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.
Let f and g be functions satisfying $f(x+y) = f(x)f(y), f(1) = 7$ and $g(x+y) = g(xy), g(1) = 1$, for all $x, y \in \mathbb{N}$. If $\sum_{x=1}^{n} \left(\frac{f(x)}{g(x)}\right) = 19607$, then n is equal to :
Let S be the set of the first 11 natural numbers. Then the number of elements in $A = \{B \subseteq S : n(B) \geq 2$ and the product of all elements of B is even is ________.