Question

If a complex number $z$ satisfies $|2x+10+10i|\le 5\sqrt{3}-5$ then the least principal argument of $z$ is

• A. $-\frac{5\pi }{6}$
• B. $-\frac{11\pi }{12}$
• C. $-\frac{3\pi }{4}$
• D. $-\frac{2\pi }{3}$

Answer 1

The correct option is A $-\frac{5\pi }{6}$

$|2x+10+10i|\le 5\sqrt{3}-5$

$|z+5+5i|\le \frac{5\sqrt{3}-5}{2}$

This equation represents a circle with centre $A(-5,-5)$ and radius

$\frac{5\sqrt{3}-5}{2}$. The circle is in third quadrant.

Draw a tangent from origin to the circle meeting the circle at $B$ as shown in figure.

Let $\angle BOA=\theta$.

$\sin \theta =\frac{AB}{OA}$

$OA=\sqrt{5^2+5^2}=5\sqrt{2}$

$\sin \theta =\frac{\frac{5\sqrt{3}-5}{2}}{5\sqrt{2}}=\frac{\sqrt{3}-1}{2\sqrt{2}}=\sin {15}^{\circ }$

$\therefore \theta =\frac{\pi }{12}$

Argument of $A=(\pi +\frac{\pi }{4})=\frac{5\pi }{4}=-\frac{3\pi }{4}$

Least principal argument of $z$ = Argument of $B$ $=-\frac{3\pi }{4}-\frac{\pi }{12}=-\frac{10\pi }{12}=-\frac{5\pi }{6}$