If from a point $P (z_{1})$ on the curve $| z | = 2$ , two tangents are drawn to the curve $|z| = 1$ meeting at $Q (z_{2})$ and $R (z_{3})$ then $(\frac{4}{z_{1}} + \frac{1}{z_{2}} + \frac{1}{z_{3}}) (\frac{4}{3 z_{1}} + \frac{1}{3 z_{2}} + \frac{1}{3 z_{3}}) = ......$

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Detailed Solution

$z_{1} {\bar{z}}_{1} = {| z_{1} |}^{2} = 4, z_{2} {\bar{z}}_{2} = {| z_{2} |}^{2} = 1, z_{3} {\bar{z}}_{3} = {| z_{3} |}^{2} = 1$

P lies on $| z | = 2 \Rightarrow | z_{1} | = 2 \Rightarrow O P = 2$

$Q, R$ lies on $| z | = 1 \Rightarrow | z_{2} | = 1, | z_{3} | = 1 \Rightarrow O Q = 1, O R = 1$

$\begin{array}{l} P Q = \sqrt{2^{2} - 1^{2}} = \sqrt{3} \\ \tan θ = \frac{O Q}{P Q} = \frac{1}{\sqrt{3}} \Rightarrow θ = \frac{π}{6} \end{array}$

$∴ Δ P Q R$ is an equilateral triangle

Circumcentre of $Δ P Q R = \frac{z_{1} + z_{2} + z_{3}}{3}$

which lies on $| z | = 1 \Rightarrow |\frac{z_{1} + z_{2} + z_{3}}{3}| = 1$

$| z_{1} + z_{2} + z_{3} | = 3 \Rightarrow (z_{1} + z_{2} + z_{3}) ({\bar{z}}_{1} + {\bar{z}}_{2} + {\bar{z}}_{3}) = 9$

$\begin{array}{l} \Rightarrow (\frac{4}{{\bar{z}}_{1}} + \frac{1}{{\bar{z}}_{2}} + \frac{1}{{\bar{z}}_{3}}) (\frac{4}{{\bar{z}}_{1}} + \frac{1}{{\bar{z}}_{2}} + \frac{1}{{\bar{z}}_{3}}) = 9 \\ ∴ (\frac{4}{{\bar{z}}_{1}} + \frac{1}{{\bar{z}}_{2}} + \frac{1}{{\bar{z}}_{3}}) (\frac{4}{3 {\bar{z}}_{1}} + \frac{1}{3 {\bar{z}}_{2}} + \frac{1}{3 {\bar{z}}_{3}}) = 3 \end{array}$