Question

If $|z-25i|\le 15$, then $\mid$ maximum amp $(z)-$ minimum amp $(z)\mid$ is equal to

• A. ${\sin }^{-1}\left(\frac{3}{5}\right)-{\cos }^{-1}\left(\frac{3}{5}\right)$
• B. $\frac{\pi }{2}+{\cos }^{-1}\left(\frac{3}{5}\right)$
• C. $\pi -2{\cos }^{-1}\left(\frac{3}{5}\right)$
• D. ${\cos }^{-1}\left(\frac{3}{5}\right)$

Answer 1

Answer: (C)

We have, max. amp $(z)=\text{amp}\left(z_2\right)$,
$\min .\text{amp}(z)=\text{amp}\left(z_1\right).$

Now, amp $\left(z_1\right)={\theta }_1={\cos }^{-1}\left(\frac{15}{25}\right)={\cos }^{-1}\left(\frac{3}{5}\right)$
and, $\text{amp}\left(z_2\right)=\frac{\pi }{2}+{\theta }_2$
$=\frac{\pi }{2}+{\sin }^{-1}\left(\frac{15}{25}\right)$
$=\frac{\pi }{2}+{\sin }^{-1}\left(\frac{3}{5}\right)$
$\therefore |\max \cdot \text{amp}(z)-\min .\text{amp}(z)|$
$=\left|\frac{\pi }{2}+{\sin }^{-1}\frac{3}{5}-{\cos }^{-1}\frac{3}{5}\right|$
$=\left|\frac{\pi }{2}+\frac{\pi }{2}--{\cos }^{-1}\frac{3}{5}-{\cos }^{-1}\frac{3}{5}\right|$
$=\pi -2{\cos }^{-1}\left(\frac{3}{5}\right)$