Question

A point $P\left(z_1\right)$ lies on the curve $|z|=2$. A pair of tangents from the point $P$ is drawn to the curve $|z|=1$ meeting it at points $Q\left(z_2\right)$ and $R\left(z_2\right)$. Then

• A. $(\frac{4}{z_1}+\frac{1}{z_2}+\frac{1}{z_3})(\frac{4}{\overline{z_1}}+\frac{1}{\overline{z_2}}+\frac{1}{\overline{z_3}})=9$
• B. $\arg \left(\frac{z_2}{z_3}\right)=\frac{2\pi }{3}$
• C. point with complex representation $\left(\frac{z_1+z_2+z_3}{3}\right)$ lies on the curve $|z|=1$
• D. point with complex representation $\left(\frac{z_1}{2}\right)$ lies on the curve $|z|=1$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $(\frac{4}{z_1}+\frac{1}{z_2}+\frac{1}{z_3})(\frac{4}{\overline{z_1}}+\frac{1}{\overline{z_2}}+\frac{1}{\overline{z_3}})=9$
B $\arg \left(\frac{z_2}{z_3}\right)=\frac{2\pi }{3}$
C point with complex representation $\left(\frac{z_1+z_2+z_3}{3}\right)$ lies on the curve $|z|=1$
D point with complex representation $\left(\frac{z_1}{2}\right)$ lies on the curve $|z|=1$


Since, radius is always perpendicular to tengent. Hence, $\Delta OPQ$ and $\Delta OPR$ are right angled triangle.
$P\left(z_1\right)$ lies on the curve $|z|=2$
$\therefore |OP|=2$
Similarly, $Q\left(z_2\right)$ and $R\left(z_2\right)$ lie on the curve $|z|=1$
$\therefore |OQ|=|OR|=1$

In $\Delta OPQ$,
$\sin \theta =\frac{|OQ|}{|OP|}=\frac{1}{2}$
$\Rightarrow \theta =\frac{\pi }{6}\Rightarrow \angle QPR=\frac{\pi }{3}$
$\Rightarrow \angle POQ=\angle POR=\frac{\pi }{3}$

We know that,
$\arg \left(\frac{z_2}{z_3}\right)=\arg \left(z_2\right)-\arg \left(z_3\right)$
$\Rightarrow \arg \left(\frac{z_2}{z_3}\right)$ is the angle between $Q\left(z_2\right)$ and $R\left(z_2\right).$
$\therefore \arg \left(\frac{z_2}{z_3}\right)=\angle POQ+\angle POR=\frac{2\pi }{3}$

$\Rightarrow △PQR$ is equilateral.

Circumcircle of $△PQR$ has $OP$ as diameter.
$\Rightarrow$ Circumcentre is $\left(\frac{z_1}{2}\right)$ and $\left|\frac{z_1}{2}\right|=1$

Since, circumcentre and centroid coincide for equilateral triangle.
$\therefore \left|\frac{z_1+z_2+z_3}{3}\right|=1$
$\Rightarrow |z_1+z_2+z_3|=3$

$\left|\frac{z_1}{2}\right|=1\Rightarrow \frac{z_1\overline{z_1}}{4}=1\text{(squaring both the sides)}$
$\Rightarrow \frac{4}{z_1}=\overline{z_1}\text{and}\frac{4}{\overline{z_1}}=z_1$
Similarly, $z_2\overline{z_2}=1=z_3\overline{z_3}$

$(\frac{4}{z_1}+\frac{1}{z_2}+\frac{1}{z_3})(\frac{4}{\overline{z_1}}+\frac{1}{\overline{z_2}}+\frac{1}{\overline{z_3}})=(\overline{z_1}+\overline{z_2}+\overline{z_3})(z_1+z_2+z_3)=|z_1+z_2+z_3{|}^2=9$