Let z be a complex number such that |z| = 1. If (2+k²z)/(k+z̄) = kz, k ∈ R, then the maximum distanc

Q: Wait, why do we reject the negative root here?

Actually, in this specific problem, the algebra leads us to a unique solution for $k$. We do not have a quadratic equation that yields two roots; the $k^2z$ terms cancel out perfectly, leaving us with a linear relationship. Always trust the algebra—if the terms cancel, they cancel!

Q: Why can't we just use energy conservation?

I see the confusion! You might be thinking of physics problems. In complex numbers, we do not have energy conservation, but we have something just as powerful: geometric invariance. We are dealing with distances in the Argand plane, not forces in a field.

Q: How do I know if the maximum distance is $d+r$ or $d-r$?

Great question! The distance from a point to a circle varies. If you draw a line from your point $P$ through the center $C$, it will intersect the circle at two points. The closer one is $d-r$ (the minimum distance), and the farther one is $d+r$ (the maximum distance). Since we want the maximum, we add the radius.

Question

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

Options:

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

(B) 2

(C) 3

(D) $\sqrt{3} + 1$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the complex plane! Today, we are going to unravel a problem that looks like a tangled mess of algebra but is actually a beautiful, elegant dance of geometry.

We are given a complex number $z$ such that $|z| = 1$. This is not just a constraint; it is a golden key.

In the world of complex numbers, whenever you see $|z| = 1$, your intuition should immediately scream: $z\bar{z} = 1$. Because the modulus squared of any complex number is the number multiplied by its conjugate, and since the modulus is $1$, its square is also $1$.

Keep this identity, $z\bar{z} = 1$, in your back pocket. It is the secret weapon that will simplify our entire journey.

### The Algebraic Dance

Now, let us face the main equation:

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

It looks intimidating, but the most effective way to dismantle a fraction is to cross-multiply. Let us multiply both sides by the denominator, $(k+\bar{z})$:

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

Now, let us expand the right side:

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

Look at that! The term $k^2z$ appears on both sides of the equation. Like a perfectly choreographed performance, they cancel each other out, leaving us with the remarkably simple equation:

$$2 = k(z\bar{z})$$

Since we know from our earlier identity that $z\bar{z} = 1$, the equation collapses into $2 = k(1)$, which means **$k = 2$**. We have successfully cracked the code!

### From Algebra to the Argand Plane

Now that we have found $k = 2$, we can locate our point $P$. The problem defines $P$ as $k + ik^2$.

Substituting $k = 2$, we get $P = 2 + i(2^2) = 2 + 4i$. In the Cartesian coordinate system of the complex plane, this is the point $(2, 4)$.

Next, let us look at the circle given by $|z - (1 + 2i)| = 1$. This is the standard form of a circle in the complex plane, $|z - z_0| = r$, where $z_0$ is the center and $r$ is the radius.

Here, the center $C$ is at $1 + 2i$, or $(1, 2)$, and the radius $r$ is $1$.

### The Final Leap

We are almost there. We need the maximum distance from point $P(2, 4)$ to the circle centered at $C(1, 2)$ with radius $r = 1$.

The distance from $P$ to the center $C$ is calculated as:

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

The maximum distance is simply the distance to the center plus the radius:

$$D_{max} = d + r = \sqrt{5} + 1$$

And there it is! A problem that started with a daunting fraction ends with a simple, elegant geometric result. The final answer is **$\sqrt{5} + 1$**.

Animated Solution for Mathematics - Complex Numbers: Let $z$ be a complex number such that $∣ z ∣ = 1$ . If $\frac{2 + k ^{2} z}{k + z ˉ} = k z, k \in R$ , then the maximum distance of $k + i k^{2}$ from the circle $∣ z - (1 + 2 i) ∣ = 1$ is:

Select Answer:

Visualized Solution

Analyze the Condition $∣ z ∣ = 1$

Complex Conjugate Property

The Main Equation

Cross-Multiplying

Expanding the Equation

Simplifying the Equation

Finding the Value of $k$

Locating Point $P$

Analyzing the Given Circle

Distance to the Center

Calculating Distance $d$

Maximum Distance Concept

Final Maximum Distance

The Sigma Insight: Geometrical Applications of Complex Numbers

Analyzing the Setup

The Algebraic Dance

From Algebra to the Argand Plane

The Final Leap

Similar Questions

Let $C$ be the set of all complex numbers. Let $S_{1} = {z \in C : ∣ z - 2∣ \leq 1}$ and $S_{2} = {z \in C : z (1 + i) + \overset{z}{ˉ} (1 - i) \geq 4}$ . Then, the maximum value of $∣ z - \frac{5}{2} ∣^{2}$ for $z \in S_{1} \cap S_{2}$ is equal to :

Let $S = {z \in C : ∣ z - 1∣ = 1 and (2 - 1) (z + \overset{z}{ˉ}) - i (z - \overset{z}{ˉ}) = 22}$ . Let $z_{1}, z_{2} \in S$ be such that $∣ z_{1} ∣ = max_{z \in S} ∣ z ∣$ and $∣ z_{2} ∣ = min_{z \in S} ∣ z ∣$ . Then $∣ 2 z_{1} - z_{2} ∣^{2}$ equals :

Let z be those complex number which satisfy $∣ z + 5∣ \leq 4$ and $z (1 + i) + \overset{z}{ˉ} (1 - i) \geq - 10$ , $i = - 1$ If the maximum value of $∣ z + 1 ∣^{2}$ is $α + β 2$ , then the value of $(α + β)$ is

Let $z$ be a complex number such that $∣ \frac{z - i}{z + 2 i} ∣ = 1$ and $∣ z ∣ = \frac{5}{2}$ . Then the value of $∣ z + 3 i ∣$ is:

For $α, β, z \in C$ and $λ > 1$ , if $λ - 1$ is the radius of the circle $∣ z - α ∣^{2} + ∣ z - β ∣^{2} = 2 λ$ , then $∣ α - β ∣$ is equal to

If for the complex numbers $z$ satisfying $∣ z - 2 - 2 i ∣ \leq 1$ , the maximum value of $∣3 i z + 6∣$ is attained at $a + ib$ , then $a + b$ is equal to .

If $z$ is a complex number such that $∣ z ∣ \geq 1$ , then the minimum value of $∣ z + \frac{1}{2} (3 + 4 i) ∣$ is:

The number of complex numbers z, satisfying $∣ z ∣ = 1$ and $∣ \frac{z}{z} + \frac{z}{z} ∣ = 1$ is: