Question

Let z be a complex number satisfying |z – 5i| $\le$1 such that arg(z) is minimum. Then z is equal to

• A. $\frac{2\sqrt{6}}{5}+\frac{24i}{5}$
• B. $\frac{24}{5}+\frac{2\sqrt{6i}}{5}$
• C. $\frac{6\sqrt{2}}{5}+\frac{24i}{5}$
• D. $i$

Answer 1

The correct option is A $\frac{2\sqrt{6}}{5}+\frac{24i}{5}$


$Sin\angle AOC=\frac{1}{5}sin(90-\theta )=\frac{1}{5}cos\theta =\frac{1}{5}$
$OA=\sqrt{O{C}^2-C{A}^2}=\sqrt{24}sin\theta =\frac{\sqrt{24}}{5}$. Now the co - ordinates of A are
$(\sqrt{24}cos\theta ,\sqrt{24}sin\theta )(\frac{\sqrt{24}}{5},\frac{24}{5})=\frac{\sqrt{24}}{5}+i\left(\frac{24}{5}\right)$