Question

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

• A. $co{s}^{-1}\left(\frac{3}{5}\right)$
• B. $\pi -2co{s}^{-1}\left(\frac{3}{5}\right)$
• C. $\frac{\pi }{2}+co{s}^{-1}\left(\frac{3}{5}\right)$
• D. $si{n}^{-1}\left(\frac{3}{5}\right)-co{s}^{-1}\left(\frac{3}{5}\right)$

Answer 1

Answer: (B)

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

We have
$maxamp\left(z\right)=amp\left(z_2\right),minamp\left(z\right)=amp\left(z_1\right)$

Now $amp\left(z_1\right)={\theta }_1=co{s}^{-1}\left(\frac{15}{25}\right)=co{s}^{-1}\left(\frac{3}{5}\right)$
$amp\left(z_2\right)=\frac{\pi }{2}+{\theta }_2=\frac{\pi }{2}+si{n}^{-1}\left(\frac{15}{25}\right)=\frac{\pi }{2}+si{n}^{-1}\left(\frac{3}{5}\right)$
$\therefore$ |max amp(z) - min amp(z)|
$=|\frac{\pi }{2}+si{n}^{-1}\frac{3}{5}-co{s}^{-1}\frac{3}{5}|$
$=|\frac{\pi }{2}+\frac{\pi }{2}co{s}^{-1}\frac{3}{5}-co{s}^{-1}\frac{3}{5}|=\pi -2co{s}^{-1}\frac{3}{5}$