Question ID: #590
Let $z$ be a complex number such that $|z|=1$. If $\frac{2+k^{2}z}{k+\overline{z}}=kz$, $k \in \mathbb{R}$, then the maximum distance of $k+ik^{2}$ from the circle $|z-(1+2i)|=1$ is:
- (1) $\sqrt{5}+1$
- (2) 2
- (3) 3
- (4) $\sqrt{3}+1$
Solution:
Simplify the given equation:
$$\frac{2+k^2 z}{k+\bar{z}} = kz$$
$$2 + k^2 z = kz(k+\bar{z})$$
$$2 + k^2 z = k^2 z + k|z|^2$$
Since $|z|=1$, we have $|z|^2 = 1$:
$$2 + k^2 z = k^2 z + k$$
$$2 = k \Rightarrow k = 2$$
The point is $P = k + i k^2 = 2 + 4i$.
The circle is given by $|z – (1+2i)| = 1$.
Center $C = (1, 2)$ and Radius $r = 1$.
We need the maximum distance of point $P(2, 4)$ from the circle.
Maximum distance = Distance(PC) + radius.
Calculate distance $PC$:
$$PC = \sqrt{(2-1)^2 + (4-2)^2} = \sqrt{1^2 + 2^2} = \sqrt{1+4} = \sqrt{5}$$
Max Distance = $\sqrt{5} + 1$
Ans. (1)
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