ExamBro
ExamBro
JEE Mains · Maths · STD 11 - 4.1 complex nubers

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

  1. A \(\sqrt{5}+1\)
  2. B 2
  3. C 3
  4. D \(\sqrt{3}+1\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\sqrt{5}+1\)

Step-by-step Solution

Detailed explanation

\(\begin{aligned} & \frac{2+\mathrm{k}^2 \mathrm{z}}{\mathrm{k}+\overline{\mathrm{z}}}=\mathrm{kz} \\ & |\mathrm{z}|^2 \mathrm{k}=2 \\ & \mathrm{k}=2\end{aligned}\) point \(\mathrm{p}(2,4)\); center \((1,2)\) distance from circle \((x-1)^2+(y-2)^2=1\) is max. if…
Same subject
Explore more questions on app