If |z-25i| leq 15,then the value of text{Maxi | vedclass

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(B) The given inequality $|z - 25i| \leq 15$ represents a disk in the complex plane with center at $(0, 25)$ and radius $r = 15$.Let the tangents from

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$2 \cos^{-1}\left(\frac{4}{5}\right)$

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(B) The given inequality $|z - 25i| \leq 15$ represents a disk in the complex plane with center at $(0, 25)$ and radius $r = 15$.Let the tangents from the origin to the circle touch the circle at points $P$ and $Q$.The angle $	heta$ between the $y$-axis and the line segment from the origin to the center $(0, 25)$ is given by $\sin 	heta = \frac{r}{d} = \frac{15}{25} = \frac{3}{5}$,where $d = 25$ is the distance from the origin to the center.Thus,$	heta = \sin^{-1}\left(\frac{3}{5}ight)$.The argument of $z$ ranges from $\frac{\pi}{2} - 	heta$ to $\frac{\pi}{2} + 	heta$.Therefore,$	ext{Maximum } \arg(z) = \frac{\pi}{2} + 	heta$ and $	ext{Minimum } \arg(z) = \frac{\pi}{2} - 	heta$.The difference is $(\frac{\pi}{2} + 	heta) - (\frac{\pi}{2} - 	heta) = 2	heta = 2 \sin^{-1}\left(\frac{3}{5}ight)$.Since $\sin^{-1}\left(\frac{3}{5}ight) = \cos^{-1}\left(\frac{4}{5}ight)$,the difference is $2 \cos^{-1}\left(\frac{4}{5}ight)$.

If $|z-25i| \leq 15$,then the value of $\text{Maximum } \arg(z) - \text{Minimum } \arg(z)$ is equal to

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