Let z be a complex number such that |z| = 1. If , then the maximum distance of from the circle |z – (1 + 2i)| = 1 is : [2025]

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Solution

Q.
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 + 2i)| = 1 is : [2025]

1 $\sqrt{3} + 1$

2 3

3 $\sqrt{5} + 1$

4 2

Ans.
(3)

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

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

$\Rightarrow | z |^{2} k = 2$ [ $∵ z \bar{z} = | z |^{2}$ ]

$\Rightarrow k = 2$ [ $∵ | z | = 1$ ]

The centre of the given circle is (1, 2) and its radius is 1.

Now, $k + i k^{2} = 2 + 4 i$ .

Maximum distance = $O P + r = \sqrt{1 + 4} + 1 = \sqrt{5} + 1$ .

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