Question

If $|z-25 i| \leq 15$, 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, $\text{amp}\left(z_{2}\right)=\frac{\pi}{2}+\theta_{2}$
$=\frac{\pi}{2}+\sin ^{-1}\left(\frac{15}{25}\right)$
$=\frac{\pi}{2}+\sin ^{-1}\left(\frac{3}{5}\right)$
$\therefore |\max \cdot \text{amp}(z)-\min . \text{amp}(z)|$
$=\left|\frac{\pi}{2}+\sin ^{-1} \frac{3}{5}-\cos ^{-1} \frac{3}{5}\right|$
$=\left|\frac{\pi}{2}+\frac{\pi}{2}--\cos ^{-1} \frac{3}{5}-\cos ^{-1} \frac{3}{5}\right|$
$=\pi-2 \cos ^{-1}\left(\frac{3}{5}\right)$