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Complex Numbers question

2025 · 2 Apr · Shift 1 · Q34

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Mathematics
Complex Numbers
2025 · 2 Apr · Shift 1 · Q34

Complex Numbers question

2025 · 2 Apr · Shift 1 · Q34

JEE MainMathematicsComplex NumbersMCQ+4 / −1

Let

z

be a complex number such that

|z|=1

. If

\frac{2+\mathrm{k}^2 z}{\mathrm{k}+\bar{z}}=\mathrm{k} z, \mathrm{k} \in \mathbf{R}

, then the maximum distance of

\mathrm{k}+i \mathrm{k}^2

from the circle

|z-(1+2 i)|=1

is :

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

View written solutionFree

Correct answer: A

Use the given condition on $z$

We are given

|z|=1

and

\frac{2+k^2 z}{k+\bar z}=kz, \qquad k\in\mathbb R.

Since $|z|=1$ , we use

\bar z=\frac1z \quad (z\neq 0).

So the equation becomes

\frac{2+k^2 z}{k+1/z}=kz.

Multiply numerator and denominator appropriately:

2+k^2 z = kz\left(k+\frac1z\right)=k^2 z+k.

Hence,

2+k^2 z=k^2 z+k \implies 2=k.

So the only possible real value is

k=2.

Find the point whose distance is asked

The point is

k+i k^2 = 2+i(4)=2+4i.

Interpret “maximum distance from the circle”

The circle is

|z-(1+2i)|=1,

so its center is

C=(1,2)

and radius is

r=1.

The point is

P=(2,4).

Distance from $P$ to the center:

PC=| (2+4i)-(1+2i) |=|1+2i|=\sqrt{1^2+2^2}=\sqrt5.

For a fixed point outside/inside a circle, the maximum distance to a point on the circle is

PC+r.

Therefore,

\text{maximum distance}=\sqrt5+1.

Check options

A: $\sqrt5+1$ ✅
B: $3$ ❌
C: $\sqrt3+1$ ❌
D: $2$ ❌

So the correct option is

\boxed{\sqrt5+1}.

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