JEE MAIN - Mathematics (2025 - 2nd April Morning Shift - No. 9)
Let $z$ be a complex number such that $|z|=1$. If $\frac{2+\mathrm{k}^2 z}{\mathrm{k}+\bar{z}}=\mathrm{k} z, \mathrm{k} \in \mathbf{R}$, then the maximum distance of $\mathrm{k}+i \mathrm{k}^2$ from the circle $|z-(1+2 i)|=1$ is :
$\sqrt{5}+1$
3
$\sqrt{3}+1$
2
Explanation
_2nd_April_Morning_Shift_en_9_1.png)
$$\begin{aligned} &\begin{aligned} & \frac{2+k^2 z}{k+\bar{z}}=k z \\ & \Rightarrow 2+k^2 z=k^2 z+k \bar{z} \\ & \Rightarrow 2+k|z|^2 \quad\left(z \bar{z}=|z|^2,|z|=1\right) \\ & \Rightarrow 2=k \\ & \therefore k+k^2 i=2+4 i \end{aligned}\\ &\therefore \quad \text { The maximum distance is }\\ &\begin{aligned} & =\sqrt{(4-2)+(2-1)^2}+\text { radius } \\ & =\sqrt{(2)^2+(1)^2}+1 \\ & =\sqrt{5}+1 \end{aligned} \end{aligned}$$
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