Jee Mains

Let z be a complex number such that $|\mathrm{z}|=1$. If $\frac{2+\mathrm{k}^{2} \mathrm{z}}{\mathrm{k}+\overline{\mathrm{z}}}=\mathrm{kz}, \mathrm{k} \in \mathrm{R}$, then the maximum distance of $\mathrm{k}+\mathrm{ik}^{2}$ from the circle $|\mathrm{z}-(1+2 \mathrm{i})|=1$ is

Tagged: ,

Log In Register Lost Password
Viewing 3 posts - 1 through 3 (of 3 total)
  • Author
    Posts
  • 30 October, 2026 at 8:11 am #12478
    UPLIFTJEE
    Keymaster

    Let z be a complex number such that $|\mathrm{z}|=1$. If $\frac{2+\mathrm{k}^{2} \mathrm{z}}{\mathrm{k}+\overline{\mathrm{z}}}=\mathrm{kz}, \mathrm{k} \in \mathrm{R}$, then the maximum distance of $\mathrm{k}+\mathrm{ik}^{2}$ from the circle $|\mathrm{z}-(1+2 \mathrm{i})|=1$ is
    [JEE Main 02 April 2025 S I]
    (A) $\sqrt{5}+1$
    (B) 3
    (C) 2
    (D) $\sqrt{3}+1$

    30 October, 2026 at 8:11 am #12479
    UPLIFTJEE
    Keymaster

    Solution

    $\frac{2+\mathrm{k}^{2} \mathrm{z}}{\mathrm{k}+\overline{\mathrm{z}}}=\mathrm{kz} \Rightarrow 2+\mathrm{k}^{2} \mathrm{z}=\mathrm{k}^{2} \mathrm{z}+\mathrm{kz} \overline{\mathrm{z}}$

    \[ \begin{aligned} & \Rightarrow|\mathrm{z}|^{2} \mathrm{k}=2 \Rightarrow \mathrm{k}=2(\because|\mathrm{z}|=1) \\ & \mathrm{k}+\mathrm{ik}^{2}=2+4 \mathrm{i}, \operatorname{Point}(2,4) \end{aligned} \]

    circle $|\mathrm{z}-1-2 \mathrm{i}|=1$, centre $\mathrm{C}(1,2)$, radius $\mathrm{r}=1$
    Required maximum distance $=\mathrm{PQ}=\mathrm{PC}+\mathrm{r}=\sqrt{1+4}+1=\sqrt{5}+1$

    30 October, 2026 at 8:11 am #12480
    UPLIFTJEE
    Keymaster

    Answer

    A

  • Author
    Posts
Log In Register Lost Password
Viewing 3 posts - 1 through 3 (of 3 total)
  • You must be logged in to reply to this topic.