JEE Mathematics 2025 Question with Solution

Given: $|z| = 1$ and

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

Find: The maximum distance of $k + ik^2$ from the circle $|z - (1 + 2i)| = 1$.

Since $|z| = 1$, we use

$$z\bar{z} = 1$$

Cross-multiplying the given equation,

$$2 + k^2 z = kz(k + \bar{z})$$

Expanding the right-hand side,

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

Using $z\bar{z} = 1$,

$$2 + k^2 z = k^2 z + k$$

Cancelling $k^2 z$ from both sides,

$$2 = k$$

So,

$$k = 2$$

Now the point represented by $k + ik^2$ is

$$P = 2 + 4i$$

The circle $|z - (1 + 2i)| = 1$ has center

$$C_0 = 1 + 2i$$

and radius

$$r = 1$$

Distance from $P$ to the center is

$$d = |P - C_0| = |(2 + 4i) - (1 + 2i)| = |1 + 2i|$$

Therefore,

$$d = \sqrt{1^2 + 2^2} = \sqrt{5}$$

The maximum distance from a point to a circle equals distance from the center plus the radius:

$$\text{Maximum Distance} = d + r = \sqrt{5} + 1$$

Therefore, the correct option is A, and the maximum distance is $\sqrt{5} + 1$.

Given: $|z| = 1$ and the relation involving $k$.

Find: The farthest distance of $k + ik^2$ from the circle $|z - (1 + 2i)| = 1$.

Use the identity $z\bar{z} = 1$ directly in the simplified equation from the solution working to get

$$k = 2$$

Hence the point is

$$(\Re, \Im) = (k, k^2) = (2, 4)$$

The circle has center $(1, 2)$ and radius $1$. So the center-to-point distance is

$$\sqrt{(2 - 1)^2 + (4 - 2)^2} = \sqrt{5}$$

Thus the farthest point on the circle is one radius farther along the same line, giving

$$\sqrt{5} + 1$$

Therefore, the correct option is A.