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If $|z-25 i| \leq 15$, then Maximum $\arg (z)-$ Minimum $\arg (z)$ is equal to
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Verified Answer
The correct answer is:
$2 \cos ^{-1}\left(\frac{4}{5}\right)$
$$
\because \cos \theta=\frac{15}{25}=\frac{3}{5}
$$
$\therefore \operatorname{Min} \arg (z)=\cos ^{-1}\left(\frac{3}{5}\right)$
$\operatorname{Max} \arg (z)=\pi-\cos ^{-1}\left(\frac{3}{5}\right)=\frac{\pi}{2}+\sin ^{-1}\left(\frac{3}{5}\right)$
$\therefore$ difference $=\frac{\pi}{2}+\sin ^{-1}\left(\frac{3}{5}\right)-\cos ^{-1}\left(\frac{3}{5}\right)=2 \sin ^{-1}\left(\frac{3}{5}\right)=2 \cos ^{-1}\left(\frac{4}{5}\right)$
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