If |z-25i| leq 15, then the value of text{Max | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The given inequality $|z - 25i| \leq 15$ represents a disk in the complex plane with center at $(0, 25)$ and radius $r = 15$.Let the tangents from

Vedclass · Accepted Answer

$2 \cos^{-1}\left(\frac{4}{5}\right)$

Vedclass · Answer

(B) The given inequality $|z - 25i| \leq 15$ represents a disk in the complex plane with center at $(0, 25)$ and radius $r = 15$.Let the tangents from the origin to the circle touch the circle at points $P$ and $Q$.The angle $	heta$ between the $y$-axis and the line segment from the origin to the center $(0, 25)$ is given by $\sin 	heta = \frac{r}{d} = \frac{15}{25} = \frac{3}{5}$, where $d = 25$ is the distance from the origin to the center.Thus, $	heta = \sin^{-1}\left(\frac{3}{5}ight)$.The argument of $z$ ranges from $\frac{\pi}{2} - 	heta$ to $\frac{\pi}{2} + 	heta$.Therefore, $	ext{Maximum } \arg(z) = \frac{\pi}{2} + 	heta$ and $	ext{Minimum } \arg(z) = \frac{\pi}{2} - 	heta$.The difference is $(\frac{\pi}{2} + 	heta) - (\frac{\pi}{2} - 	heta) = 2	heta = 2 \sin^{-1}\left(\frac{3}{5}ight)$.Since $\sin^{-1}\left(\frac{3}{5}ight) = \cos^{-1}\left(\frac{4}{5}ight)$, the difference is $2 \cos^{-1}\left(\frac{4}{5}ight)$.

If $|z-25i| \leq 15$, then the value of $\text{Maximum } \arg(z) - \text{Minimum } \arg(z)$ is equal to

Similar Questions

The solutions of the equation in $z$,$|z|^2 - (z + \bar{z}) + i(z - \bar{z}) + 2 = 0$ are $(i = \sqrt{-1})$.

Let $z_1 = 2 + 3i$ and $z_2 = 3 + 4i$. The set $S = \{ z \in \mathbb{C} : |z - z_1|^2 - |z - z_2|^2 = |z_1 - z_2|^2 \}$ represents a

If $z$ is any complex number satisfying $|z - 3 - 2i| \leq 2$,then the minimum value of $|2z - 6 + 5i|$ is

What are the polar coordinates of the point $(-\sqrt{3}, 1)$?

If the amplitude of $z-2-3i$ is $\pi/4$,then the locus of $z=x+iy$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module