If |z - 25i| le 15, then |max .amp(z) - min . | vedclass

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Question

(b)We have
$max\ amp(z)=amp$$({z_2}),$ $min\ amp (z)=amp$$({z_1})$
Now $amp({z_1}) = {	heta _1} = {\cos ^{ - 1}}\left( {\frac{{15}}{{25}}} ight)

Vedclass · Accepted Answer

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

Vedclass · Answer

(b)We have
$max\ amp(z)=amp$$({z_2}),$ $min\ amp (z)=amp$$({z_1})$
Now $amp({z_1}) = {	heta _1} = {\cos ^{ - 1}}\left( {\frac{{15}}{{25}}} ight) = {\cos ^{ - 1}}\left( {\frac{3}{5}} ight)$
$amp({z_2}) = \frac{\pi }{2} + {	heta _2} = \frac{\pi }{2} + {\sin ^{ - 1}}\left( {\frac{{15}}{{25}}} ight) = \frac{\pi }{2} + {\sin ^{ - 1}}\left( {\frac{3}{5}} ight)$
$|\max \,\,amp(z) - \min \,\,amp(z)|$
$ = \left| {\frac{\pi }{2} + {{\sin }^{ - 1}}\frac{3}{5} - {{\cos }^{ - 1}}\frac{3}{5}} ight|$
$ = \left| {\frac{\pi }{2} + \frac{\pi }{2} - {{\cos }^{ - 1}}\frac{3}{5} - {{\cos }^{ - 1}}\frac{3}{5}} ight| = \pi - 2{\cos ^{ - 1}}\frac{3}{5}$

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

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