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Q.
If $|z-25 i| \leq 15$, then the least positive value of arg $z$ is
Complex Numbers and Quadratic Equations
Solution:
Since $|z-25 i| \leq 15$, therefore, distance between $z$ and $25 i$ is less than or equal to $15$ .
Thus, point $z$ will lie in the interior and boundary of the circle whose centre is $(0,25)$ and radius is $15$ .
Let $O P$ be tangent to the circle at point $P$.
Let $\angle P O X=\theta$.
Then, $\angle O C P=\theta$
Now, $O C=25, C P=15$
$\therefore O P=20 .$
Now, $\tan \theta=\frac{O P}{C P}=\frac{20}{15}=\frac{4}{3}$
$\therefore $ Least positive value of $\arg z=\theta=\tan ^{-1}\left(\frac{4}{3}\right)$
