Let z be a complex number satisfying |z - 5i| | vedclass

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(C) The condition $|z - 5i| \le 1$ represents a disk centered at $C(0, 5)$ with radius $r = 1$.To minimize $	ext{amp } z$,the line $OA$ must be tange

Vedclass · Accepted Answer

$\frac{2\sqrt{6}}{5} + \frac{24}{5}i$

Vedclass · Answer

(C) The condition $|z - 5i| \le 1$ represents a disk centered at $C(0, 5)$ with radius $r = 1$.To minimize $	ext{amp } z$,the line $OA$ must be tangent to the circle at point $A$.In $	riangle OAC$,$\angle OAC = 90^\circ$ (radius is perpendicular to the tangent).$OC = 5$ and $AC = 1$.In $	riangle OAC$,$\sin(\angle AOC) = \frac{AC}{OC} = \frac{1}{5}$.Let $\alpha = \angle AOC$. Then $\sin \alpha = \frac{1}{5}$,so $\cos \alpha = \sqrt{1 - (\frac{1}{5})^2} = \frac{\sqrt{24}}{5} = \frac{2\sqrt{6}}{5}$.The angle of $z$ is $	heta = 90^\circ - \alpha$. Thus,$\cos 	heta = \sin \alpha = \frac{1}{5}$ and $\sin 	heta = \cos \alpha = \frac{2\sqrt{6}}{5}$.The distance $OA = \sqrt{OC^2 - AC^2} = \sqrt{25 - 1} = \sqrt{24} = 2\sqrt{6}$.Thus,$z = OA(\cos 	heta + i \sin 	heta) = 2\sqrt{6}(\frac{1}{5} + i\frac{2\sqrt{6}}{5}) = \frac{2\sqrt{6}}{5} + i\frac{4 	imes 6}{5} = \frac{2\sqrt{6}}{5} + \frac{24}{5}i$.

Let $z$ be a complex number satisfying $|z - 5i| \le 1$ such that $\text{amp } z$ is minimum. Then $z$ is equal to

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