If |z - 25i| le 15,then |max text{amp}(z) - m | vedclass

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Question

(B) The given inequality $|z - 25i| \le 15$ represents a disk centered at $25i$ with radius $15$. Let $O$ be the origin $(0,0)$ and $C$ be the center

Vedclass · Accepted Answer

$\pi - 2\cos^{-1}\left(\frac{3}{5}\right)$

Vedclass · Answer

(B) The given inequality $|z - 25i| \le 15$ represents a disk centered at $25i$ with radius $15$. Let $O$ be the origin $(0,0)$ and $C$ be the center $(0, 25)$. The tangents from the origin to the circle touch at points $z_1$ and $z_2$. In the right-angled triangle formed by the origin,the center of the circle,and a point of tangency,the hypotenuse is $25$ and the opposite side to the angle $\alpha$ at the origin is $15$. Thus,$\sin \alpha = \frac{15}{25} = \frac{3}{5}$,which implies $\alpha = \sin^{-1}\left(\frac{3}{5}ight)$. The argument of the center is $\frac{\pi}{2}$. The range of arguments is $\left[\frac{\pi}{2} - \alpha, \frac{\pi}{2} + \alphaight]$. Therefore,$\max 	ext{amp}(z) = \frac{\pi}{2} + \alpha$ and $\min 	ext{amp}(z) = \frac{\pi}{2} - \alpha$. The difference is $(\frac{\pi}{2} + \alpha) - (\frac{\pi}{2} - \alpha) = 2\alpha = 2\sin^{-1}\left(\frac{3}{5}ight)$. Using the identity $\sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2}$,we have $2\sin^{-1}\left(\frac{3}{5}ight) = 2\left(\frac{\pi}{2} - \cos^{-1}\left(\frac{3}{5}ight)ight) = \pi - 2\cos^{-1}\left(\frac{3}{5}ight)$.

If $|z - 25i| \le 15$,then $|\max \text{amp}(z) - \min \text{amp}(z)| = $

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