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JEE Main 20252 AprilMorning Shift
Exam
JEE Main
Year
2025
Date
2 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Complex Number

JEE Main 2025 · 2 April · Morning Shift · Mathematics · Complex Number

Let z be a complex number such that |z|=1 . If 2+k^2 z k+ z =kz, k...

Let zz be a complex number such that ∣z∣=1|z|=1. If 2+k2zk+z‾=kz,k∈R\frac{2+\mathrm{k}^2 \mathrm{z}}{\mathrm{k}+\overline{\mathrm{z}}}=\mathrm{kz}, \mathrm{k} \in \mathbf{R}, then the maximum distance of k+ik2\mathrm{k}+\mathrm{ik}^2 from the circle ∣z−(1+2i)∣=1|\mathrm{z}-(1+2 \mathrm{i})|=1 is:

Options

  1. A5+1\sqrt{5}+1
  2. B2
  3. C3
  4. D3+1\sqrt{3}+1

Correct answer

A. 5+1\sqrt{5}+1

Step-by-step solution

2+k2zk+z‾=kz∣z∣2k=2k=2\begin{aligned} & \frac{2+\mathrm{k}^2 \mathrm{z}}{\mathrm{k}+\overline{\mathrm{z}}}=\mathrm{kz} \\ & |\mathrm{z}|^2 \mathrm{k}=2 \\ & \mathrm{k}=2\end{aligned}
point p(2,4)\mathrm{p}(2,4); center (1,2)(1,2)
distance from circle
(x−1)2+(y−2)2=1(x-1)^2+(y-2)^2=1 is max.
if (OP+r)=1+4+1=5+1(\mathrm{OP}+\mathrm{r})=\sqrt{1+4}+1=\sqrt{5}+1
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