Q. The complex numbers $z$ having positive argument and satisfying $|2 - 3i| < |z|$, is
A) $\frac{12}{5} + \frac{16}{5}i$
B) $\frac{4}{5} + \frac{6}{5}i$
C) $\frac{6}{5} - \frac{5}{2}i$
D) None of these
Explanation
(a)
Let $z = \frac{12}{5} + \frac{16}{5}i$
$\because\ \tan{\theta = \frac{16}{12} = \frac{4}{3} > 0}$
$\therefore\ \ \ \ \ \theta > 0$
And $|z| = \sqrt{\left( \frac{12}{5} \right)^{2} + \left( \frac{16}{5} \right)^{2}} = \frac{1}{5}\sqrt{144 + 256} = 4$
Now, $|2 - 3i| = \sqrt{4 + 9} = \sqrt{13}$
$\therefore\ \ \ \ \ |2 - 3i| < |z|$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.