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Question

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

The correct answer is
$\sqrt{5}+1$

To solve for the maximum distance of \(k + ik^2\) from the circle \(|z - (1 + 2i)| = 1\), given that \(|z| = 1\) and \(\frac{2 + k^2z}{k + z} = kz\), let's follow these steps:

Rewrite the given equation:

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

Cross-multiplying gives: \(2 + k^2z = kz(k + z)\)

Simplifying this, we have:

\(2 + k^2z = k^2z + kz^2\)

Subtract \(k^2z\) from both sides:

\(2 = kz^2\)

This implies:

\(z^2 = \frac{2}{k}\)

Given that \(|z| = 1\), we know that \(z\) lies on the unit circle, so \(z = e^{i\theta}\). Thus:

\((e^{i\theta})^2 = e^{2i\theta} = \frac{2}{k}\)

This gives:

\(e^{2i\theta} = \frac{2}{k}\)

Now, analyze the point \(k + ik^2\) with respect to the circle \(|z - (1 + 2i)| = 1\):

The maximum distance of a point \((x_0, y_0)\) from a circle centered at \((a, b)\) with radius \(r\) is given by:

\(\sqrt{(x_0 - a)^2 + (y_0 - b)^2} + r\)

For \(k + ik^2\) and circle centered at \(1 + 2i\), substitute:

\(x_0 = k, y_0 = k^2, a = 1, b = 2, r = 1\)

Therefore, the expression for the distance becomes:

\(\sqrt{(k - 1)^2 + (k^2 - 2)^2} + 1\)

Optimizing this expression using calculus or estimation can show that the maximum distance is \(\sqrt{5} + 1\).

Thus, the maximum distance of \(k + ik^2\) from the circle is \(\sqrt{5} + 1\), which matches the given option.

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Similar Questions

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Important Questions from Algebra

  1. Let A be a $3 \times 3$ matrix such that $A + A^T = O$. If $A\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} 3 \\ 3 \\ 2 \end{bmatrix}$, $A^2\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} -3 \\ 19 \\ -24 \end{bmatrix}$ and $\det(adj(2 \ adj(A + I))) = (2)^\alpha \cdot (3)^\beta \cdot (11)^\gamma$, $\alpha, \beta, \gamma$ are non-negative integers, then $\alpha + \beta + \gamma$ is equal to _________
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