re}(\mathrm{a}+\overline{\mathrm{z}})>\operatorname{lm}(\overline{\mathrm{a}}+\mathrm{z})\}$

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  • 22 October, 2026 at 6:00 pm #12384
    UPLIFTJEE
    Keymaster

    For $\mathrm{a} \in \mathrm{C}$, let $\mathrm{A}=\{\mathrm{z} \in \mathrm{C}: \operatorname{Re}(\mathrm{a}+\overline{\mathrm{z}})>\operatorname{lm}(\overline{\mathrm{a}}+\mathrm{z})\}$ and $\mathrm{B}=\{\mathrm{z} \in \mathrm{C}: \operatorname{Re}(\mathrm{a}+\overline{\mathrm{z}})<\operatorname{lm}(\overline{\mathrm{a}}+\mathrm{z})\}$. Then among the two statements: (S1) : If Re (a), Im (a) > 0, then the set A contains all the real numbers
    (S2) : If Re (a), Im (a) < 0, then the set B contains all the real numbers [JEE (Main) 11 April 2023 SII] (A) both are true (B) only (S1) is true (C) only (S2) is true (D) both are false

    22 October, 2026 at 6:00 pm #12385
    UPLIFTJEE
    Keymaster

    Solution

    Let $\mathrm{a}=\mathrm{x}_{1}+\mathrm{iy}_{1} \mathrm{z}=\mathrm{x}+\mathrm{iy} \operatorname{Now} \operatorname{Re}(\mathrm{a}+\overline{\mathrm{z}})>\operatorname{IM}(\overline{\mathrm{a}}+\mathrm{z}) \therefore \mathrm{x}_{1}+\mathrm{x}>-\mathrm{y}_{1}+\mathrm{y}$
    $\mathrm{x}_{1}=2, \mathrm{y}_{1}=10, \mathrm{x}=-12, \mathrm{y}=0$ Given inequality is not valid for these values S1 is false. Now $\operatorname{Re}(\mathrm{a}+\overline{\mathrm{z}})<\operatorname{IM}(\overline{\mathrm{a}}+\mathrm{z})$ $\mathrm{x}_{1}+\mathrm{x}<-\mathrm{y}_{1}+\mathrm{y} \Rightarrow \mathrm{x}_{1}=-2, \mathrm{y}_{1}=-10, \mathrm{x}=12, \mathrm{y}=0$ Given inequality is not valid for these values. S2 is false

    22 October, 2026 at 6:00 pm #12386
    UPLIFTJEE
    Keymaster

    Answer

    D

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