Let z be a complex number such that z =1 . If frac 2+k 2 z k+bar z =k z, k in R , then the maximum distance of k+i k 2 from the circle z-(1+2 i) =1 is

Mathematics · Complex Numbers

JEE Main 2025 — 2 April, Morning Shift — Question 41

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

Option A:
$\sqrt{5}+1$
Correct
Option B:
$3$
Option C:
$\sqrt{3}+1$
Option D:
$2$

Answer: A

Step-by-step solution

$\frac{2+k^{2} z}{k+\bar{z}}=k z$

$\Rightarrow 2+k^{2} z=k^{2} z+k z \bar{z}$

$\Rightarrow 2+k|z|^{2} \quad\left(z \bar{z}=|z|^{2},|z|=1\right)$

$\Rightarrow 2=k$

$\therefore k+k^{2} i=2+4 i$

The maximum distance is

$=\sqrt{(4-2)+(2-1)^{2}}+$ radius

$=\sqrt{(2)^{2}+(1)^{2}}+1$

$=\sqrt{5}+1$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 2 April, Morning Shift
Subject: Mathematics
Chapter: Complex Numbers
Topic: Geometry of Complex Numbers