Question

lf $|z-25i|\le 15$, then $|max(arg\left(z\left)\right)-min\left(\arg \right(z\left)\right)\right|=$

• A. $2{\cos }^{-1}\frac{3}{5}$
• B. $2{\cos }^{-1}\frac{4}{5}$
• C. $\frac{\pi }{2}+{\cos }^{-1}\frac{3}{5}$
• D. ${\sin }^{-1}\frac{3}{5}-{\cos }^{-1}\frac{3}{5}$

Answer 1

The correct option is A $2{\cos }^{-1}\frac{3}{5}$

Let $z=x+iy$
Hence the above expression reduces to
$x^2+(y-25{)}^2={15}^2$
Hence
$x=15cos\theta$
$y=25+15sin\theta$.
Hence
$arg\left(z\right)=ta{n}^{-1}\left(\frac{25+15sin\theta }{15cos\theta }\right)$
$=ta{n}^{-1}\left(\frac{5+3sin\theta }{3cos\theta }\right)$.
Now
Let $k=\frac{5+3sin\theta }{3cos\theta }$
Hence
$\frac{dk}{d\theta }=\frac{3cos\theta \left(3cos\theta \right)+3sin\theta (5+3sin\theta )}{9co{s}^2\theta }=0$
Hence
$9co{s}^2\theta +9si{n}^2\theta +15sin\theta =0$
Or
$sin\theta =\frac{-9}{15}=\frac{-3}{5}$.
Hence
$cos\theta =\pm \frac{4}{5}$.
Hence
$arg(z{)}_{max}=ta{n}^{-1}(\frac{5+3\frac{-3}{5}}{3.\frac{4}{5}})$
$=ta{n}^{-1}\left(\frac{25-9}{12}\right)$
$=ta{n}^{-1}\left(\frac{4}{3}\right)$
Similarly
$arg(z{)}_{min}=ta{n}^{-1}(\frac{-4}{3})=-ta{n}^{-1}(\frac{4}{3})$.
Hence
$arg(z{)}_{max}-arg(z{)}_{min}=2ta{n}^{-1}\left(\frac{4}{3}\right)$
$=2co{s}^{-1}\left(\frac{3}{5}\right)$