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Question 21 — JEE (Main) 30 Jan 2024 (S I)

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  • 7 October, 2026 at 11:17 pm #11796
    Om Sharma
    Keymaster

    If $\mathrm{z}=\mathrm{x}+\mathrm{iy}, \mathrm{xy} \neq 0$, satisfies the equation $\mathrm{z}^{2}+\mathrm{i} \overline{\mathrm{z}}=0$, then $\left|\mathrm{z}^{2}\right|$ is equal to [JEE (Main) 30 Jan 2024 S I]
    (A) 9
    (B) 1
    (C) 4
    (D) $\frac{1}{4}$

    7 October, 2026 at 11:17 pm #11797
    Om Sharma
    Keymaster

    Solution

    $\mathrm{z}^{2}=-\mathrm{i} \overline{\mathrm{z}} \Rightarrow\left|\mathrm{z}^{2}\right|=|\mathrm{i} \overline{\mathrm{z}}|=|\mathrm{i} \| \overline{\mathrm{z}}|=|\mathrm{z}| \Rightarrow|\mathrm{z}|^{2}-|\mathrm{z}|=0 \Rightarrow|\mathrm{z}|(|\mathrm{z}|-1)=0$
    $|z|=0$ (not acceptable as $\mathrm{xy} \neq 0$ ) $\therefore|\mathrm{z}|=1 \Rightarrow|\mathrm{z}|^{2}=1$

    7 October, 2026 at 11:17 pm #11798
    Om Sharma
    Keymaster

    Answer

    B

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