سوال ۲۶ کنکور ارشد مهندسی کامپیوتر ۱۴۰۵

Locate the two point sets on the complex plane.

$z^{4}=1$ gives the four 4th roots of unity, all on the unit circle: $z \in \{\,1,\; i,\; -1,\; -i\,\}, \qquad |z| = 1,$ at angles $0^\circ, 90^\circ, 180^\circ, 270^\circ$.

$w^{4}=-4 = 4\,e^{i\pi}$, so $|w| = 4^{1/4} = \sqrt{2}$ and $w = \sqrt{2}\,e^{i(\pi/4 + k\pi/2)}$: $w \in \{\,1+i,\; -1+i,\; -1-i,\; 1-i\,\}, \qquad |w| = \sqrt{2},$ the four points $(\pm 1, \pm 1)$, at angles $45^\circ, 135^\circ, 225^\circ, 315^\circ$.

Maximize the distance. Using the law of cosines in $\mathbb{C}$: $|z-w|^{2} = |z|^{2} + |w|^{2} - 2\operatorname{Re}(z\bar w) = 1 + 2 - 2\sqrt{2}\cos\theta = 3 - 2\sqrt{2}\cos\theta,$ where $\theta$ is the angle between $z$ and $w$. Every $z$-angle is a multiple of $90^\circ$ and every $w$-angle is $45^\circ$ plus a multiple of $90^\circ$, so $\theta$ is always an odd multiple of $45^\circ$. The smallest attainable cosine is $\cos 135^\circ = -\tfrac{\sqrt{2}}{2}$, giving $|z-w|^{2}_{\max} = 3 - 2\sqrt{2}\left(-\tfrac{\sqrt{2}}{2}\right) = 3 + 2 = 5.$

So $\max|z-w| = \sqrt{5}$. Concretely take $z=1$, $w=-1-i$: $|1-(-1-i)| = |2+i| = \sqrt{4+1} = \sqrt{5}.$

Sanity check. By the triangle inequality $|z-w| \le |z|+|w| = 1+\sqrt{2} \approx 2.414 < \sqrt{6}\approx 2.449$, so option 4 is impossible.

The answer is $\sqrt{5}$ (option 3).

Exam trick: For two points at fixed radii $r_1, r_2$ separated by angle $\theta$, $\text{dist}^2 = r_1^2 + r_2^2 - 2r_1 r_2\cos\theta$; maximize by choosing the pair whose discrete angles come closest to being opposite ($\theta$ near $180^\circ$).